2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan,...

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V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 1 2.COMPLEX NUMBERS POWERS OF IMAGINARY UNIT 2 = โˆ’1, 3 = โˆ’, 4 =1 Division algorithm : = 4() + = () 4+ = () 4 () = ( 4 ) () โŸน = 0, = 1 ; = 1 , = ; = 2, = โˆ’1; = 3, = โˆ’ โˆš = โˆš โˆš is valid only if at least one of a, b is non-negative. For โˆˆ , 2 >0 EXERCISE 2.1 Simplify the following: 1. 1947 + 1950 1947 รท 4 = 4(486) + 3 ; 1950 = 4(487) + 2 1947 + 1950 = ( 4 ) 486 () 3 + ( 4 ) 487 () 2 = โˆ’ โˆ’ 1 = โˆ’1 โˆ’ 2. 1948 โˆ’ โˆ’1869 1948 รท 4 = 4(487) + 0 ; 1869 รท 4 = 4(467) + 1 1948 โˆ’ โˆ’1869 = 1948 โˆ’ 1 1869 = ( 4 ) 487 () 0 โˆ’ 1 ( 4 ) 467 1 =1โˆ’ 1 =1โˆ’( 1 ร— โˆ’ โˆ’ ) =1โˆ’( โˆ’ โˆ’ 2 ) = 1 + (โˆต โˆ’ 2 = 1) 3.โˆ‘ 12 =1 โˆ‘ 12 =1 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + + 8 + 9 + 10 + 11 + 12 = 1 + 2 + 3 + 4 + 4 + ( 2 ) 3 + ( 2 ) 3 + ( 2 ) 4 + ( 3 ) 3 + ( 2 ) 5 + ( 2 ) 5 + ( 2 ) 6 =โˆ’1โˆ’+1+โˆ’1โˆ’+1+โˆ’1โˆ’+1 =0 4. 59 + 1 59 59 รท 4 = 4(14) + 3 59 + 1 59 = ( 4 ) 14 3 + 1 ( 4 ) 14 3 = โˆ’ + 1 โˆ’ =+ 1 โˆ’ = โˆ’ + 1 =โˆ’ =0 5.. 2 . 3 . 4 โ€ฆ.. 2000 = (1+2+3โ€ฆ..+2000) = 2000ร—2001 2 = 1000ร—2001 = 1000 . 2001 = ( 4 ) 250 ( 4 ) 500 1 (โˆต 2001 รท 4 = 4(500) + 1) = 1.1. = 6.โˆ‘ +50 10 =1 โˆ‘ +50 10 =1 = 51 + 52 + 53 +โ‹ฏ+ 60 = 50 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 ) = ( 4 ) 12 2 ( โˆ’ 1 โˆ’ + 1 + โˆ’ 1 โˆ’ + 1 + โˆ’ 1) = โˆ’1( โˆ’ 1) =1โˆ’ . Simplify the following: (i) i 7 (ii)i 1729 (iii)i โˆ’1924 +i 2018 (iv) โˆ‘ i n 102 n=1 (v)i. i 2 .i 3 .i 4 โ€ฆ..i 40 COMPLEX NUMBERS Rectangular Form Of a Complex Number: A complex number is of the form = + , , โˆˆ . is called the real part and is called the imaginary part of the complex number. Two complex numbers 1 = + , 2 = + are said to be equal if and only if ( 1 ) = ( 2 ) and ( 1 ) = ( 2 ). ie. = = Scalar multiplication of complex numbers: = + โˆˆ โ„Ž = ( + ) = + Addition of complex numbers: If 1 = + , 2 = + , โ„Ž 1 + 2 = ( + ) + ( + ) Subtraction of complex numbers: If 1 = + , 2 = + , โ„Ž 1 โˆ’ 2 = ( โˆ’ ) + ( โˆ’ ) Multiplication of complex numbers: If 1 = + , 2 = + , โ„Ž 1 2 = ( โˆ’ ) + ( + ) Multiplication of a complex z by i successively gives a 90 0 counter clockwise rotation successively about the origin. EXERCISE 2.2 1. Evaluate the following if = 5 โˆ’ 2 and = โˆ’1 + 3 () + () โˆ’ ()2 + 3 () () 2 + 2 + 2 ()( + ) 2 Padasal www.Padasal www.Padasal www.Padasal www.Padasal www.Padasal www.Padasal www.Padasal www Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www i.Org i.Org i.Org i.Org i.Org i.Org i.Org i.Org adasa www.Padasa www.Padasa www.Padasa www.Padasa www.Padasa www.Padasa www.Padasa www. adasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www. adasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www. adasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www. adasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www. adasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www. adasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www. adasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www. adasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www. adasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www.Padasalai.Org www. ai.Org ai.Org ai.Org ai.Org ai.Org ai.Org ai.Org ai.Org Padasalai TrbTnpsc

Transcript of 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan,...

Page 1: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 1

2.COMPLEX NUMBERS

POWERS OF IMAGINARY UNIT

๐‘–2 = โˆ’1, ๐‘–3 = โˆ’๐‘–, ๐‘–4 = 1 Division algorithm : ๐‘› = 4(๐‘ž) + ๐‘Ÿ ๐‘–๐‘› = (๐‘–)4๐‘ž+๐‘Ÿ = (๐‘–)4๐‘ž(๐‘–)๐‘Ÿ = (๐‘–4)๐‘ž(๐‘–)๐‘Ÿ โŸน ๐‘–๐‘“ ๐‘Ÿ = 0, ๐‘–๐‘› = 1 ; ๐‘–๐‘“ ๐‘Ÿ = 1 , ๐‘–๐‘› = ๐‘–; ๐‘–๐‘“ ๐‘Ÿ = 2, ๐‘–๐‘› = โˆ’1; ๐‘–๐‘“ ๐‘Ÿ = 3, ๐‘–๐‘› = โˆ’๐‘–

โˆš๐‘Ž๐‘ = โˆš๐‘Žโˆš๐‘ is valid only if at least one of a, b is non-negative. For ๐‘ฆ โˆˆ ๐‘…, ๐‘ฆ2 > 0

EXERCISE 2.1 Simplify the following: 1.๐‘–1947 + ๐‘–1950 1947 รท 4 = 4(486) + 3 ; 1950 = 4(487) + 2 ๐‘–1947 + ๐‘–1950 = (๐‘–4)486(๐‘–)3 + (๐‘–4)487(๐‘–)2 = โˆ’๐‘– โˆ’ 1 = โˆ’1 โˆ’ ๐‘– 2.๐‘–1948 โˆ’ ๐‘–โˆ’1869 1948 รท 4 = 4(487) + 0 ; 1869 รท 4 = 4(467) + 1

๐‘–1948 โˆ’ ๐‘–โˆ’1869 = ๐‘–1948 โˆ’1

๐‘–1869= (๐‘–4)487(๐‘–)0 โˆ’

1

(๐‘–4)467๐‘–1= 1 โˆ’

1

๐‘–= 1 โˆ’ (

1

๐‘–ร—โˆ’๐‘–

โˆ’๐‘–)

= 1 โˆ’ (โˆ’๐‘–

โˆ’๐‘–2) = 1 + ๐‘– (โˆต โˆ’๐‘–2 = 1)

3.โˆ‘ ๐‘–๐‘›12๐‘›=1

โˆ‘ ๐‘–๐‘›12๐‘›=1 = ๐‘–1 + ๐‘–2 + ๐‘–3 + ๐‘–4 + ๐‘–5 + ๐‘–6 + ๐‘–7 + +๐‘–8 + ๐‘–9 + ๐‘–10 + ๐‘–11 + ๐‘–12 = ๐‘–1 + ๐‘–2 + ๐‘–3 + ๐‘–4 + ๐‘–4๐‘– + (๐‘–2)3 + (๐‘–2)3๐‘– + (๐‘–2)4 + (๐‘–3)3 + (๐‘–2)5 + (๐‘–2)5๐‘– + (๐‘–2)6 = ๐‘– โˆ’ 1 โˆ’ ๐‘– + 1 + ๐‘– โˆ’ 1 โˆ’ ๐‘– + 1 + ๐‘– โˆ’ 1 โˆ’ ๐‘– + 1 = 0

4.๐‘–59 +1

๐‘–59

59 รท 4 = 4(14) + 3

๐‘–59 +1

๐‘–59= (๐‘–4)14๐‘–3 +

1

(๐‘–4)14๐‘–3

= โˆ’๐‘– +1

โˆ’๐‘–= ๐‘– +

1

โˆ’๐‘–

๐‘–

๐‘–

= โˆ’๐‘– +๐‘–

1= ๐‘– โˆ’ ๐‘–

= 0 5.๐‘–. ๐‘–2. ๐‘–3. ๐‘–4โ€ฆ . . ๐‘–2000

= ๐‘–(1+2+3โ€ฆ..+2000)

= ๐‘–2000ร—2001

2 = ๐‘–1000ร—2001 = ๐‘–1000. ๐‘–2001 = (๐‘–4)250(๐‘–4)500๐‘–1 (โˆต 2001 รท 4 = 4(500) + 1) = 1.1. ๐‘– = ๐‘–

6.โˆ‘ ๐‘–๐‘›+5010๐‘›=1

โˆ‘ ๐‘–๐‘›+5010๐‘›=1 = ๐‘–51 + ๐‘–52 + ๐‘–53 +โ‹ฏ+ ๐‘–60

= ๐‘–50(๐‘–1 + ๐‘–2 + ๐‘–3 + ๐‘–4 + ๐‘–5 + ๐‘–6 + ๐‘–7 + ๐‘–8 + ๐‘–9 + ๐‘–10) = (๐‘–4)12๐‘–2(๐‘– โˆ’ 1 โˆ’ ๐‘– + 1 + ๐‘– โˆ’ 1 โˆ’ ๐‘– + 1 + ๐‘– โˆ’ 1) = โˆ’1(๐‘– โˆ’ 1) = 1 โˆ’ ๐‘–

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ Simplify the following: (i) i7 (ii)i1729 (iii)iโˆ’1924 + i2018 (iv)โˆ‘ in102

n=1 (v)i. i2. i3. i4โ€ฆ . . i40 COMPLEX NUMBERS

Rectangular Form Of a Complex Number: A complex number is of the form ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ, ๐‘ฅ, ๐‘ฆ โˆˆ ๐‘…. ๐‘ฅ is called the real part and ๐‘ฆ is called the imaginary part of the complex number.

Two complex numbers ๐‘ง1 = ๐‘Ž + ๐‘–๐‘, ๐‘ง2 = ๐‘ + ๐‘–๐‘‘ are said to be equal if and only if ๐‘…๐‘’(๐‘ง1) = ๐‘…๐‘’(๐‘ง2) and ๐ผ๐‘š(๐‘ง1) = ๐ผ๐‘š(๐‘ง2). ie. ๐‘Ž = ๐‘ ๐‘Ž๐‘›๐‘‘ ๐‘ = ๐‘‘

Scalar multiplication of complex numbers: ๐ผ๐‘“ ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ ๐‘Ž๐‘›๐‘‘ ๐‘˜ โˆˆ๐‘… ๐‘กโ„Ž๐‘’๐‘› ๐‘˜๐‘ง = ๐‘˜(๐‘ฅ + ๐‘–๐‘ฆ) = ๐‘˜๐‘ฅ + ๐‘˜๐‘ฆ๐‘–

Addition of complex numbers: If ๐‘ง1 = ๐‘Ž + ๐‘–๐‘, ๐‘ง2 = ๐‘ + ๐‘–๐‘‘, ๐‘กโ„Ž๐‘’๐‘› ๐‘ง1 + ๐‘ง2 =(๐‘Ž + ๐‘) + ๐‘–(๐‘ + ๐‘‘)

Subtraction of complex numbers: If ๐‘ง1 = ๐‘Ž + ๐‘–๐‘, ๐‘ง2 = ๐‘ + ๐‘–๐‘‘, ๐‘กโ„Ž๐‘’๐‘› ๐‘ง1 โˆ’๐‘ง2 = (๐‘Ž โˆ’ ๐‘) + ๐‘–(๐‘ โˆ’ ๐‘‘)

Multiplication of complex numbers: If ๐‘ง1 = ๐‘Ž + ๐‘–๐‘, ๐‘ง2 = ๐‘ + ๐‘–๐‘‘, ๐‘กโ„Ž๐‘’๐‘› ๐‘ง1๐‘ง2 = (๐‘Ž๐‘ โˆ’ ๐‘๐‘‘) + ๐‘–(๐‘Ž๐‘‘ + ๐‘๐‘)

Multiplication of a complex z by i successively gives a 900counter clockwise rotation successively about the origin.

EXERCISE 2.2 1. Evaluate the following if ๐‘ง = 5 โˆ’ 2๐‘– and ๐‘ค = โˆ’1 + 3๐‘– (๐‘–)๐‘ง + ๐‘ค (๐‘–๐‘–)๐‘ง โˆ’ ๐‘–๐‘ค (๐‘–๐‘–๐‘–)2๐‘ง + 3๐‘ค (๐‘–๐‘ฃ)๐‘ง๐‘ค (๐‘ฃ)๐‘ง2 + 2๐‘ง๐‘ค + ๐‘ค2 (๐‘ฃ๐‘–)(๐‘ง + ๐‘ค)2

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Padasalai TrbTnpsc

Page 2: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 2

(๐‘–)๐‘ง + ๐‘ค = 5 โˆ’ 2๐‘– โˆ’ 1 + 3๐‘– = 4 + ๐‘– (๐‘–๐‘–)๐‘ง โˆ’ ๐‘–๐‘ค = 5 โˆ’ 2๐‘– โˆ’ ๐‘–(โˆ’1 + 3๐‘–) = 5 โˆ’ 2๐‘– + ๐‘– โˆ’ 3๐‘–2 = 5 โˆ’ 2๐‘– + ๐‘– + 3 = 8 โˆ’ ๐‘–

โˆต ๐‘–2 = โˆ’1 (๐‘–๐‘–๐‘–)2๐‘ง + 3๐‘ค = 2(5 โˆ’ 2๐‘– ) + 3(โˆ’1 + 3๐‘–) = 10 โˆ’ 4๐‘– โˆ’ 3 + 9๐‘– = 7 + 5๐‘– (๐‘–๐‘ฃ) ๐‘ง1๐‘ง2 = (๐‘Ž๐‘ โˆ’ ๐‘๐‘‘) + ๐‘–(๐‘Ž๐‘‘ + ๐‘๐‘); ๐‘Ž = 5, ๐‘ = โˆ’2, ๐‘ = โˆ’1, ๐‘‘ = 3

๐‘ง๐‘ค = (5 โˆ’ 2๐‘–)(โˆ’1 + 3๐‘–) = [(5)(โˆ’1) โˆ’ (โˆ’2)(3)] + ๐‘–[(5)(3) + (โˆ’2)(โˆ’1)] = [โˆ’5 + 6] + ๐‘–[15 + 2] = 1 + 17๐‘–

(๐‘ฃ)๐‘ง2 = (5 โˆ’ 2๐‘–)2 = 25 โˆ’ 20๐‘– + 4๐‘–2 = 25 โˆ’ 20๐‘– โˆ’ 4 = 21 โˆ’ 20๐‘– 2๐‘ง๐‘ค = 2(5 โˆ’ 2๐‘– )(โˆ’1 + 3๐‘–) = 2[1 + 17๐‘–] from (๐‘–๐‘ฃ) 2๐‘ง๐‘ค = 2 + 34๐‘– ๐‘ค2 = (โˆ’1 + 3๐‘–)2 = 1 โˆ’ 6๐‘– + 9๐‘–2 = 1 โˆ’ 6๐‘– โˆ’ 9 = โˆ’8 โˆ’ 6๐‘– ๐‘ง2 + 2๐‘ง๐‘ค + ๐‘ค2 = 21 โˆ’ 20๐‘– + 2 + 34๐‘– โˆ’ 8 โˆ’ 6๐‘– = 15 + 8๐‘– (๐‘ฃ๐‘–)(๐‘ง + ๐‘ค)2 = (4 + ๐‘–)2 from(๐‘–)

= 16 + 8๐‘– + ๐‘–2 = 16 + 8๐‘– โˆ’ 1 = 15 + 8๐‘– 2. Given the complex number ๐‘ง = 2 + 3๐‘–, represent the complex numbers in Argand diagram. (๐‘–)๐‘ง, ๐‘–๐‘ง ๐‘Ž๐‘›๐‘‘ ๐‘ง + ๐‘–๐‘ง (๐‘–๐‘–)๐‘ง, โˆ’๐‘–๐‘ง ๐‘Ž๐‘›๐‘‘ ๐‘ง โˆ’ ๐‘–๐‘ง (๐‘–)๐‘ง = 2 + 3๐‘– = (2,3) ๐‘–๐‘ง = ๐‘–(2 + 3๐‘–) = 2๐‘– + 3๐‘–2 = โˆ’3 + 2๐‘– = (โˆ’3,2) ๐‘ง + ๐‘–๐‘ง = 2 + 3๐‘– + ๐‘–(2 + 3๐‘–) = 2 + 3๐‘– โˆ’ 3 + 2๐‘– = โˆ’1 + 5๐‘– = (โˆ’1,5)

(๐‘–)๐‘ง = 2 + 3๐‘– = (2,3) โˆ’๐‘–๐‘ง = โˆ’๐‘–(2 + 3๐‘–) = โˆ’2๐‘– โˆ’ 3๐‘–2 = 3 โˆ’ 2๐‘– = (3,โˆ’2) ๐‘ง โˆ’ ๐‘–๐‘ง = 2 + 3๐‘– โˆ’ ๐‘–(2 + 3๐‘–) = 2 + 3๐‘– + 3 โˆ’ 2๐‘– = 5 + ๐‘– = (5,1)

3. Find the values of the real numbers ๐‘ฅ ๐‘Ž๐‘›๐‘‘ ๐‘ฆ, if the complex numbers (3 โˆ’ ๐‘–)๐‘ฅ โˆ’(2 โˆ’ ๐‘–)๐‘ฆ + 2๐‘– + 5 and 2๐‘ฅ + (โˆ’1 + 2๐‘–)๐‘ฆ + 3 + 2๐‘– are equal. ๐‘ง1 = (3 โˆ’ ๐‘–)๐‘ฅ โˆ’ (2 โˆ’ ๐‘–)๐‘ฆ + 2๐‘– + 5 = 3๐‘ฅ โˆ’ ๐‘ฅ๐‘– โˆ’ 2๐‘ฆ + ๐‘ฆ๐‘– + 2๐‘– + 5 = (3๐‘ฅ โˆ’ 2๐‘ฆ + 5) + ๐‘–(โˆ’๐‘ฅ + ๐‘ฆ + 2)

๐‘ง2 = 2๐‘ฅ + (โˆ’1 + 2๐‘–)๐‘ฆ + 3 + 2๐‘– = 2๐‘ฅ โˆ’ ๐‘ฆ + 2๐‘ฆ๐‘– + 3 + 2๐‘– = (2๐‘ฅ โˆ’ ๐‘ฆ + 3) + ๐‘–(2๐‘ฆ + 2)

Given : ๐‘ง1 = ๐‘ง2 โŸน (3๐‘ฅ โˆ’ 2๐‘ฆ + 5) + ๐‘–(โˆ’๐‘ฅ + ๐‘ฆ + 2) = (2๐‘ฅ โˆ’ ๐‘ฆ + 3) + ๐‘–(2๐‘ฆ + 2) Equating the real and imaginary parts on both sides, 3๐‘ฅ โˆ’ 2๐‘ฆ + 5 = 2๐‘ฅ โˆ’ ๐‘ฆ + 3 3๐‘ฅ โˆ’ 2๐‘ฆ + 5 โˆ’ 2๐‘ฅ + ๐‘ฆ โˆ’ 3 = 0

๐‘ฅ โˆ’ ๐‘ฆ = โˆ’2 โ†’ (1)

โˆ’๐‘ฅ + ๐‘ฆ + 2 = 2๐‘ฆ + 2 โˆ’๐‘ฅ + ๐‘ฆ + 2 โˆ’ 2๐‘ฆ โˆ’ 2 = 0 โˆ’๐‘ฅ โˆ’ ๐‘ฆ = 0 โ†’ (2)

(1) + (2) โŸน โˆ’2๐‘ฆ = โˆ’2 โŸน ๐‘ฆ = 1

Sub. ๐‘ฆ = โˆ’1 in (1) โŸน ๐‘ฅ โˆ’ 1 = โˆ’2 โŸน ๐‘ฅ = โˆ’1

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ Find the values of the real numbers ๐‘ฅ ๐‘Ž๐‘›๐‘‘ ๐‘ฆ, if the complex numbers (2 + ๐‘–)๐‘ฅ + (1 โˆ’ ๐‘–)๐‘ฆ + 2๐‘– โˆ’ 3 and ๐‘ฅ + (โˆ’1 + 2๐‘–)๐‘ฆ + 1 + ๐‘– are equal. [๐ด๐‘›๐‘ : ๐‘ฅ = 2 ๐‘Ž๐‘›๐‘‘ ๐‘ฆ = 1]

PROPERTIES OF COMPLEX NUMBERS Properties of complex numbers under addition:

I. Closure property: โˆ€ ๐‘ง1, ๐‘ง2 โˆˆ โ„‚ , (๐‘ง1 + ๐‘ง2) โˆˆ โ„‚ II. Commutative property: โˆ€ ๐‘ง1, ๐‘ง2 โˆˆ โ„‚, ๐‘ง1 + ๐‘ง2 = ๐‘ง2 + ๐‘ง1

III. Associative property: โˆ€ ๐‘ง1, ๐‘ง2, ๐‘ง3 โˆˆ โ„‚, (๐‘ง1 + ๐‘ง2) + ๐‘ง3 = ๐‘ง1 + (๐‘ง2 + ๐‘ง3) IV. Additive identity: โˆ€ ๐‘ง โˆˆ โ„‚, ๐‘ง + 0 = 0 + ๐‘ง = ๐‘ง โŸน 0 = 0 + 0๐‘– is an additive

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Padasalai TrbTnpsc

Page 3: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 3

identity. V. Additive inverse: โˆ€ ๐‘ง โˆˆ โ„‚ ,โˆ’ ๐‘ง โˆˆ โ„‚ โ‹บ ๐‘ง + (โˆ’๐‘ง) = (โˆ’๐‘ง) + ๐‘ง = 0 โˆˆ โ„‚ , ๐‘ ๐‘œ โˆ’ ๐‘ง

is an additive identity. Properties of complex numbers under multiplication:

I. Closure property: โˆ€ ๐‘ง1, ๐‘ง2 โˆˆ โ„‚ , (๐‘ง1. ๐‘ง2) โˆˆ โ„‚ II. Commutative property: โˆ€ ๐‘ง1, ๐‘ง2 โˆˆ โ„‚, ๐‘ง1. ๐‘ง2 = ๐‘ง2. ๐‘ง1

III. Associative property: โˆ€ ๐‘ง1, ๐‘ง2, ๐‘ง3 โˆˆ โ„‚, (๐‘ง1. ๐‘ง2). ๐‘ง3 = ๐‘ง1. (๐‘ง2. ๐‘ง3) IV. Multiplicative identity: โˆ€ ๐‘ง โˆˆ โ„‚, ๐‘ง. 1 = 1. ๐‘ง = ๐‘ง โŸน 1 = 1 + 0๐‘– is a

multiplicative identity.

V. Multiplicative inverse: โˆ€ ๐‘ง โˆˆ โ„‚ โˆƒ1

๐‘ง ๐‘ง โˆˆ โ„‚ โ‹บ ๐‘ง.

1

๐‘ง=

1

๐‘ง. ๐‘ง = 1 + ๐‘–0 โˆˆ โ„‚ , ๐‘ ๐‘œ

1

๐‘ง is

multiplicative identity.

If ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ then its multiplicative inverse is ๐‘งโˆ’1 = (๐‘ฅ

๐‘ฅ2+๐‘ฆ2) + ๐‘– (โˆ’

๐‘ฆ

๐‘ฅ2+๐‘ฆ2)

Distributive property: โˆ€ ๐‘ง1, ๐‘ง2, ๐‘ง3 โˆˆ โ„‚, (๐‘ง1 + ๐‘ง2). ๐‘ง3 = ๐‘ง1๐‘ง3 + ๐‘ง2๐‘ง3 (or) o โˆ€ ๐‘ง1, ๐‘ง2, ๐‘ง3 โˆˆ โ„‚ , ๐‘ง1. (๐‘ง2 + ๐‘ง3) = ๐‘ง1๐‘ง2 + ๐‘ง1๐‘ง3

Complex numbers obey the laws of indices:

(๐‘–)๐‘ง๐‘š๐‘ง๐‘› = ๐‘ง๐‘š+๐‘› (๐‘–๐‘–)๐‘ง๐‘š

๐‘ง๐‘›= ๐‘ง๐‘šโˆ’๐‘›

(๐‘–๐‘–๐‘–)(๐‘ง๐‘š)๐‘› = ๐‘ง๐‘š๐‘ง๐‘› (๐‘–๐‘ฃ)(๐‘ง1๐‘ง2)๐‘š = ๐‘ง1

๐‘š๐‘ง2๐‘š

EXERCISE 2.3 1.If ๐‘ง1 = 1 โˆ’ 3๐‘–, ๐‘ง2 = โˆ’4๐‘– and ๐‘ง3 = 5, show that (๐‘–)(๐‘ง1 + ๐‘ง2) + ๐‘ง3 = ๐‘ง1 + (๐‘ง2 + ๐‘ง3) (๐‘–๐‘–)(๐‘ง1๐‘ง2)๐‘ง3 = ๐‘ง1(๐‘ง2๐‘ง2) (๐‘–)(๐‘ง1 + ๐‘ง2) + ๐‘ง3 = (1 โˆ’ 3๐‘– โˆ’ 4๐‘–) + 5 = 1 โˆ’ 7๐‘– + 5 = 6 โˆ’ 7๐‘– โ†’ (1) ๐‘ง1 + (๐‘ง2 + ๐‘ง3) = 1 โˆ’ 3๐‘– + (โˆ’4๐‘– + 5) = 1 โˆ’ 3๐‘– โˆ’ 4๐‘– + 5 = 6 โˆ’ 7๐‘– โ†’ (2) From (1) & (2), (๐‘ง1 + ๐‘ง2) + ๐‘ง3 = ๐‘ง1 + (๐‘ง2 + ๐‘ง3) (๐‘–๐‘–)(๐‘ง1๐‘ง2)๐‘ง3 = [(1 โˆ’ 3๐‘–)(โˆ’4๐‘–)]5 = (โˆ’4๐‘– + 12๐‘–

2)5 = (โˆ’4๐‘– โˆ’ 12)5 = โˆ’60 โˆ’ 20๐‘– โ†’ (1)

๐‘ง1(๐‘ง2๐‘ง2) = (1 โˆ’ 3๐‘–)[(โˆ’4๐‘–)(5)] = (1 โˆ’ 3๐‘–)[โˆ’20๐‘–] = โˆ’20๐‘– + 60๐‘–2 = โˆ’60 โˆ’ 20๐‘– โ†’ (2)

From (1) & (2), (๐‘ง1๐‘ง2)๐‘ง3 = ๐‘ง1(๐‘ง2๐‘ง2) 2.If ๐‘ง1 = 3, ๐‘ง2 = โˆ’7๐‘– and ๐‘ง3 = 5 + 4๐‘–, show that (๐‘–)๐‘ง1. (๐‘ง2 + ๐‘ง3) = ๐‘ง1๐‘ง2 + ๐‘ง1๐‘ง3 (๐‘–๐‘–)(๐‘ง1 + ๐‘ง2). ๐‘ง3 = ๐‘ง1๐‘ง3 + ๐‘ง2๐‘ง3 (๐‘–)๐‘ง1. (๐‘ง2 + ๐‘ง3) = 3(โˆ’7๐‘– + 5 + 4๐‘–) = 3(5 โˆ’ 3๐‘–) = 15 โˆ’ 9๐‘– โ†’ (1) ๐‘ง1๐‘ง2 + ๐‘ง1๐‘ง3 = (3)(โˆ’7๐‘–) + (3)(5 + 4๐‘–) = โˆ’21๐‘– + 15 + 12๐‘– = 15 โˆ’ 9๐‘– โ†’ (2)

From (1) & (2), ๐‘ง1. (๐‘ง2 + ๐‘ง3) = ๐‘ง1๐‘ง2 + ๐‘ง1๐‘ง3 (๐‘–๐‘–)(๐‘ง1 + ๐‘ง2). ๐‘ง3 = (3 โˆ’ 7๐‘–)(5 + 4๐‘–)

= 15 + 12๐‘– โˆ’ 35๐‘– โˆ’ 28๐‘–2 = 15 + 12๐‘– โˆ’ 35๐‘– + 28 = 43 โˆ’ 23๐‘– โ†’ (1) ๐‘ง1๐‘ง3 + ๐‘ง2๐‘ง3 = (3)(5 + 4๐‘–) + (โˆ’7๐‘–)(5 + 4๐‘–)

= 15 + 12๐‘– โˆ’ 35๐‘– โˆ’ 28๐‘–2 = 15 + 12๐‘– โˆ’ 35๐‘– + 28 = 43 โˆ’ 23๐‘– โ†’ (2) From (1) & (2), (๐‘ง1 + ๐‘ง2). ๐‘ง3 = ๐‘ง1๐‘ง3 + ๐‘ง2๐‘ง3 3.If ๐‘ง1 = 2 + 5๐‘–, ๐‘ง2 = โˆ’3 โˆ’ 4๐‘– and ๐‘ง3 = 1 + ๐‘–, find the additive and multiplicative inverse of ๐‘ง1, ๐‘ง2 and ๐‘ง3 Additive inverse of ๐‘ง is โˆ’๐‘ง. So, Additive inverse of ๐‘ง1 = 2 + 5๐‘– is โˆ’๐‘ง1 = โˆ’2 โˆ’ 5๐‘–, Additive inverse of ๐‘ง2 = โˆ’3 โˆ’ 4๐‘– is- ๐‘ง2 = 3 + 4๐‘– Additive inverse of ๐‘ง3 = 1 + ๐‘– is โˆ’๐‘ง3 = โˆ’1 โˆ’ ๐‘–

Multiplicative inverse of z = x + iy is zโˆ’1 = (x

x2+y2) + i (โˆ’

y

x2+y2)

Multiplicative inverse of ๐‘ง1 = 2 + 5๐‘– is z1โˆ’1 = (

2

22+52) + i (โˆ’

5

22+52) =

2

29โˆ’

5

29๐‘–

Multiplicative inverse of ๐‘ง2 = โˆ’3 โˆ’ 4๐‘– is๐‘ง2โˆ’1 = (

โˆ’3

(โˆ’3)2+(โˆ’4)2) + i (โˆ’

โˆ’4

(โˆ’3)2+(โˆ’4)2)

= โˆ’3

25+

4

25๐‘–

Multiplicative inverse of ๐‘ง3 = 1 + ๐‘– is zโˆ’1 = (1

12+12) + i (โˆ’

1

12+12) =

1

2โˆ’1

2๐‘–

CONJUGATE OF A COMPLEX NUMBER The conjugate of the complex number z = x + iy is z = x โˆ’ iy. ๐‘ง๐‘ง = ๐‘ฅ2 + ๐‘ฆ2 The conjugate is useful in division of complex numbers.

PROPERTIES OF COMPLEX CONJUGATES: z1 + z2 = z1 + z2 z1 โˆ’ z2 = z1 โˆ’ z2 z1. z2 = z1. z2

(z1

z2)

=

z1

z2 , z2 โ‰  0

Re(z) =z+z

2

Im(z) =zโˆ’z

2i

(zn) = (z)n z is real if and only if z = z z is purely imaginary if and only if z = โˆ’z z = z

EXERCISE 2.4

1.Write the following in the rectangular form:

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Padasalai TrbTnpsc

Page 4: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 4

(๐‘–)(5 + 9๐‘–) + (2 โˆ’ 4๐‘–) (๐‘–๐‘–)10โˆ’5๐‘–

6+2๐‘– (๐‘–๐‘–๐‘–)3๏ฟฝ๏ฟฝ +

1

2โˆ’๐‘–

(๐‘–)(5 + 9๐‘–) + (2 โˆ’ 4๐‘–) = 5 + 9๐‘– + 2 โˆ’ 4๐‘– = 5 โˆ’ 9๐‘– + 2 + 4๐‘– = 7 โˆ’ 5๐‘–

(๐‘–๐‘–)10โˆ’5๐‘–

6+2๐‘–=

10โˆ’5๐‘–

6+2๐‘–ร—6โˆ’2๐‘–

6โˆ’2๐‘–=

60โˆ’20๐‘–โˆ’30๐‘–+10๐‘–2

62+22

=60โˆ’50๐‘–โˆ’10

40=

50โˆ’50๐‘–

40=

5

4โˆ’5

4๐‘–

(๐‘–๐‘–๐‘–)3๏ฟฝ๏ฟฝ +1

2โˆ’๐‘–= โˆ’3๐‘– +

1

2โˆ’๐‘–ร—2+๐‘–

2+๐‘–= โˆ’3๐‘– +

2+๐‘–

22+12= โˆ’3๐‘– +

2

5+1

5๐‘–

=2

5+โˆ’15๐‘–+๐‘–

5=

2

5โˆ’14

5๐‘–

2.If z = x + iy, find the following in rectangular form:

(๐‘–)๐‘…๐‘’ (1

๐‘ง) (๐‘–๐‘–)๐‘…๐‘’(๐‘–๐‘ง) (๐‘–๐‘–๐‘–)๐ผ๐‘š(3๐‘ง + 4๐‘ง โˆ’ 4๐‘–)

(๐‘–)1

๐‘ง=

1

x+iyร—xโˆ’iy

xโˆ’iy=

xโˆ’iy

๐‘ฅ2+๐‘ฆ2=

๐‘ฅ

๐‘ฅ2+๐‘ฆ2โˆ’

๐‘ฆ

๐‘ฅ2+๐‘ฆ2๐‘–

โˆด ๐‘…๐‘’ (1

๐‘ง) = ๐‘…๐‘’ (

๐‘ฅ

๐‘ฅ2+๐‘ฆ2โˆ’

๐‘ฆ

๐‘ฅ2+๐‘ฆ2๐‘– ) =

๐‘ฅ

๐‘ฅ2+๐‘ฆ2

(๐‘–๐‘–)๐‘–๐‘ง = ๐‘–(x + iy ) = ๐‘–(๐‘ฅ โˆ’ ๐‘–๐‘ฆ) = ๐‘–๐‘ฅ โˆ’ ๐‘–2๐‘ฆ = ๐‘ฆ + ๐‘–๐‘ฅ โˆด ๐‘…๐‘’(๐‘–๐‘ง) = ๐‘…๐‘’(โˆ’๐‘ฆ + ๐‘–๐‘ฅ) = ๐‘ฆ (๐‘–๐‘–๐‘–)3๐‘ง + 4๐‘ง โˆ’ 4๐‘– = 3(x + iy) + 4x + iy โˆ’ 4๐‘– = 3๐‘ฅ + 3๐‘ฆ๐‘– + 4(๐‘ฅ โˆ’ ๐‘–๐‘ฆ) โˆ’ 4๐‘– = 3๐‘ฅ + 3๐‘ฆ๐‘– + 4๐‘ฅ โˆ’ 4๐‘ฆ๐‘– โˆ’ 4๐‘– = 7๐‘ฅ โˆ’ ๐‘ฆ๐‘– โˆ’ 4๐‘– = 7๐‘ฅ + ๐‘–(โˆ’๐‘ฆ โˆ’ 4)

โˆด ๐ผ๐‘š(3๐‘ง + 4๐‘ง โˆ’ 4๐‘–) = ๐ผ๐‘š(7๐‘ฅ + ๐‘–(โˆ’๐‘ฆ โˆ’ 4)) = โˆ’๐‘ฆ โˆ’ 4

3. If ๐‘ง1 = 2 โˆ’ ๐‘– and ๐‘ง2 = โˆ’4 + 3๐‘–, find the inverse of ๐‘ง1๐‘ง2 and z1

z2

๐‘ง1๐‘ง2 = (2 โˆ’ ๐‘– )(โˆ’4 + 3๐‘–) = โˆ’8 + 6๐‘– + 4๐‘– โˆ’ 3๐‘–2 = โˆ’8 + 10๐‘– + 3 = โˆ’5 + 10๐‘–

Inverse of ๐‘ง1๐‘ง2 = (x

x2+y2) + i (โˆ’

y

x2+y2) = (

โˆ’5

(โˆ’5)2+102) + i (โˆ’

10

(โˆ’5)2+102)

=โˆ’5

125โˆ’

10

125๐‘– = โˆ’

1

25โˆ’

2

25๐‘–

z1

z2=

2โˆ’๐‘–

โˆ’4+3๐‘–=

2โˆ’๐‘–

โˆ’4+3๐‘–ร—โˆ’4โˆ’3๐‘–

โˆ’4โˆ’3๐‘–=

โˆ’8โˆ’6๐‘–+4๐‘–+3๐‘–2

(โˆ’4)2+32=

โˆ’8โˆ’2๐‘–โˆ’3

25 = โˆ’

11

25โˆ’

2

25๐‘–

Inverse of z1

z2= (

x

x2+y2) + i (โˆ’

y

x2+y2)

= (โˆ’11 25โ„

(โˆ’11 25โ„ )2+(โˆ’2 25โ„ )

2) + i (โˆ’โˆ’2 25โ„

(โˆ’11 25โ„ )2+(โˆ’2 25โ„ )

2)

= โˆ’11

25ร—125

625

+2

25ร—125

625

๐‘– = โˆ’11

5+2

5๐‘–

4. The complex numbers ๐‘ข, ๐‘ฃ and ๐‘ค are related by 1

๐‘ข=

1

๐‘ฃ+

1

๐‘ค. If ๐‘ฃ = 3 โˆ’ 4๐‘– and

๐‘ค = 4 + 3๐‘–, find ๐‘ข in the rectangular form. 1

๐‘ข=

1

๐‘ฃ+

1

๐‘คโŸน

1

๐‘ข=

1

3โˆ’4๐‘–+

1

4+3๐‘–

= (1

3โˆ’4๐‘–ร—3+4๐‘–

3+4๐‘–) + (

1

4+3๐‘–ร—4โˆ’3๐‘–

4โˆ’3๐‘–)

=3+4๐‘–

32+42+

4โˆ’3๐‘–

42+32

=1

25(3 + 4๐‘– + 4 โˆ’ 3๐‘–)

โŸน1

๐‘ข=

7+๐‘–

25

๐‘ข =25

7+๐‘–=

25

7+๐‘–ร—7โˆ’๐‘–

7โˆ’๐‘–=

25(7โˆ’๐‘–)

72+12=

25(7โˆ’๐‘–)

50=

7โˆ’๐‘–

2=

7

2โˆ’1

2๐‘–

5.Prove the following properties:

(i)z is real if and only if z = z (๐‘–๐‘–)Re(z) =z+z

2 (๐‘–๐‘–๐‘–)Im(z) =

zโˆ’z

2i

(i)z is real โŸน z = x + 0i z = ๐‘ฅ + 0๐‘– = ๐‘ฅ โˆ’ 0๐‘– = ๐‘ฅ โŸน z is real if and only if z = z (๐‘–๐‘–) ๐ฟ๐‘’๐‘ก ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ ; ๐‘…๐‘’(๐‘ง) = ๐‘ฅ โ†’ (1) โŸน ๐‘ง = ๐‘ฅ โˆ’ ๐‘–๐‘ฆ

๐‘ง + ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ + ๐‘ฅ โˆ’ ๐‘–๐‘ฆ = 2๐‘ฅ โŸน๐‘ง+๏ฟฝ๏ฟฝ

2= ๐‘ฅ โ†’ (2)

From (1) & (2), Re(z) =z+z

2

(๐‘–๐‘–๐‘–) ๐ฟ๐‘’๐‘ก ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ ; ๐ผ๐‘š(๐‘ง) = ๐‘ฆ โ†’ (1) โŸน ๐‘ง = ๐‘ฅ โˆ’ ๐‘–๐‘ฆ

๐‘ง โˆ’ ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ โˆ’ ๐‘ฅ + ๐‘–๐‘ฆ = 2๐‘ฆ๐‘– โŸน๐‘งโˆ’๏ฟฝ๏ฟฝ

2๐‘–= ๐‘ฆ โ†’ (2)

From (1) & (2), Im(z) =zโˆ’z

2i

6.Find the least value of the positive integer n for which (โˆš3 + ๐‘–)๐‘›

(๐‘–)๐‘Ÿ๐‘’๐‘Ž๐‘™ (๐‘–๐‘–)๐‘๐‘ข๐‘Ÿ๐‘’๐‘™๐‘ฆ ๐‘–๐‘š๐‘Ž๐‘”๐‘–๐‘›๐‘Ž๐‘Ÿ๐‘ฆ

(โˆš3 + ๐‘–)3= 3โˆš3 + 9๐‘– โˆ’ 3โˆš3 โˆ’ ๐‘– = 8๐‘–

(โˆš3 + ๐‘–)6= ((โˆš3 + ๐‘–)

3)2

= (8๐‘–)2 = โˆ’64

(๐‘–)๐‘Šโ„Ž๐‘’๐‘› ๐‘› = 6, (โˆš3 + ๐‘–)๐‘›๐‘–๐‘  ๐‘Ÿ๐‘’๐‘Ž๐‘™

(๐‘–๐‘–)๐‘Šโ„Ž๐‘’๐‘› ๐‘› = 3, (โˆš3 + ๐‘–)๐‘›๐‘–๐‘  ๐‘๐‘ข๐‘Ÿ๐‘’๐‘™๐‘ฆ ๐‘–๐‘š๐‘Ž๐‘”๐‘–๐‘›๐‘Ž๐‘Ÿ๐‘ฆ

7.Show that (๐‘–)(2 + ๐‘–โˆš3)10โˆ’ (2 โˆ’ ๐‘–โˆš3)

10 is purely imaginary.

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Padasalai TrbTnpsc

Page 5: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 5

(๐‘–๐‘–) (19โˆ’7๐‘–

9+๐‘–)12+ (

20โˆ’5๐‘–

7โˆ’6๐‘–)12

is real.

(๐‘–) ๐‘ง = (2 + ๐‘–โˆš3)10โˆ’ (2 โˆ’ ๐‘–โˆš3)

10

๐‘ง = (2 + ๐‘–โˆš3)10โˆ’ (2 โˆ’ ๐‘–โˆš3)

10 = (2 + ๐‘–โˆš3)

10 โˆ’ (2 โˆ’ ๐‘–โˆš3)

10

= (2 + ๐‘–โˆš3 )

10โˆ’ (2 โˆ’ ๐‘–โˆš3

)10= (2 โˆ’ ๐‘–โˆš3)

10โˆ’ (2 + ๐‘–โˆš3)

10

= โˆ’{(2 + ๐‘–โˆš3)10โˆ’ (2 โˆ’ ๐‘–โˆš3)

10} = โˆ’๐‘ง

โˆด (2 + ๐‘–โˆš3)10โˆ’ (2 โˆ’ ๐‘–โˆš3)

10 is purely imaginary.

(๐‘–๐‘–) 19โˆ’7๐‘–

9+๐‘–=

19โˆ’7๐‘–

9+๐‘–ร—9โˆ’๐‘–

9โˆ’๐‘–=

171โˆ’19๐‘–โˆ’63๐‘–+7๐‘–2

92+12=

164โˆ’82๐‘–

82=

82(2โˆ’๐‘–)

82= 2 โˆ’ ๐‘–

20โˆ’5๐‘–

7โˆ’6๐‘–=

20โˆ’5๐‘–

7โˆ’6๐‘–ร—7+6๐‘–

7+6๐‘–=

140+120๐‘–โˆ’35๐‘–โˆ’30๐‘–2

72+62=

170+85๐‘–

85= 2 + ๐‘–

๐‘ง = (19โˆ’7๐‘–

9+๐‘–)12+ (

20โˆ’5๐‘–

7โˆ’6๐‘–)12= (2 โˆ’ ๐‘–)12 โˆ’ (2 + ๐‘–)12

๐‘ง = (2 โˆ’ ๐‘–)12 โˆ’ (2 + ๐‘–)12 = (2 โˆ’ ๐‘–)12 โˆ’ (2 + ๐‘–)12 = (2 + ๐‘–)12 โˆ’ (2 โˆ’ ๐‘–)12 = ๐‘ง

โˆด (19โˆ’7๐‘–

9+๐‘–)12+ (

20โˆ’5๐‘–

7โˆ’6๐‘–)12

is real.

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ– Show that (๐‘–)(2 + ๐‘–โˆš3)10+ (2 โˆ’ ๐‘–โˆš3)

10 is real and

(๐‘–๐‘–) (19+9๐‘–๐‘–

5โˆ’3๐‘–)15โˆ’ (

8+๐‘–

1+2๐‘–)15

is purely imaginary.

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ‘ Write 3+4๐‘–

5โˆ’12๐‘– in the x + iy form , hence find its real and imaginary

parts. 3+4๐‘–

5โˆ’12๐‘–=

3+4๐‘–

5โˆ’12๐‘–ร—5+12๐‘–

5+12๐‘–=

15+36๐‘–+20๐‘–โˆ’48

52+122=

โˆ’33+56๐‘–

169= โˆ’

33

169+

56

169๐‘–

Real part = โˆ’33

169 and imaginary part=

56

169

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ’ Simplify (1+๐‘–

1โˆ’๐‘–)3โˆ’ (

1โˆ’๐‘–

1+๐‘–)3

1+๐‘–

1โˆ’๐‘–=

1+๐‘–

1โˆ’๐‘–ร—1+๐‘–

1+๐‘–=

1+๐‘–+๐‘–+๐‘–2

12+12=

1+2๐‘–โˆ’1

2=

2๐‘–

2= ๐‘–

1โˆ’๐‘–

1+๐‘–= (

1+๐‘–

1โˆ’๐‘–)โˆ’1= โˆ’๐‘–

(1+๐‘–

1โˆ’๐‘–)3โˆ’ (

1โˆ’๐‘–

1+๐‘–)3= (๐‘–)3 โˆ’ (โˆ’๐‘–)3 = โˆ’๐‘– โˆ’ ๐‘– = โˆ’2๐‘–

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ“ If ๐‘ง+3

๐‘งโˆ’5๐‘–=

1+4๐‘–

2, find the complex number z.

๐‘ง+3

๐‘งโˆ’5๐‘–=

1+4๐‘–

2

โŸน 2(๐‘ง + 3) = (1 + 4๐‘–)(๐‘ง โˆ’ 5๐‘–) โŸน 2๐‘ง + 6 = ๐‘ง โˆ’ 5๐‘– + 4๐‘ง๐‘– โˆ’ 20๐‘–2 โŸน 2๐‘ง โˆ’ ๐‘ง โˆ’ 4๐‘ง๐‘– = โˆ’5๐‘– + 20 โˆ’ 6 โŸน ๐‘งโˆ’ 4๐‘ง๐‘– = 14 โˆ’ 5๐‘– โŸน ๐‘ง(1 โˆ’ 4๐‘–) = 14 โˆ’ 5๐‘–

โŸน ๐‘ง =14โˆ’5๐‘–

1โˆ’4๐‘–

=14โˆ’5๐‘–

1โˆ’4๐‘–ร—1+4๐‘–

1+4๐‘–=

14+56๐‘–โˆ’5๐‘–โˆ’20๐‘–2

12+42=

14+51๐‘–+20

17=

34+51๐‘–

17=

17(2+3๐‘–)

17= 2 + 3๐‘–

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ” If ๐‘ง1 = 3 โˆ’ 2๐‘– and ๐‘ง2 = 6 + 4๐‘–, find z1

z2.

z1

z2=

3โˆ’2๐‘–

6+4๐‘–=

3โˆ’2๐‘–

6+4๐‘–ร—6โˆ’4๐‘–

6โˆ’4๐‘–=

18โˆ’12๐‘–โˆ’12๐‘–โˆ’8

62+42=

10โˆ’24๐‘–

52 =

2(5โˆ’12๐‘–)

52=

5โˆ’12๐‘–

26

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ• Find ๐‘งโˆ’1, if ๐‘ง = (2 + 3๐‘–)(1 โˆ’ ๐‘–). ๐‘ง = (2 + 3๐‘–)(1 โˆ’ ๐‘–) = 2 โˆ’ 2๐‘– + 3๐‘– โˆ’ 3๐‘–2 = 2 + ๐‘– + 3 = 5 + ๐‘–

๐‘งโˆ’1 = (๐’™

๐’™๐Ÿ+๐’š๐Ÿ) + ๐’Š (โˆ’

๐’š

๐’™๐Ÿ+๐’š๐Ÿ) = (

๐Ÿ“

๐Ÿ“๐Ÿ+๐Ÿ๐Ÿ) + i (โˆ’

1

52+12) =

5

26โˆ’

1

26i

MODULUS OF A COMPLEX NUMBER

If ๐‘ง = x + iy , then the modulus of z is |๐‘ง| = โˆš๐‘ฅ2 + ๐‘ฆ2 ๐‘ง๐‘ง = |๐‘ง|2 PROPERTIES OF MODULUS OF A COMPLEX NUMBER:

|๐‘| = |๏ฟฝ๏ฟฝ| |๐‘ง1 + ๐‘ง2| โ‰ค |๐‘ง1| + |๐‘ง2| ( Triangle inequality ) |๐‘ง1 โˆ’ ๐‘ง2| โ‰ฅ |๐‘ง1| โˆ’ |๐‘ง2| |๐‘ง1๐‘ง2| = |๐‘ง1||๐‘ง2|

|๐‘ง1

๐‘ง2| =

|๐‘ง1|

|๐‘ง2|, ๐‘ง2 โ‰  0

|๐‘ง๐‘›| = |๐‘ง|๐‘› , ๐‘› is an integer. ๐‘…๐‘’(๐‘ง) โ‰ค |๐‘ง| ๐ผ๐‘š(๐‘ง) โ‰ค |๐‘ง|

||๐‘ง1| โˆ’ |๐‘ง2|| โ‰ค |๐‘ง1 + ๐‘ง2| โ‰ค |๐‘ง1| + |๐‘ง2|

The distance between the two points ๐‘ง1 and ๐‘ง2 in the complex plane is

|๐‘ง1 โˆ’ ๐‘ง2| (or) โˆš(๐‘ฅ1 โˆ’ ๐‘ฅ2)2 + (๐‘ฆ1 โˆ’ ๐‘ฆ2)

2 Formula for finding the square root of the complex number ๐‘ง = ๐‘Ž + ๐‘–๐‘ is

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Padasalai TrbTnpsc

Page 6: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 6

โˆš๐‘Ž + ๐‘–๐‘ = ยฑ(โˆš|๐‘ง|+๐‘Ž

2+ ๐‘–

๐‘

|๐‘|โˆš|๐‘ง|โˆ’๐‘Ž

2),

If b is negative ๐‘

|๐‘|= โˆ’1, x and y have different signs.

If b is positive ๐‘

|๐‘|= 1, x and y have same signs.

Exercise 2.5

1.Find the modulus of the following complex numbers:

(๐‘–)2๐‘–

3+4๐‘– (๐‘–๐‘–)

2โˆ’๐‘–

1+๐‘–+1โˆ’2๐‘–

1โˆ’๐‘– (๐‘–๐‘–๐‘–)(1 โˆ’ ๐‘–)10 (๐‘–๐‘ฃ)2๐‘–(3 โˆ’ 4๐‘–)(4 โˆ’ 3๐‘–)

|๐‘ง| = โˆš๐‘ฅ2 + ๐‘ฆ2

(๐‘–) |2๐‘–

3+4๐‘–| =

|2๐‘–|

|3+4๐‘–|=

โˆš02+22

โˆš32+42=

โˆš4

โˆš25=

2

5

(๐‘–๐‘–) |2โˆ’๐‘–

1+๐‘–+1โˆ’2๐‘–

1โˆ’๐‘–| = |

(1โˆ’๐‘–)(2โˆ’๐‘–)+(1+๐‘–)(1โˆ’2๐‘–)

(1+๐‘–)(1โˆ’๐‘–)| = |

2โˆ’๐‘–โˆ’2๐‘–+๐‘–2+1โˆ’2๐‘–+๐‘–โˆ’2๐‘–2

12+12|

= |2โˆ’๐‘–โˆ’2๐‘–โˆ’1+1โˆ’2๐‘–+2

2| = |

4โˆ’4๐‘–

2| = |2 โˆ’ 2๐‘–| = โˆš22 + 22 = โˆš8 = 2โˆš2

(๐‘–๐‘–๐‘–)|(1 โˆ’ ๐‘–)10| = |1 โˆ’ ๐‘–|10 = (โˆš12 + 12)10= โˆš2

10= 25 = 32

(๐‘–๐‘ฃ)|2๐‘–(3 โˆ’ 4๐‘–)(4 โˆ’ 3๐‘–)| = |2๐‘–||3 โˆ’ 4๐‘–||4 โˆ’ 3๐‘–| = โˆš02 + 22โˆš32 + 42โˆš42 + 32

= โˆš4โˆš25โˆš25 = 2.5.5 = 50

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ— If ๐‘ง1 = 3 + 4๐‘– , ๐‘ง2 = 5 โˆ’ 12๐‘– and ๐‘ง3 = 6 + 8๐‘–, find |๐‘ง1|, |๐‘ง2|, |๐‘ง3|, |๐‘ง1 + ๐‘ง2|, |๐‘ง2 โˆ’ ๐‘ง3| and |๐‘ง1 + ๐‘ง3|.

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐ŸŽ Find the following : (๐‘–) |2+๐‘–

โˆ’1+2๐‘–| (๐‘–๐‘–)|(1 + ๐‘–) (2 + 3๐‘–)(4๐‘– โˆ’ 3)|

(๐‘–๐‘–๐‘–) |๐‘–(2+๐‘–)3

(1+๐‘–)2|

2.For ant two complex numbers ๐‘ง1 and ๐‘ง2, such that |๐‘ง1| = |๐‘ง2| = 1 and ๐‘ง1๐‘ง2 โ‰  โˆ’1

then show that ๐‘ง1+๐‘ง2

1+๐‘ง1๐‘ง2 is a real number.

Given |๐‘ง1| = |๐‘ง2| = 1

|๐‘ง1| = 1 โŸน |๐‘ง1|2 = 1 โŸน ๐‘ง1๐‘ง1 = 1 โŸน ๐‘ง1 =

1

๐‘ง1

๐‘™๐‘™๐‘™๐‘™๐‘ฆ ๐‘ง2 =1

๐‘ง2

Let ๐‘ค =๐‘ง1+๐‘ง2

1+๐‘ง1๐‘ง2โŸน ๏ฟฝ๏ฟฝ = (

๐‘ง1+๐‘ง2

1+๐‘ง1๐‘ง2)

=

๐‘ง1 +๐‘ง2

1+๐‘ง1๐‘ง2 =

๐‘ง1 +๐‘ง2

1+๐‘ง1 ๐‘ง2 =

1

๐‘ง1+1

๐‘ง2

1+1

๐‘ง1

1

๐‘ง2

=

๐‘ง1+๐‘ง2๐‘ง1๐‘ง21+๐‘ง1๐‘ง2๐‘ง1๐‘ง2

=๐‘ง1+๐‘ง2

1+๐‘ง1๐‘ง2= ๐‘ค

Since ๏ฟฝ๏ฟฝ = ๐‘ค,๐‘ค is purely real. 3.Which one of the points 10 โˆ’ 8๐‘–, 11 + 6๐‘– is closest to 1 + ๐‘–.

Distance between the two points ๐‘ง1 and ๐‘ง2 = โˆš(๐‘ฅ1 โˆ’ ๐‘ฅ2)2 + (๐‘ฆ1 โˆ’ ๐‘ฆ2)

2 ๐‘ง = 1 + ๐‘–, ๐‘ง1 = 10 โˆ’ 8๐‘–

Distance between the two points ๐‘ง and ๐‘ง1 = โˆš(1 โˆ’ 10)2 + (1 + 8)2

= โˆš81 + 81 = โˆš162 = โˆš2 ร— 81 = 9โˆš2 ๐‘ง = 1 + ๐‘–, ๐‘ง2 = 11 + 6๐‘–

Distance between the two points ๐‘ง and ๐‘ง2 = โˆš(1 โˆ’ 11)2 + (1 โˆ’ 6)2

= โˆš100 + 125 = โˆš125 = โˆš5 ร— 25 = 5โˆš5

โˆต 5โˆš5 < 9โˆš2, 11 + 6๐‘– is the closest point to 1 + ๐‘–

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ Which one of the points ๐‘–, โˆ’2 + ๐‘– ๐‘Ž๐‘›๐‘‘ 3 is farthest from the origin?

4.If |๐‘ง| = 3, show that 7 โ‰ค |๐‘ง + 6 โˆ’ 8๐‘–| โ‰ค 13. Let ๐‘ง1 = ๐‘ง ๐‘Ž๐‘›๐‘‘ ๐‘ง2 = 6 โˆ’ 8๐‘–

W.K.T, ||๐‘ง1| โˆ’ |๐‘ง2|| โ‰ค |๐‘ง1 + ๐‘ง2| โ‰ค |๐‘ง1| + |๐‘ง2|

โŸน |3 โˆ’ โˆš62 + 82| โ‰ค |๐‘ง + 6 โˆ’ 8๐‘–| โ‰ค 3 + โˆš62 + 82

โŸน |3 โˆ’ โˆš100| โ‰ค |๐‘ง + 6 โˆ’ 8๐‘–| โ‰ค 3 + โˆš100

โŸน |3 โˆ’ 10| โ‰ค |๐‘ง + 6 โˆ’ 8๐‘–| โ‰ค 3 + 10 โŸน |โˆ’7| โ‰ค |๐‘ง + 6 โˆ’ 8๐‘–| โ‰ค 13 โŸน 7 โ‰ค |๐‘ง + 6 โˆ’ 8๐‘–| โ‰ค 13

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ‘ If |๐‘ง| = 2, show that 3 โ‰ค |๐‘ง + 3 + 4๐‘–| โ‰ค 7.

5. If |๐‘ง| = 1, show that 2 โ‰ค |๐‘ง2 โˆ’ 3| โ‰ค 4. Let ๐‘ง1 = ๐‘ง

2 ๐‘Ž๐‘›๐‘‘ ๐‘ง2 = โˆ’3

W.K.T, ||๐‘ง1| โˆ’ |๐‘ง2|| โ‰ค |๐‘ง1 + ๐‘ง2| โ‰ค |๐‘ง1| + |๐‘ง2|

โŸน ||๐‘ง|2 โˆ’ |โˆ’3|| โ‰ค |๐‘ง2 โˆ’ 3| โ‰ค |๐‘ง|2 + |โˆ’3|

โŸน |12 โˆ’ 3| โ‰ค |๐‘ง2 โˆ’ 3| โ‰ค 12 + 3 โŸน |โˆ’2| โ‰ค |๐‘ง2 โˆ’ 3| โ‰ค 4 โŸน 2 โ‰ค |๐‘ง2 โˆ’ 3| โ‰ค 4

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Padasalai TrbTnpsc

Page 7: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 7

6.If |๐‘ง โˆ’2

๐‘ง| = 2, show that the greatest and least value of |๐‘ง| are โˆš3 + 1 and

โˆš3 โˆ’ 1 respectively.

|๐‘ง โˆ’2

๐‘ง| = 2 โŸน 2 = |๐‘ง โˆ’

2

๐‘ง| โ‰ฅ ||๐‘ง| โˆ’ |

2

๐‘ง||

โŸน ||๐‘ง| โˆ’ |2

๐‘ง|| โ‰ค 2 โŸน โˆ’2 โ‰ค |๐‘ง| โˆ’ |

2

๐‘ง| โ‰ค 2 โˆต |๐‘ฅ โˆ’ ๐‘Ž| โ‰ค ๐‘Ÿ โŸน โˆ’๐‘Ÿ โ‰ค ๐‘ฅ โˆ’ ๐‘Ž โ‰ค ๐‘Ÿ

โŸนโˆ’2 โ‰ค |๐‘ง| โˆ’2

|๐‘ง|โ‰ค 2 โŸน โˆ’2 โ‰ค

|๐‘ง|2โˆ’2

|๐‘ง|โ‰ค 2 โŸน โˆ’2|๐‘ง| โ‰ค |๐‘ง|2 โˆ’ 2 โ‰ค 2|๐‘ง| โ†’ (1)

From (1) โŸน โˆ’2|๐‘ง| โ‰ค |๐‘ง|2 โˆ’ 2 โŸน 2|๐‘ง| โˆ’ 2|๐‘ง| โ‰ค |๐‘ง|2 โˆ’ 2 + 2|๐‘ง| โŸน 0 โ‰ค |๐‘ง|2 + 2|๐‘ง| โˆ’ 2 โŸน |๐‘ง|2 + 2|๐‘ง| โˆ’ 2 โ‰ฅ 0

โŸน [|๐‘ง| โˆ’ (โˆš3 + 1)][|๐‘ง| โˆ’ (โˆš3 โˆ’ 1)] โ‰ฅ 0

โŸน |๐‘ง| โ‰ฅ (โˆš3 + 1) ; |๐‘ง| โ‰ค (โˆ’โˆš3 โˆ’ 1)

โŸน |๐‘ง| โ‰ฅ (โˆš3 โˆ’ 1) โ†’ (2)

From (1) โŸน |๐‘ง|2 โˆ’ 2 โ‰ค 2|๐‘ง| โŸน |๐‘ง|2 โˆ’ 2|๐‘ง| โˆ’ 2 โ‰ค 0

โŸน [|๐‘ง| โˆ’ (1 + โˆš3)][|๐‘ง| โˆ’ (1 โˆ’ โˆš3)] โ‰ค 0

โŸน (1 โˆ’ โˆš3) โ‰ค |๐‘ง| โ‰ค (1 + โˆš3)

โŸน |๐‘ง| โ‰ค (1 + โˆš3) โ†’ (3)

From (2) & (3), โˆš3 โˆ’ 1 โ‰ค |๐‘ง| โ‰ค โˆš3 + 1

Hence the greatest and least value of |๐‘ง| are โˆš3 + 1 and โˆš3 โˆ’ 1 respectively.

7.If ๐‘ง1, ๐‘ง2 ๐‘Ž๐‘›๐‘‘ ๐‘ง3 are three complex numbers such that |๐‘ง1| = 1, |๐‘ง2| = 2, |๐‘ง3| = 3 and |๐‘ง1 + ๐‘ง2 + ๐‘ง3| = 1, show that |9๐‘ง1๐‘ง2 + 4๐‘ง1๐‘ง3 + ๐‘ง2๐‘ง3| = 6.

|๐‘ง1| = 1 โŸน |๐‘ง1|2 = 1 โŸน ๐‘ง1๐‘ง1 = 1 โŸน ๐‘ง1 =

1

๐‘ง1 โˆต ๐‘ง๐‘ง = |๐‘ง|2

|๐‘ง2| = 2 โŸน |๐‘ง2|2 = 4 โŸน ๐‘ง2๐‘ง2 = 4 โŸน ๐‘ง2 =

4

๐‘ง2

|๐‘ง3| = 3 โŸน |๐‘ง3|2 = 9 โŸน ๐‘ง3๐‘ง3 = 9 โŸน ๐‘ง3 =

9

๐‘ง3

๐‘ง1 + ๐‘ง2 + ๐‘ง3 =1

๐‘ง1 +

4

๐‘ง2 +

9

๐‘ง3

โŸน ๐‘ง1 + ๐‘ง2 + ๐‘ง3 =๐‘ง2 ๐‘ง3 +4๐‘ง1 ๐‘ง3 +9๐‘ง1 ๐‘ง2

๐‘ง1๐‘ง2๐‘ง3

โŸน |๐‘ง1 + ๐‘ง2 + ๐‘ง3| = |๐‘ง2 ๐‘ง3 +4๐‘ง1 ๐‘ง3 +9๐‘ง1 ๐‘ง2

๐‘ง1๐‘ง2๐‘ง3 |

โŸน |๐‘ง1 + ๐‘ง2 + ๐‘ง3| =|๐‘ง2 ๐‘ง3 +4๐‘ง1 ๏ฟฝ๏ฟฝ3+9๐‘ง1 ๐‘ง2 |

|๐‘ง1 ๐‘ง2 ๐‘ง3 |

โŸน |๐‘ง1 + ๐‘ง2 + ๐‘ง3| =|9๐‘ง1๐‘ง2+4๐‘ง1๐‘ง3+๐‘ง2๐‘ง3|

|๐‘ง1๐‘ง2๐‘ง3| โˆต z = z

โŸน |๐‘ง1 + ๐‘ง2 + ๐‘ง3| =|9๐‘ง1๐‘ง2+4๐‘ง1๐‘ง3+๐‘ง2๐‘ง3|

|๐‘ง1||๐‘ง2||๐‘ง3| โˆต |๐‘ง1๐‘ง2| = |๐‘ง1||๐‘ง2|

โŸน 1 =|9๐‘ง1๐‘ง2+4๐‘ง1๐‘ง3+๐‘ง2๐‘ง3|

1.2.3

โŸน |9๐‘ง1๐‘ง2 + 4๐‘ง1๐‘ง3 + ๐‘ง2๐‘ง3| = 6

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ If ๐‘ง1, ๐‘ง2 ๐‘Ž๐‘›๐‘‘ ๐‘ง3 are three complex numbers such that |๐‘ง1| =

|๐‘ง2| = |๐‘ง3| = |๐‘ง1 + ๐‘ง2 + ๐‘ง3| = 1, find the value of |1

๐‘ง1+

1

๐‘ง2+

1

๐‘ง3|.

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ“ If ๐‘ง1, ๐‘ง2 ๐‘Ž๐‘›๐‘‘ ๐‘ง3 are three complex numbers such that |๐‘ง1| =

|๐‘ง2| = |๐‘ง3| = ๐‘Ÿ > 0 and ๐‘ง1 + ๐‘ง2 + ๐‘ง3 โ‰  0. Prove that |๐‘ง1๐‘ง2+๐‘ง2๐‘ง3+๐‘ง3๐‘ง1

๐‘ง1+๐‘ง2+๐‘ง3| = ๐‘Ÿ

|๐‘ง1| = ๐‘Ÿ โŸน |๐‘ง1|2 = ๐‘Ÿ2 โŸน ๐‘ง1๐‘ง1 = ๐‘Ÿ

2 โŸน ๐‘ง1 =๐‘Ÿ2

๐‘ง1 โˆต ๐‘ง๐‘ง = |๐‘ง|2

|๐‘ง2| = ๐‘Ÿ โŸน |๐‘ง2|2 = ๐‘Ÿ2 โŸน ๐‘ง2๐‘ง2 = ๐‘Ÿ

2 โŸน ๐‘ง2 =๐‘Ÿ2

๐‘ง2

|๐‘ง3| = ๐‘Ÿ โŸน |๐‘ง3|2 = ๐‘Ÿ2 โŸน ๐‘ง3๐‘ง3 = ๐‘Ÿ

2 โŸน ๐‘ง3 =๐‘Ÿ2

๐‘ง3

๐‘ง1 + ๐‘ง2 + ๐‘ง3 =๐‘Ÿ2

๐‘ง1 +๐‘Ÿ2

๐‘ง2 +๐‘Ÿ2

๐‘ง3 = ๐‘Ÿ2 (

๐‘ง2 ๐‘ง3 +๐‘ง1 ๐‘ง3 +๐‘ง1 ๐‘ง2

๐‘ง1๐‘ง2๐‘ง3 )

โŸน |๐‘ง1 + ๐‘ง2 + ๐‘ง3| = |๐‘Ÿ2| |๐‘ง2 ๐‘ง3 +๐‘ง1 ๐‘ง3 +๐‘ง1 ๐‘ง2

๐‘ง1๐‘ง2๐‘ง3 | = ๐‘Ÿ2

|๐‘ง2 ๐‘ง3 +๐‘ง1 ๐‘ง3 +๐‘ง1 ๐‘ง2 |

|๐‘ง1๐‘ง2๐‘ง3 |= ๐‘Ÿ2

|๐‘ง1๐‘ง2+๐‘ง2๐‘ง3+๐‘ง3๐‘ง1|

|๐‘ง1||๐‘ง2||๐‘ง3|

โŸน |๐‘ง1 + ๐‘ง2 + ๐‘ง3| = ๐‘Ÿ2 |๐‘ง1๐‘ง2+๐‘ง2๐‘ง3+๐‘ง3๐‘ง1|

๐‘Ÿ.๐‘Ÿ.๐‘ŸโŸน |๐‘ง1 + ๐‘ง2 + ๐‘ง3| =

|๐‘ง1๐‘ง2+๐‘ง2๐‘ง3+๐‘ง3๐‘ง1|

๐‘Ÿ

โŸน |๐‘ง1๐‘ง2+๐‘ง2๐‘ง3+๐‘ง3๐‘ง1

๐‘ง1+๐‘ง2+๐‘ง3| = ๐‘Ÿ

8.If the area of the triangle formed by the vertices ๐‘ง, ๐‘–๐‘ง ๐‘Ž๐‘›๐‘‘ ๐‘ง + ๐‘–๐‘ง is 50 square units, find the value of |๐‘ง|.

Let ๐‘ง = ๐‘Ž + ๐‘–๐‘ = (๐‘Ž, ๐‘) โŸน |๐‘ง| = โˆš๐‘Ž2 + ๐‘2 Vertices : ๐‘ง = (๐‘Ž, ๐‘) ๐‘–๐‘ง = ๐‘–(๐‘Ž + ๐‘–๐‘) = ๐‘–๐‘Ž + ๐‘–2๐‘ = โˆ’๐‘ + ๐‘–๐‘Ž = (โˆ’๐‘, ๐‘Ž) ๐‘ง + ๐‘–๐‘ง = ๐‘Ž + ๐‘–๐‘ โˆ’ ๐‘ + ๐‘–๐‘Ž = (๐‘Ž โˆ’ ๐‘) + ๐‘–(๐‘Ž + ๐‘) = (๐‘Ž โˆ’ ๐‘, ๐‘Ž + ๐‘)

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Padasalai TrbTnpsc

Page 8: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 8

Area of the triangle = 50 sq. units

โŸน1

2|๐‘Ž โˆ’๐‘ ๐‘Ž โˆ’ ๐‘๐‘ ๐‘Ž ๐‘Ž + ๐‘

๐‘Ž๐‘| = 50

โŸน (๐‘Ž2 โˆ’ ๐‘Ž๐‘ โˆ’ ๐‘2 + ๐‘Ž๐‘ โˆ’ ๐‘2) โˆ’ (โˆ’๐‘2 + ๐‘Ž2 โˆ’ ๐‘Ž๐‘ + ๐‘Ž2 + ๐‘Ž๐‘) = 50 ร— 2 โŸน ๐‘Ž2 โˆ’ 2๐‘2 + ๐‘2 โˆ’ 2๐‘Ž2 = 100 โŸนโˆ’๐‘Ž2 โˆ’ ๐‘2 = 100 โŸนโˆ’(๐‘Ž2 + ๐‘2) = ยฑ100 โŸน ๐‘Ž2 + ๐‘2 = โˆ“100

โŸนโˆš๐‘Ž2 + ๐‘2 = โˆ“10 โŸน |๐‘ง| = 10 ( โˆต ๐‘Ž2 + ๐‘2 is not negative) NOTE: This can also done by using distance formula, find the length of base and

height and use ๐‘จ๐’“๐’†๐’‚ ๐’๐’‡ ๐’•๐’‰๐’† โˆ†=๐Ÿ

๐Ÿ๐’ƒ๐’‰

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ’ Show that the points 1,โˆ’1

2+ ๐‘–

โˆš3

2 and โˆ’

1

2โˆ’ ๐‘–

โˆš3

2 are the vertices

of an equilateral triangle.

Let ๐‘ง1 = 1, ๐‘ง2 = โˆ’1

2+ ๐‘–

โˆš3

2 and ๐‘ง3 = โˆ’

1

2โˆ’ ๐‘–

โˆš3

2

The length of the sides of the triangle are

|๐‘ง1 โˆ’ ๐‘ง2| = |1 โˆ’ (โˆ’1

2+ ๐‘–

โˆš3

2)| = |

3

2โˆ’ ๐‘–

โˆš3

2| = โˆš(

3

2)2+ (

โˆš3

2)2

= โˆš9

4+3

4= โˆš

12

4= โˆš3

|๐‘ง2 โˆ’ ๐‘ง3| = |(โˆ’1

2+ ๐‘–

โˆš3

2) โˆ’ (โˆ’

1

2โˆ’ ๐‘–

โˆš3

2)| = |โˆ’

1

2+ ๐‘–

โˆš3

2+1

2+ ๐‘–

โˆš3

2| = โˆš(

2โˆš3

2)2

= โˆš3

|๐‘ง3 โˆ’ ๐‘ง1| = |(โˆ’1

2โˆ’ ๐‘–

โˆš3

2) โˆ’ 1| = |โˆ’

3

2โˆ’ ๐‘–

โˆš3

2| = โˆš(

3

2)2+ (

โˆš3

2)2

= โˆš9

4+3

4= โˆš3

Since all the three sides are equal, the given points form an equilateral triangle.

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ– Given the complex number = 3 + 2๐‘– , represent the complex numbers ๐‘ง, ๐‘–๐‘ง and ๐‘ง + ๐‘–๐‘ง in one Argand diagram. Show that these complex numbers form the vertices of an isosceles right triangle. Given : ๐‘ง = 3 + 2๐‘– = ๐ด(3,2) ๐‘–๐‘ง = ๐‘–(3 + 2๐‘–) = โˆ’2 + 3๐‘– = ๐ต(โˆ’2,3) ๐‘ง + ๐‘–๐‘ง = 3 + 2๐‘– โˆ’ 2 + 3๐‘– = 1 + 5๐‘– = ๐ถ(1,5) The length of the sides of the triangle are

๐ด๐ต = โˆš(๐‘ฅ1 โˆ’ ๐‘ฅ2)2 + (๐‘ฆ1 โˆ’ ๐‘ฆ2)

2 = โˆš(3 + 2)2 + (2 โˆ’ 3)2 = โˆš25 + 1 = โˆš26

๐ต๐ถ = โˆš(๐‘ฅ1 โˆ’ ๐‘ฅ2)2 + (๐‘ฆ1 โˆ’ ๐‘ฆ2)

2 = โˆš(โˆ’2 โˆ’ 1)2 + (3 โˆ’ 5)2 = โˆš9 + 4 = โˆš13

๐ถ๐ด = โˆš(๐‘ฅ1 โˆ’ ๐‘ฅ2)2 + (๐‘ฆ1 โˆ’ ๐‘ฆ2)

2 = โˆš(1 โˆ’ 3)2 + (5 โˆ’ 2)2 = โˆš4 + 9 = โˆš13 ๐ด๐ต2 = 26 , ๐ต๐ถ2 = 13, ๐ถ๐ด2 = 13

Since (๐‘–)๐ต๐ถ = ๐ถ๐ด = โˆš13 and (๐‘–๐‘–)๐ต๐ถ2 + ๐ถ๐ด2 = 13 + 13 = 26 = ๐ด๐ต2, So the given vertices form an isosceles right triangle.

9.Show that the equation ๐‘ง3 + 2๐‘ง = 0 has five solutions. ๐‘ง3 + 2๐‘ง = 0 โŸน ๐‘ง3 = โˆ’2๐‘ง โŸน |๐‘ง3| = |โˆ’2||๐‘ง| โŸน |๐‘ง3| = 2|๐‘ง| โŸน |๐‘ง|3 โˆ’ 2|๐‘ง| = 0

โŸน |๐‘ง|(|๐‘ง|2 โˆ’ 2) = 0 โŸน {|๐‘ง| = 0 โŸน ๐‘ง = 0 ๐‘–๐‘  ๐‘œ๐‘›๐‘’ ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘œ๐‘™๐‘ข๐‘ก๐‘–๐‘œ๐‘›

|๐‘ง|2 โˆ’ 2 = 0 โŸน ๐‘ง๐‘ง โˆ’ 2 = 0 โŸน ๐‘ง =2

๐‘ง

Put ๐‘ง =2

๐‘ง in ๐‘ง3 + 2๐‘ง = 0 โŸน ๐‘ง3 + 2(

2

๐‘ง) = 0 โŸน ๐‘ง4 + 4 = 0, which provides four

solution. Hence ๐‘ง3 + 2๐‘ง = 0 has five solutions.

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ” Show that the equation ๐‘ง2 = ๐‘ง has four solutions.

10.Find the square root of (๐‘–)4 + 3๐‘– (๐‘–๐‘–) โˆ’ 6 + 8๐‘– (๐‘–๐‘–๐‘–) โˆ’ 5 โˆ’ 12๐‘–.

โˆš๐‘Ž + ๐‘–๐‘ = ยฑ(โˆš|๐‘ง| + ๐‘Ž

2+ ๐‘–

๐‘

|๐‘|โˆš|๐‘ง| โˆ’ ๐‘Ž

2)

(๐‘–)4 + 3๐‘–

๐‘Ž = 4, ๐‘ = 3, |๐‘ง| = โˆš42 + 32 = โˆš25 = 5

โˆš4 + 3๐‘– = ยฑ(โˆš5+4

2+ ๐‘–

3

3โˆš5โˆ’4

2) = ยฑ(

3

โˆš2+ ๐‘–

1

โˆš2)

(๐‘–๐‘–) โˆ’ 6 + 8๐‘–

๐‘Ž = โˆ’6, ๐‘ = 8, |๐‘ง| = โˆš62 + 82 = โˆš100 = 10

โˆšโˆ’6 + 8๐‘– = ยฑ(โˆš10โˆ’6

2+ ๐‘–

8

8โˆš10+6

2) = ยฑ(โˆš2 + ๐‘–โˆš8) = ยฑ(โˆš2 + ๐‘–2โˆš2)

(๐‘–๐‘–๐‘–) โˆ’ 5 โˆ’ 12๐‘– 4

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Padasalai TrbTnpsc

Page 9: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 9

๐‘Ž = โˆ’5, ๐‘ = โˆ’12, |๐‘ง| = โˆš52 + 122 = โˆš169 = 13

โˆšโˆ’5 โˆ’ 12๐‘– = ยฑ(โˆš13โˆ’5

2+ ๐‘–

โˆ’12

12โˆš13+5

2) = ยฑ(โˆš4 โˆ’ ๐‘–โˆš9) = ยฑ(2 โˆ’ 3๐‘–)

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ• Find the square root of 6 + 8๐‘–

GEOMETRY AND LOCUS OF COMPLEX NUMBERS

A circle is defined as the locus of a point which moves in a plane such that its distance from a fixed point in that plane is always a constant. The fixed point is the centre and the constant distant is the radius of the circle.

Equation of complex form of a circle is |๐‘ง โˆ’ ๐‘ง0| = ๐‘Ÿ. |๐‘ง โˆ’ ๐‘ง0| < ๐‘Ÿ represents the points interior of the circle. |๐‘ง โˆ’ ๐‘ง0| > ๐‘Ÿ represents the points exterior of the circle. ๐‘ฅ2 + ๐‘ฆ2 = ๐‘Ÿ2 represents a circle with centre at the origin and the radius r

units.

๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2

If z =a+ib

๐‘+๐‘–๐‘‘then Re(Z) =

๐‘Ž๐‘+๐‘๐‘‘

๐‘2+๐‘‘2

If z =a+ib

๐‘+๐‘–๐‘‘ then Im(Z) =

๐‘๐‘โˆ’๐‘Ž๐‘‘

๐‘2+๐‘‘2

EXERCISE 2.6

1.If ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ is a complex number such that |๐‘งโˆ’4๐‘–

๐‘ง+4๐‘–| = 1. Show that the locus of z is

real axis. Given: ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ

|๐‘งโˆ’4๐‘–

๐‘ง+4๐‘–| = 1 โŸน |๐‘ง โˆ’ 4๐‘–| = |๐‘ง + 4๐‘–|

โŸน |๐‘ฅ + ๐‘–๐‘ฆ โˆ’ 4๐‘–| = |๐‘ฅ + ๐‘–๐‘ฆ + 4๐‘–| โŸน |๐‘ฅ + ๐‘–(๐‘ฆ โˆ’ 4)| = |๐‘ฅ + ๐‘–(๐‘ฆ + 4)|

โŸนโˆš๐‘ฅ2 + (๐‘ฆ โˆ’ 4)2 = โˆš๐‘ฅ2 + (๐‘ฆ โˆ’ 4)2 Squaring on both sides, โŸน ๐‘ฅ2 + (๐‘ฆ โˆ’ 4)2 = ๐‘ฅ2 + (๐‘ฆ โˆ’ 4)2 โŸน ๐‘ฅ2 + ๐‘ฆ2 โˆ’ 8๐‘ฆ + 16 = ๐‘ฅ2 + ๐‘ฆ2 + 8๐‘ฆ + 16 โŸน 8๐‘ฆ = 0 โŸน ๐‘ฆ = 0

โˆด The locus of z is real axis.

2. If ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ is a complex number such that ๐ผ๐‘š (2๐‘ง+1

๐‘–๐‘ง+1) = 0 , show that the locus

of z is 2๐‘ฅ2 + 2๐‘ฆ2 + ๐‘ฅ โˆ’ 2๐‘ฆ = 0. Given: ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ 2๐‘ง+1

๐‘–๐‘ง+1=

2(๐‘ฅ+๐‘–๐‘ฆ)+1

๐‘–(๐‘ฅ+๐‘–๐‘ฆ)+1 =

(2๐‘ฅ+1)+๐‘–2๐‘ฆ

(1โˆ’๐‘ฆ)+๐‘–๐‘ฅ Im(Z) =

๐‘๐‘โˆ’๐‘Ž๐‘‘

๐‘2+๐‘‘2

Here ๐‘Ž = 2๐‘ฅ + 1 , ๐‘ = 2๐‘ฆ, ๐‘ = 1 โˆ’ ๐‘ฆ, ๐‘‘ = ๐‘ฅ

๐ผ๐‘š (2๐‘ง+1

๐‘–๐‘ง+1) = 0 โŸน

2๐‘ฆ(1โˆ’๐‘ฆ)โˆ’๐‘ฅ(2๐‘ฅ+1)

(1โˆ’๐‘ฆ)2+๐‘ฅ2= 0 โŸน 2๐‘ฆ โˆ’ 2๐‘ฆ2 โˆ’ 2๐‘ฅ2 โˆ’ ๐‘ฅ = 0

โŸน 2๐‘ฅ2 + 2๐‘ฆ2 + ๐‘ฅ โˆ’ 2๐‘ฆ = 0 3.Obtain the Cartesian form of the locus of ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ in each of the following cases: (๐‘–)[๐‘…๐‘’(๐‘–๐‘ง)]2 = 3 (๐‘–๐‘–)๐ผ๐‘š[(1 โˆ’ ๐‘–)๐‘ง + 1] = 0 (๐‘–๐‘–๐‘–)|๐‘ง + ๐‘–| = |๐‘ง โˆ’ 1| (๐‘–๐‘ฃ)๐‘ง = ๐‘งโˆ’1 Given: ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ (๐‘–)๐‘–๐‘ง = ๐‘–(๐‘ฅ + ๐‘–๐‘ฆ) = ๐‘ฆ โˆ’ ๐‘–๐‘ฅ

[๐‘…๐‘’(๐‘–๐‘ง)]2 = 3 โŸน [๐‘…๐‘’(๐‘ฆ โˆ’ ๐‘–๐‘ฅ )]2 = 3 โŸน ๐‘ฆ2 = 3

(๐‘–๐‘–)(1 โˆ’ ๐‘–)๐‘ง + 1 = (1 โˆ’ ๐‘–)(๐‘ฅ + ๐‘–๐‘ฆ) + 1 = ๐‘ฅ + ๐‘–๐‘ฆ โˆ’ ๐‘–๐‘ฅ + ๐‘ฆ + 1 = (๐‘ฅ + ๐‘ฆ + 1) + ๐‘–(๐‘ฆ โˆ’ ๐‘ฅ)

๐ผ๐‘š[(1 โˆ’ ๐‘–)๐‘ง + 1] = 0 โŸน ๐ผ๐‘š[(๐‘ฅ + ๐‘ฆ + 1) + ๐‘–(๐‘ฆ โˆ’ ๐‘ฅ)] = 0 โŸน ๐‘ฆโˆ’ ๐‘ฅ = 0 (๐‘œ๐‘Ÿ)๐‘ฅ โˆ’ ๐‘ฆ = 0

(๐‘–๐‘–๐‘–) |๐‘ง + ๐‘–| = |๐‘ง โˆ’ 1| โŸน |๐‘ฅ + ๐‘–๐‘ฆ + ๐‘–| = |๐‘ฅ + ๐‘–๐‘ฆ โˆ’ 1| โŸน |๐‘ฅ + ๐‘–(๐‘ฆ + 1)| = |(๐‘ฅ โˆ’ 1) + ๐‘–๐‘ฆ|

โŸนโˆš(๐‘ฅ)2 + (๐‘ฆ + 1)2 = โˆš(๐‘ฅ โˆ’ 1)2 + (๐‘ฆ)2 Squaring on both sides, โŸน (๐‘ฅ)2 + (๐‘ฆ + 1)2 = (๐‘ฅ โˆ’ 1)2 + (๐‘ฆ)2 โŸน ๐‘ฅ2 + ๐‘ฆ2 + 2๐‘ฆ + 1 = ๐‘ฅ2 โˆ’ 2๐‘ฅ + 1 + ๐‘ฆ2 โŸน 2๐‘ฅ + 2๐‘ฆ = 0 โŸน ๐‘ฅ + ๐‘ฆ = 0

(๐‘–๐‘ฃ)๐‘ง = ๐‘งโˆ’1 โŸน ๐‘ง =1

๐‘งโŸน ๐‘ง๐‘ง = 1 โŸน (๐‘ฅ + ๐‘–๐‘ฆ)(๐‘ฅ โˆ’ ๐‘–๐‘ฆ) = 1 โŸน ๐‘ฅ2 + ๐‘ฆ2 = 1

4.Show that the following equations represent a circle, and find its centre and

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Padasalai TrbTnpsc

Page 10: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 10

radius: (๐‘–)|๐‘ง โˆ’ 2 โˆ’ ๐‘–| = 3 (๐‘–๐‘–)|2๐‘ง + 2 โˆ’ 4๐‘–| = 2 (๐‘–๐‘–๐‘–)|3๐‘ง โˆ’ 6 + 12๐‘–| = 8

๐ธ๐‘ž๐‘ข๐‘Ž๐‘ก๐‘–๐‘œ๐‘› ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘๐‘–๐‘Ÿ๐‘๐‘™๐‘’ ๐‘–๐‘  |๐‘ง โˆ’ ๐‘ง0| = ๐‘Ÿ

(๐‘–)|๐‘ง โˆ’ 2 โˆ’ ๐‘–| = 3 โŸน |๐‘ง โˆ’ (2 + ๐‘–)| = 3 Centre = 2 + ๐‘– = (2,1) ; Radius = 3 ๐‘ข๐‘›๐‘–๐‘ก๐‘ 

(๐‘–๐‘–)|2๐‘ง + 2 โˆ’ 4๐‘–| = 2 โŸน 2 |๐‘ง + 1 โˆ’ 2๐‘–| = 2 โŸน |๐‘ง โˆ’ (โˆ’1 + 2๐‘–)| = 1 Centre = โˆ’1 + 2๐‘– = (โˆ’1,2) ; Radius = 1 ๐‘ข๐‘›๐‘–๐‘ก

(๐‘–๐‘–๐‘–)|3๐‘ง โˆ’ 6 + 12๐‘–| = 8 โŸน 3|๐‘ง โˆ’ 2 + 4๐‘–| = 8 โŸน |๐‘ง โˆ’ (2 โˆ’ 4๐‘–)| =8

3

Centre = 2 โˆ’ 4๐‘– = (2,โˆ’4) ; Radius =8

3 ๐‘ข๐‘›๐‘–๐‘ก๐‘ 

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ— Show that |3๐‘ง โˆ’ 5 + ๐‘–| = 4 represent a circle, and find its centre and radius. 5. Obtain the Cartesian form of the locus of ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ in each of the following cases: (๐‘–)|๐‘ง โˆ’ 4| = 16 (๐‘–๐‘–)|๐‘ง โˆ’ 4|2 โˆ’ |๐‘ง โˆ’ 1|2 = 16 Given: ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ (๐‘–)|๐‘ง โˆ’ 4| = 16 โŸน |๐‘ฅ + ๐‘–๐‘ฆ โˆ’ 4| = 16

โŸน |(๐‘ฅ โˆ’ 4) + ๐‘–๐‘ฆ| = 16

โŸนโˆš(๐‘ฅ โˆ’ 4)2 + ๐‘ฆ2 = 16 Squaring on both sides, โŸน (๐‘ฅ โˆ’ 4)2 + ๐‘ฆ2 = 162 โŸน ๐‘ฅ2 โˆ’ 8๐‘ฅ + 16 + ๐‘ฆ2 = 256 โŸน ๐‘ฅ2 + ๐‘ฆ2 โˆ’ 8๐‘ฅ โˆ’ 240 = 0 is the locus of z.

(๐‘–๐‘–)|๐‘ง โˆ’ 4|2 โˆ’ |๐‘ง โˆ’ 1|2 = 16 โŸน |๐‘ฅ + ๐‘–๐‘ฆ โˆ’ 4|2 โˆ’ |๐‘ฅ + ๐‘–๐‘ฆ โˆ’ 1|2 = 16 โŸน |(๐‘ฅ โˆ’ 4) + ๐‘–๐‘ฆ|2 โˆ’ |(๐‘ฅ โˆ’ 1) + ๐‘–๐‘ฆ|2 = 16

โŸน (โˆš(๐‘ฅ โˆ’ 4)2 + ๐‘ฆ2)2โˆ’โˆš(๐‘ฅ โˆ’ 1)2 + ๐‘ฆ2

2= 16

โŸน (๐‘ฅ โˆ’ 4)2 + ๐‘ฆ2 โˆ’ [(๐‘ฅ โˆ’ 1)2 + ๐‘ฆ2] = 16 โŸน ๐‘ฅ2 โˆ’ 8๐‘ฅ + 16 + ๐‘ฆ2 โˆ’ ๐‘ฅ2 + 2๐‘ฅ โˆ’ 1 โˆ’ ๐‘ฆ2 = 16 โŸนโˆ’6๐‘ฅ โˆ’ 1 = 0 โŸน 6๐‘ฅ + 1 = 0 is the locus of z.

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ Obtain the Cartesian form of the locus of ๐‘ง in each of the following cases: (๐‘–)|๐‘ง| = |๐‘ง โˆ’ ๐‘–| (๐‘–๐‘–)|2๐‘ง โˆ’ 3 โˆ’ ๐‘–| = 3

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐ŸŽ Show that |๐‘ง + 2 โˆ’ ๐‘–| < 2 represents interior points of a circle. Find its centre and radius.

Consider |๐‘ง + 2 โˆ’ ๐‘–| = 2 โŸน |๐‘ง โˆ’ (โˆ’2 + ๐‘–)| = 2. Centre = โˆ’2 + ๐‘– = (โˆ’2,1) ; Radius = 2 ๐‘ข๐‘›๐‘–๐‘ก๐‘ 

โˆด |๐‘ง + 2 โˆ’ ๐‘–| < 2 represents all points inside the circle.

POLAR AND EULER FORM OF A COMPLEX NUMBER Polar form of ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ is ๐‘ง = ๐‘Ÿ(cos ๐œƒ + ๐‘– sin ๐œƒ) (๐‘œ๐‘Ÿ) ๐‘ง = ๐‘Ÿ๐‘๐‘–๐‘ ๐œƒ

๐‘€๐‘œ๐‘‘๐‘ข๐‘™๐‘ข๐‘  ๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2

๐›ผ = ๐‘ก๐‘Ž๐‘›โˆ’1 (๐‘ฆ

๐‘ฅ)

arg ๐‘ง = ๐œƒ ;

{

๐œƒ = ๐›ผ ; ๐‘–๐‘“(+,+) ๐œƒ ๐‘™๐‘–๐‘’๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐ผ๐‘ ๐‘ก ๐‘ž๐‘ข๐‘Ž๐‘‘๐‘Ÿ๐‘Ž๐‘›๐‘ก

๐œƒ = ๐œ‹ โˆ’ ๐›ผ ; ๐‘–๐‘“(โˆ’,+) ๐œƒ ๐‘™๐‘–๐‘’๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ 2๐‘›๐‘‘ ๐‘ž๐‘ข๐‘Ž๐‘‘๐‘Ÿ๐‘Ž๐‘›๐‘ก

๐œƒ = โˆ’๐œ‹ + ๐›ผ ; ๐‘–๐‘“(โˆ’,โˆ’) ๐œƒ ๐‘™๐‘–๐‘’๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ 3๐‘Ÿ๐‘‘ ๐‘ž๐‘ข๐‘Ž๐‘‘๐‘Ÿ๐‘Ž๐‘›๐‘ก

๐œƒ = โˆ’๐›ผ ; ๐‘–๐‘“(+,โˆ’) ๐œƒ ๐‘™๐‘–๐‘’๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ 4๐‘กโ„Ž ๐‘ž๐‘ข๐‘Ž๐‘‘๐‘Ÿ๐‘Ž๐‘›๐‘ก

In general arg ๐‘ง =๐ด๐‘Ÿ๐‘” ๐‘ง + 2๐‘›๐œ‹ , ๐‘› โˆˆ ๐‘ง. Properties of arguments:

๐‘Ž๐‘Ÿ๐‘”(๐‘ง1๐‘ง2) = ๐‘Ž๐‘Ÿ๐‘”(๐‘ง1) + ๐‘Ž๐‘Ÿ๐‘”(๐‘ง2)

๐‘Ž๐‘Ÿ๐‘” (๐‘ง1

๐‘ง2) = ๐‘Ž๐‘Ÿ๐‘”(๐‘ง1) โˆ’ ๐‘Ž๐‘Ÿ๐‘”(๐‘ง2)

๐‘Ž๐‘Ÿ๐‘”(๐‘ง๐‘›) = ๐‘› ๐‘Ž๐‘Ÿ๐‘”๐‘ง Some of the principal argument and argument:

๐‘ง 1 ๐‘– โˆ’1 โˆ’๐‘–

๐ด๐‘Ÿ๐‘” ๐‘ง

0

๐œ‹

2

๐œ‹

โˆ’๐œ‹

2

๐‘Ž๐‘Ÿ๐‘”๐‘ง

2๐‘›๐œ‹

2๐‘›๐œ‹ +๐œ‹

2

2๐‘›๐œ‹ + ๐œ‹

2๐‘›๐œ‹ โˆ’๐œ‹

2

Eulerโ€™s form of the complex number: ๐‘ง = ๐‘Ÿ๐‘’๐‘–๐œƒ ; ๐‘’๐‘–๐œƒ = cos ๐œƒ + ๐‘– sin ๐œƒ Properties of polar form:

๐‘ง = ๐‘Ÿ(cos ๐œƒ + ๐‘– sin ๐œƒ) โŸน ๐‘งโˆ’1 =1

๐‘Ÿ(cos๐œƒ โˆ’ ๐‘– sin ๐œƒ)

If ๐‘ง1 = ๐‘Ÿ1(cos ๐œƒ1 + ๐‘– sin ๐œƒ1), ๐‘ง2 = ๐‘Ÿ2(cos ๐œƒ2 + ๐‘– sin ๐œƒ2) ๐‘กโ„Ž๐‘’๐‘› ๐‘ง1๐‘ง2 = ๐‘Ÿ1๐‘Ÿ2[cos(๐œƒ1 + ๐œƒ2) + ๐‘– sin(๐œƒ1 + ๐œƒ2)]

If ๐‘ง1 = ๐‘Ÿ1(cos ๐œƒ1 + ๐‘– sin ๐œƒ1), ๐‘ง2 = ๐‘Ÿ2(cos ๐œƒ2 + ๐‘– sin ๐œƒ2) ๐‘กโ„Ž๐‘’๐‘› ๐‘ง1

๐‘ง2=

๐‘Ÿ1

๐‘Ÿ2[cos(๐œƒ1 โˆ’ ๐œƒ2) + ๐‘– sin(๐œƒ1 โˆ’ ๐œƒ2)]

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Padasalai TrbTnpsc

Page 11: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 11

Some of the useful polar forms :

1 = ๐‘๐‘–๐‘ (0) ; โˆ’1 = ๐‘๐‘–๐‘ (๐œ‹) ; ๐‘– = ๐‘๐‘–๐‘  (๐œ‹

2) ; โˆ’๐‘– = ๐‘๐‘–๐‘  (โˆ’

๐œ‹

2)

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ Find the modulus and principal argument of the following

complex numbers: (๐‘–)โˆš3 + ๐‘– (๐‘–๐‘–) โˆ’ โˆš3 + ๐‘– (๐‘–๐‘–๐‘–) โˆ’ โˆš3 โˆ’ ๐‘– (๐‘–๐‘ฃ)โˆš3 โˆ’ ๐‘–

(๐‘–)โˆš3 + ๐‘–

๐‘€๐‘œ๐‘‘๐‘ข๐‘™๐‘ข๐‘  ๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2 = โˆšโˆš32+ 12 = โˆš4 = 2

๐›ผ = ๐‘ก๐‘Ž๐‘›โˆ’1 |๐‘ฆ

๐‘ฅ| = ๐‘ก๐‘Ž๐‘›โˆ’1 |

1

โˆš3| =

๐œ‹

6

Since the complex number โˆš3 + ๐‘– lies in the first quadrant, principal argument

๐œƒ = ๐›ผ โŸน ๐œƒ =๐œ‹

6

(๐‘–๐‘–) โˆ’ โˆš3 + ๐‘–

๐‘€๐‘œ๐‘‘๐‘ข๐‘™๐‘ข๐‘  ๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2 = โˆš(โˆ’โˆš3)2+ 12 = โˆš4 = 2

๐›ผ = ๐‘ก๐‘Ž๐‘›โˆ’1 (๐‘ฆ

๐‘ฅ) = ๐‘ก๐‘Ž๐‘›โˆ’1 |

1

โˆ’โˆš3| =

๐œ‹

6

Since the complex number โˆ’โˆš3 + ๐‘– lies in the second quadrant, principal

argument ๐œƒ = ๐œ‹ โˆ’ ๐›ผ โŸน ๐œƒ = ๐œ‹ โˆ’๐œ‹

6=

5๐œ‹

6

(๐‘–๐‘–๐‘–) โˆ’ โˆš3 โˆ’ ๐‘–

๐‘€๐‘œ๐‘‘๐‘ข๐‘™๐‘ข๐‘  ๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2 = โˆš(โˆ’โˆš3)2+ (โˆ’1)2 = โˆš4 = 2

๐›ผ = ๐‘ก๐‘Ž๐‘›โˆ’1 (๐‘ฆ

๐‘ฅ) = ๐‘ก๐‘Ž๐‘›โˆ’1 |

โˆ’1

โˆ’โˆš3| =

๐œ‹

6

Since the complex number โˆ’โˆš3 โˆ’ ๐‘– lies in the third quadrant, principal argument

๐œƒ = โˆ’๐œ‹ + ๐›ผ โŸน ๐œƒ = โˆ’๐œ‹ +๐œ‹

6= โˆ’

5๐œ‹

6

(๐‘–๐‘ฃ)โˆš3 โˆ’ ๐‘–

๐‘€๐‘œ๐‘‘๐‘ข๐‘™๐‘ข๐‘  ๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2 = โˆšโˆš32+ (โˆ’1)2 = โˆš4 = 2

๐›ผ = ๐‘ก๐‘Ž๐‘›โˆ’1 (๐‘ฆ

๐‘ฅ) = ๐‘ก๐‘Ž๐‘›โˆ’1 |

โˆ’1

โˆš3| =

๐œ‹

6

Since the complex number โˆš3 โˆ’ ๐‘– lies in the fourth quadrant, principal argument

๐œƒ = โˆ’๐›ผ โŸน ๐œƒ = โˆ’๐œ‹

6

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ’ Find the principal argument ๐ด๐‘Ÿ๐‘” ๐‘ง, when ๐‘ง =โˆ’2

1+๐‘–โˆš3

๐‘Ž๐‘Ÿ๐‘”๐‘ง = ๐‘Ž๐‘Ÿ๐‘” (โˆ’2

1+๐‘–โˆš3) = ๐‘Ž๐‘Ÿ๐‘”(โˆ’2) โˆ’ ๐‘Ž๐‘Ÿ๐‘”(1 + ๐‘–โˆš3) โ†’ (1)

๐‘Ž๐‘Ÿ๐‘”(โˆ’2) = ๐‘ก๐‘Ž๐‘›โˆ’1 |๐‘ฆ

๐‘ฅ| = ๐‘ก๐‘Ž๐‘›โˆ’1 |

0

โˆ’2| = 0

Since the complex number โˆ’2 lies in the second quadrant, principal argument ๐œƒ = ๐œ‹ โˆ’ ๐›ผ โŸน ๐œƒ = ๐œ‹ โˆ’ 0 = ๐œ‹

๐‘Ž๐‘Ÿ๐‘”(1 + ๐‘–โˆš3) = ๐‘ก๐‘Ž๐‘›โˆ’1 |๐‘ฆ

๐‘ฅ| = ๐‘ก๐‘Ž๐‘›โˆ’1 |

โˆš3

1| =

๐œ‹

3

Since the complex number 1 + ๐‘–โˆš3 lies in the first quadrant, principal argument

๐œƒ = ๐›ผ โŸน ๐œƒ =๐œ‹

3

(1) โŸน ๐‘Ž๐‘Ÿ๐‘” (โˆ’2

1+๐‘–โˆš3) = ๐œ‹ โˆ’

๐œ‹

3=

2๐œ‹

3

EXERCISE 2.7

1.Write in polar form of the following complex numbers:

(๐‘–)2 + ๐‘–2โˆš3 (๐‘–๐‘–)3 โˆ’ ๐‘–โˆš3 (๐‘–๐‘–๐‘–) โˆ’ 2 โˆ’ ๐‘–2 (๐‘–๐‘ฃ)๐‘–โˆ’1

cos๐œ‹

3+๐‘– sin

๐œ‹

3

(๐‘–)2 + ๐‘–2โˆš3 = ๐‘Ÿ๐‘๐‘–๐‘ (๐œƒ) โ†’ (1)

๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2 = โˆš(2)2 + (2โˆš3)2= โˆš4 + 12 = โˆš16 = 4

๐›ผ = ๐‘ก๐‘Ž๐‘›โˆ’1 |๐‘ฆ

๐‘ฅ| = ๐‘ก๐‘Ž๐‘›โˆ’1 |

2โˆš3

2| =

๐œ‹

3

Principal argument lies in the first quadrant ๐œƒ = ๐›ผ =๐œ‹

3

(1) โŸน 2 + ๐‘–2โˆš3 = 4๐‘๐‘–๐‘  (๐œ‹

3) = 4๐‘๐‘–๐‘  (2๐‘˜๐œ‹ +

๐œ‹

3) , ๐‘˜ โˆˆ ๐‘ง

(๐‘–๐‘–)3 โˆ’ ๐‘–โˆš3 = ๐‘Ÿ๐‘๐‘–๐‘ (๐œƒ) โ†’ (1)

๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2 = โˆš(3)2 + (โˆ’โˆš3)2= โˆš9 + 3 = โˆš4 ร— 3 = 2โˆš3

๐›ผ = ๐‘ก๐‘Ž๐‘›โˆ’1 |๐‘ฆ

๐‘ฅ| = ๐‘ก๐‘Ž๐‘›โˆ’1 |

โˆ’โˆš3

3| = ๐‘ก๐‘Ž๐‘›โˆ’1 |

1

โˆš3| =

๐œ‹

6

Principal argument lies in the 4th quadrant ๐œƒ = โˆ’๐›ผ = โˆ’๐œ‹

6

(1) โŸน 3 โˆ’ ๐‘–โˆš3 = 2โˆš3๐‘๐‘–๐‘  (โˆ’๐œ‹

6) = 2โˆš3๐‘๐‘–๐‘  (2๐‘˜๐œ‹ โˆ’

๐œ‹

6) , ๐‘˜ โˆˆ ๐‘ง

(๐‘–๐‘–๐‘–) โˆ’ 2 โˆ’ ๐‘–2 = ๐‘Ÿ๐‘๐‘–๐‘ (๐œƒ) โ†’ (1)

๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2 = โˆš(โˆ’2)2 + (โˆ’2)2 = โˆš4 + 4 = โˆš2 ร— 4 = 2โˆš2

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Padasalai TrbTnpsc

Page 12: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 12

๐›ผ = ๐‘ก๐‘Ž๐‘›โˆ’1 |๐‘ฆ

๐‘ฅ| = ๐‘ก๐‘Ž๐‘›โˆ’1 |โˆ’

โˆ’2

2| =

๐œ‹

4

Principal argument lies in the 3rd quadrant ๐œƒ = โˆ’๐œ‹ + ๐›ผ = โˆ’๐œ‹ +๐œ‹

4= โˆ’

3๐œ‹

4

(1) โŸน โˆ’2 โˆ’ ๐‘–2 = 2โˆš2๐‘๐‘–๐‘  (โˆ’3๐œ‹

4) = 2โˆš2๐‘๐‘–๐‘  (2๐‘˜๐œ‹ โˆ’

3๐œ‹

4) , ๐‘˜ โˆˆ ๐‘ง

(๐‘–๐‘ฃ)๐‘–โˆ’1

cos๐œ‹

3+๐‘– sin

๐œ‹

3

๐‘– โˆ’ 1 = โˆ’1 + ๐‘– = ๐‘Ÿ๐‘๐‘–๐‘ (๐œƒ) โ†’ (1)

๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2 = โˆš(โˆ’1)2 + (1)2 = โˆš2

๐›ผ = ๐‘ก๐‘Ž๐‘›โˆ’1 |๐‘ฆ

๐‘ฅ| = ๐‘ก๐‘Ž๐‘›โˆ’1 |

1

โˆ’1| =

๐œ‹

4

Principal argument lies in the 2nd quadrant ๐œƒ = ๐œ‹ โˆ’ ๐›ผ โŸน ๐œƒ = ๐œ‹ โˆ’๐œ‹

4=

3๐œ‹

4

(1) โŸน ๐‘– โˆ’ 1 = โˆš2๐‘๐‘–๐‘  (3๐œ‹

4)

๐‘–โˆ’1

cos๐œ‹

3+๐‘– sin

๐œ‹

3

=โˆš2๐‘๐‘–๐‘ (

3๐œ‹

4)

๐‘๐‘–๐‘ (๐œ‹

3)= โˆš2๐‘๐‘–๐‘  (

3๐œ‹

4โˆ’๐œ‹

3) = โˆš2๐‘๐‘–๐‘  (

5๐œ‹

12) = โˆš2๐‘๐‘–๐‘  (2๐‘˜๐œ‹ +

5๐œ‹

12) , ๐‘˜ โˆˆ ๐‘ง

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ‘ Represent the complex number (๐‘–) โˆ’ 1 โˆ’ ๐‘– (๐‘–๐‘–)1 + ๐‘–โˆš3 in polar form. 2.Find the rectangular form of the complex numbers:

(๐‘–) (cos๐œ‹

6+ ๐‘– sin

๐œ‹

6) (cos

๐œ‹

12+ ๐‘– sin

๐œ‹

12) (๐‘–๐‘–)

cos๐œ‹

6โˆ’๐‘– sin

๐œ‹

6

2(cos๐œ‹

3+๐‘– sin

๐œ‹

3)

(๐‘–) (cos๐œ‹

6+ ๐‘– sin

๐œ‹

6) (cos

๐œ‹

12+ ๐‘– sin

๐œ‹

12)

= cis (๐œ‹

6+

๐œ‹

12)

= cis (3๐œ‹

12) = ๐‘๐‘–๐‘  (

๐œ‹

4)

= cos๐œ‹

4+ ๐‘– sin

๐œ‹

4

=1

โˆš2+ ๐‘–

1

โˆš2

(๐‘–๐‘–) cos

๐œ‹

6โˆ’๐‘– sin

๐œ‹

6

2(cos๐œ‹

3+๐‘– sin

๐œ‹

3)

=1

2๐‘๐‘–๐‘  (โˆ’

๐œ‹

6โˆ’๐œ‹

3) =

1

2๐‘๐‘–๐‘  (โˆ’

3๐œ‹

6) =

1

2๐‘๐‘–๐‘  (โˆ’

๐œ‹

2) =

1

2[cos (โˆ’

๐œ‹

2) + ๐‘– sin (โˆ’

๐œ‹

2)]

=1

2(cos

๐œ‹

2โˆ’ ๐‘– sin

๐œ‹

2)

=1

2(0 โˆ’ ๐‘–)

= โˆ’1

2๐‘–

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ“ Find the product 3

2(cos

๐œ‹

3+ ๐‘– sin

๐œ‹

3) . 6 (cos

5๐œ‹

6+ ๐‘– sin

5๐œ‹

6) in

rectangular form.

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ” Find the quotient 2(cos

9๐œ‹

4+๐‘– sin

9๐œ‹

4)

4(cos(โˆ’3๐œ‹

2)+๐‘– sin(

โˆ’3๐œ‹

2))

in rectangular form.

3.If (๐‘ฅ1 + ๐‘–๐‘ฆ1)(๐‘ฅ2 + ๐‘–๐‘ฆ2)โ€ฆ . . (๐‘ฅ๐‘› + ๐‘–๐‘ฆ๐‘›) = ๐‘Ž + ๐‘–๐‘ , show that (i) (๐‘ฅ12 + ๐‘ฆ1

2)(๐‘ฅ22 +

๐‘ฆ22)โ€ฆ . . (๐‘ฅ๐‘›

2 + ๐‘ฆ๐‘›2) = ๐‘Ž2 + ๐‘2 (ii) โˆ‘ ๐‘ก๐‘Ž๐‘›โˆ’1 (๐‘๐‘Ÿ

๐‘Ž๐‘Ÿ)๐‘›

๐‘Ÿ=1 = ๐‘ก๐‘Ž๐‘›โˆ’1 (๐‘

๐‘Ž) + 2๐‘˜๐œ‹, ๐‘˜๐œ–๐‘6

(๐‘–)(๐‘ฅ1 + ๐‘–๐‘ฆ1)(๐‘ฅ2 + ๐‘–๐‘ฆ2)โ€ฆ . . (๐‘ฅ๐‘› + ๐‘–๐‘ฆ๐‘›) = ๐‘Ž + ๐‘–๐‘

Taking modulus on both sides,

โŸน |(๐‘ฅ1 + ๐‘–๐‘ฆ1)(๐‘ฅ2 + ๐‘–๐‘ฆ2)โ€ฆ . . (๐‘ฅ๐‘› + ๐‘–๐‘ฆ๐‘›)| = |๐‘Ž + ๐‘–๐‘| โŸน |๐‘ฅ1 + ๐‘–๐‘ฆ1||๐‘ฅ2 + ๐‘–๐‘ฆ2| โ€ฆโ€ฆ . . |๐‘ฅ๐‘› + ๐‘–๐‘ฆ๐‘›| = |๐‘Ž + ๐‘–๐‘|

โŸนโˆš(๐‘ฅ12 + ๐‘ฆ

12)โˆš(๐‘ฅ2

2 + ๐‘ฆ22)โ€ฆ . .โˆš(๐‘ฅ๐‘›

2 + ๐‘ฆ๐‘›2) = โˆš๐‘Ž2 + ๐‘2

Squaring on both sides, โŸน (๐‘ฅ1

2 + ๐‘ฆ12)(๐‘ฅ2

2 + ๐‘ฆ22)โ€ฆ . . (๐‘ฅ๐‘›

2 + ๐‘ฆ๐‘›2) = ๐‘Ž2 + ๐‘2

(๐‘–)(๐‘ฅ1 + ๐‘–๐‘ฆ1)(๐‘ฅ2 + ๐‘–๐‘ฆ2)โ€ฆ . . (๐‘ฅ๐‘› + ๐‘–๐‘ฆ๐‘›) = ๐‘Ž + ๐‘–๐‘ Taking argument on both sides,

โŸน ๐‘Ž๐‘Ÿ๐‘”[(๐‘ฅ1 + ๐‘–๐‘ฆ1)(๐‘ฅ2 + ๐‘–๐‘ฆ2)โ€ฆ . . (๐‘ฅ๐‘› + ๐‘–๐‘ฆ๐‘›)] = ๐‘Ž๐‘Ÿ๐‘”(๐‘Ž + ๐‘–๐‘)

โŸน ๐‘ก๐‘Ž๐‘›โˆ’1 (๐‘ฆ1๐‘ฅ1) + ๐‘ก๐‘Ž๐‘›โˆ’1 (๐‘ฆ2

๐‘ฅ2) + โ‹ฏ . +๐‘ก๐‘Ž๐‘›โˆ’1 (๐‘ฆ๐‘›

๐‘ฅ๐‘›) = ๐‘ก๐‘Ž๐‘›โˆ’1 (

๐‘

๐‘Ž)

โŸนโˆ‘ ๐‘ก๐‘Ž๐‘›โˆ’1 (๐‘๐‘Ÿ๐‘Ž๐‘Ÿ)๐‘›

๐‘Ÿ=1 = ๐‘ก๐‘Ž๐‘›โˆ’1 (๐‘

๐‘Ž) + 2๐‘˜๐œ‹, ๐‘˜๐œ–๐‘6

4.If 1+๐‘ง

1โˆ’๐‘ง= cos 2๐œƒ + ๐‘– sin 2๐œƒ , show that ๐‘ง = ๐‘– tan ๐œƒ.

1+๐‘ง

1โˆ’๐‘ง= cos 2๐œƒ + ๐‘– sin 2๐œƒ

Add 1 on both sides, 1+๐‘ง

1โˆ’๐‘ง+ 1 = (1 + cos 2๐œƒ) + ๐‘– sin 2๐œƒ

โŸน1+๐‘ง+1โˆ’๐‘ง

1โˆ’๐‘ง= 2๐‘๐‘œ๐‘ 2๐œƒ + ๐‘–2 sin ๐œƒ cos ๐œƒ

โŸน2

1โˆ’๐‘ง= 2 cos ๐œƒ [cos๐œƒ + ๐‘– sin ๐œƒ]

รท ๐‘๐‘ฆ 2 โŸน1

1โˆ’๐‘ง= cos ๐œƒ [cos๐œƒ + ๐‘– sin ๐œƒ]

โŸน1

cos ๐œƒ[cos๐œƒ+๐‘– sin๐œƒ]= 1 โˆ’ ๐‘ง

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Padasalai TrbTnpsc

Page 13: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 13

โŸน ๐‘ง = 1 โˆ’1

cos ๐œƒ[cos๐œƒ+๐‘– sin๐œƒ]

โŸน ๐‘ง = 1 โˆ’1

cos ๐œƒ[cos๐œƒ+๐‘– sin๐œƒ]ร—cos๐œƒโˆ’๐‘– sin๐œƒ

cos๐œƒโˆ’๐‘– sin๐œƒ

โŸน ๐‘ง = 1 โˆ’cos๐œƒโˆ’๐‘– sin๐œƒ

cos ๐œƒ(๐‘๐‘œ๐‘ 2๐œƒ+๐‘ ๐‘–๐‘›2๐œƒ)

โŸน ๐‘ง = 1 โˆ’ [cos๐œƒ

cos๐œƒโˆ’ ๐‘–

sin๐œƒ

cos๐œƒ]

โŸน ๐‘ง = 1 โˆ’ 1 + ๐‘– tan๐œƒ โŸน ๐‘ง = ๐‘– tan๐œƒ

5.If ๐‘๐‘œ๐‘ ๐›ผ + ๐‘๐‘œ๐‘ ๐›ฝ + ๐‘๐‘œ๐‘ ๐›พ = ๐‘ ๐‘–๐‘›๐›ผ + ๐‘ ๐‘–๐‘›๐›ฝ + ๐‘ ๐‘–๐‘›๐›พ = 0 , then show that (i) ๐‘๐‘œ๐‘ 3๐›ผ + ๐‘๐‘œ๐‘ 3๐›ฝ + ๐‘๐‘œ๐‘ 3๐›พ = 3 cos(๐›ผ + ๐›ฝ + ๐›พ) (ii)๐‘ ๐‘–๐‘›3๐›ผ + ๐‘ ๐‘–๐‘›3๐›ฝ + ๐‘ ๐‘–๐‘›3๐›พ = 3sin (๐›ผ + ๐›ฝ + ๐›พ) Let ๐‘Ž = ๐‘๐‘–๐‘ ๐›ผ, ๐‘ = ๐‘๐‘–๐‘ ๐›ฝ, ๐‘ = ๐‘๐‘–๐‘ ๐›พ W.K.T, If ๐‘Ž + ๐‘ + ๐‘ = 0 ๐‘กโ„Ž๐‘’๐‘› ๐‘Ž3 + ๐‘3 + ๐‘3 = 3๐‘Ž๐‘๐‘ ๐‘Ž + ๐‘ + ๐‘ = (๐‘๐‘œ๐‘ ๐›ผ + ๐‘๐‘œ๐‘ ๐›ฝ + ๐‘๐‘œ๐‘ ๐›พ) + ๐‘–(๐‘ ๐‘–๐‘›๐›ผ + ๐‘ ๐‘–๐‘›๐›ฝ + ๐‘ ๐‘–๐‘›๐›พ) = 0 ๐‘Ž3 + ๐‘3 + ๐‘3 = 3๐‘Ž๐‘๐‘ โŸน (๐‘๐‘–๐‘ ๐›ผ)3 + (๐‘๐‘–๐‘ ๐›ฝ)3 + (๐‘๐‘–๐‘ ๐›พ)3 = 3๐‘๐‘–๐‘ ๐›ผ. ๐‘๐‘–๐‘ ๐›ฝ. ๐‘๐‘–๐‘ ๐›พ โŸน ๐‘๐‘–๐‘ (3๐›ผ) + ๐‘๐‘–๐‘ (3๐›ฝ) + ๐‘๐‘–๐‘ (3๐›พ) = 3๐‘๐‘–๐‘ (๐›ผ + ๐›ฝ + ๐›พ) โŸน (๐‘๐‘œ๐‘ 3๐›ผ + ๐‘๐‘œ๐‘ 3๐›ฝ + ๐‘๐‘œ๐‘ 3๐›พ) + ๐‘–(๐‘ ๐‘–๐‘›3๐›ผ + ๐‘ ๐‘–๐‘›3๐›ฝ + ๐‘ ๐‘–๐‘›3๐›พ) = 3(cos(๐›ผ + ๐›ฝ +๐›พ)) + ๐‘–3(sin (๐›ผ + ๐›ฝ + ๐›พ)) Equating the real and imaginary parts on both sides, ๐‘๐‘œ๐‘ 3๐›ผ + ๐‘๐‘œ๐‘ 3๐›ฝ + ๐‘๐‘œ๐‘ 3๐›พ = 3 cos(๐›ผ + ๐›ฝ + ๐›พ) ๐‘ ๐‘–๐‘›3๐›ผ + ๐‘ ๐‘–๐‘›3๐›ฝ + ๐‘ ๐‘–๐‘›3๐›พ = 3sin (๐›ผ + ๐›ฝ + ๐›พ)

6.If ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ and ๐‘Ž๐‘Ÿ๐‘” (๐‘งโˆ’๐‘–

๐‘ง+2) =

๐œ‹

4 , then show that ๐‘ฅ2 + ๐‘ฆ2 + 3๐‘ฅ โˆ’ 3๐‘ฆ + 2 = 0.

Given: ๐‘ = ๐‘ฅ + ๐‘–๐‘ฆ

๐‘Ž๐‘Ÿ๐‘” (๐‘งโˆ’๐‘–

๐‘ง+2) =

๐œ‹

4

โ‡’ ๐‘Ž๐‘Ÿ๐‘”(๐‘ง โˆ’ ๐‘–) โˆ’ ๐‘Ž๐‘Ÿ๐‘”(๐‘ง + 2) =๐œ‹

4

โ‡’ ๐‘Ž๐‘Ÿ๐‘”(๐‘ฅ + ๐‘–๐‘ฆ โˆ’ ๐‘–) โˆ’ ๐‘Ž๐‘Ÿ๐‘”(๐‘ฅ + ๐‘–๐‘ฆ + 2) =๐œ‹

4

โ‡’ ๐‘Ž๐‘Ÿ๐‘”(๐‘ฅ + ๐‘–(๐‘ฆ โˆ’ 1)) โˆ’ ๐‘Ž๐‘Ÿ๐‘”(๐‘ฅ + 2 + ๐‘–๐‘ฆ) =๐œ‹

4

โ‡’ ๐‘ก๐‘Ž๐‘›โˆ’1๐‘ฆ โˆ’ 1

๐‘ฅโˆ’ ๐‘ก๐‘Ž๐‘›โˆ’1

๐‘ฆ

๐‘ฅ + 2=๐œ‹

4

โ‡’ ๐‘ก๐‘Ž๐‘›โˆ’1๐‘ฆ โˆ’ 1๐‘ฅ

โˆ’๐‘ฆ

๐‘ฅ + 2

1 + (๐‘ฆ โˆ’ 1๐‘ฅ

ร—๐‘ฆ

๐‘ฅ + 2)=๐œ‹

4

โ‡’

((๐‘ฆ โˆ’ 1))(๐‘ฅ + 2) โˆ’ ๐‘ฅ๐‘ฆ

๐‘ฅ(๐‘ฅ + 2)

1 +๐‘ฆ2 โˆ’ ๐‘ฆ๐‘ฅ(๐‘ฅ + 2)

= tan๐œ‹

4

โŸน

๐‘ฅ๐‘ฆ + 2๐‘ฆ โˆ’ ๐‘ฅ โˆ’ 2 โˆ’ ๐‘ฅ๐‘ฆ(๐‘ฅ + 2)

๐‘ฅ2 + 2๐‘ฅ + ๐‘ฆ2 โˆ’ ๐‘ฆ(๐‘ฅ + 2)

= 1

โŸนโˆ’๐‘ฅ + 2๐‘ฆ โˆ’ 2

๐‘ฅ2 + ๐‘ฆ2 + 2๐‘ฅ โˆ’ ๐‘ฆ= 1

โŸน ๐‘ฅ2 + ๐‘ฆ2 + 2๐‘ฅ โˆ’ ๐‘ฆ + ๐‘ฅ โˆ’ 2๐‘ฆ + 2 = 0

โŸน ๐‘ฅ2 + ๐‘ฆ2 + 3๐‘ฅ โˆ’ 3๐‘ฆ + 2 = 0

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ• If ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ and ๐‘Ž๐‘Ÿ๐‘” (๐‘งโˆ’1

๐‘ง+1) =

๐œ‹

2 , then show that ๐‘ฅ2 + ๐‘ฆ2 = 1.

De MOIVREโ€™S THEOREM AND ITS APPLICATIONS De Moivreโ€™s theorem: Given any complex number cos ๐œƒ + ๐‘– sin ๐œƒ and any

integer n, (cos ๐œƒ + ๐‘– sin ๐œƒ)๐‘› = cos๐‘›๐œƒ + ๐‘– sin ๐‘›๐œƒ Some results:

(cos๐œƒ โˆ’ ๐‘– sin ๐œƒ)๐‘› = cos๐‘›๐œƒ โˆ’ ๐‘– sin ๐‘›๐œƒ (cos๐œƒ + ๐‘– sin ๐œƒ)โˆ’๐‘› = cos๐‘›๐œƒ โˆ’ ๐‘– sin ๐‘›๐œƒ (cos๐œƒ โˆ’ ๐‘– sin ๐œƒ)โˆ’๐‘› = cos๐‘›๐œƒ + ๐‘– sin ๐‘›๐œƒ sin ๐œƒ + ๐‘– cos ๐œƒ = ๐‘–(cos๐œƒ โˆ’ ๐‘– sin ๐œƒ)

Formula for finding ๐‘›๐‘กโ„Ž root of a complex number:

๐‘ง1๐‘›โ„ = ๐‘Ÿ

1๐‘›โ„ ๐‘๐‘–๐‘  (

๐œƒ+2๐‘˜๐œ‹

๐‘›) , ๐‘˜ = 0,1,โ€ฆ . , (๐‘› โˆ’ 1)

๐œ”3 = 1 โŸน 1 +๐œ” +๐œ”2 = 0 The sum of all the ๐‘›๐‘กโ„Ž roots of unity is 0.(๐‘–๐‘’. , 1 + ๐œ” + ๐œ”2 +โ‹ฏ .+๐œ”๐‘›โˆ’1 = 0)

The product of all the ๐‘›๐‘กโ„Ž roots of unity is (โˆ’1)๐‘›โˆ’1. (๐‘–๐‘’. , 1 + ๐œ” + ๐œ”2 +โ‹ฏ .+๐œ”๐‘›โˆ’1 = (โˆ’1)๐‘›โˆ’1)

All the n roots of ๐‘›๐‘กโ„Ž roots of unity are in geometrical progression. All the n roots of ๐‘›๐‘กโ„Ž roots of unity lie on the circumference of a circle whose

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Padasalai TrbTnpsc

Page 14: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 14

centre is at the origin and radius equal to 1 and these roots divide the circle into n equal parts and form a polygon of n sides.

EXERCISE 2.8

1.If ๐œ” โ‰  1 is a cube root of unity, then show that ๐‘Ž+๐‘๐œ”+๐‘๐œ”2

๐‘+๐‘๐œ”+๐‘Ž๐œ”2+๐‘Ž+๐‘๐œ”+๐‘๐œ”2

๐‘+๐‘Ž๐œ”+๐‘๐œ”2= โˆ’1.

๐‘Ž+๐‘๐œ”+๐‘๐œ”2

๐‘+๐‘๐œ”+๐‘Ž๐œ”2+๐‘Ž+๐‘๐œ”+๐‘๐œ”2

๐‘+๐‘Ž๐œ”+๐‘๐œ”2

= (๐‘Ž+๐‘๐œ”+๐‘๐œ”2

๐‘+๐‘๐œ”+๐‘Ž๐œ”2ร—๐œ”

๐œ”) + (

๐‘Ž+๐‘๐œ”+๐‘๐œ”2

๐‘+๐‘Ž๐œ”+๐‘๐œ”2ร—๐œ”2

๐œ”2)

=๐œ”(๐‘Ž+๐‘๐œ”+๐‘๐œ”2)

๐‘Ž๐œ”3+๐‘๐œ”2+๐‘๐œ”+๐œ”2(๐‘Ž+๐‘๐œ”+๐‘๐œ”2)

๐‘Ž๐œ”3+๐‘๐œ”4+๐‘๐œ”2 ๐œ”3 = 1 ; 1 + ๐œ” + ๐œ”2 = 0

=๐œ”(๐‘Ž+๐‘๐œ”+๐‘๐œ”2)

๐‘Ž+๐‘๐œ”+๐‘๐œ”2+๐œ”2(๐‘Ž+๐‘๐œ”+๐‘๐œ”2)

๐‘Ž+๐‘๐œ”+๐‘๐œ”2

= ๐œ” +๐œ”2 = โˆ’1

2.Show that (โˆš3

2+

๐‘–

2)5

+ (โˆš3

2โˆ’

๐‘–

2)5

= โˆ’โˆš3

โˆš3

2+

๐‘–

2= ๐‘Ÿ๐‘๐‘–๐‘ (๐œƒ) โ†’ (1)

๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2 = โˆš(โˆš3

2)2

+ (1

2)2= โˆš

3

4+1

4= โˆš1 = 1

๐›ผ = ๐‘ก๐‘Ž๐‘›โˆ’1 |๐‘ฆ

๐‘ฅ| = ๐‘ก๐‘Ž๐‘›โˆ’1 |

12โ„

โˆš32โ„| =

๐œ‹

6

Principal argument lies in the first quadrant ๐œƒ = ๐›ผ =๐œ‹

6

(1) โŸนโˆš3

2+

๐‘–

2= ๐‘๐‘–๐‘  (

๐œ‹

6)

โŸน (โˆš3

2+

๐‘–

2)5

= [๐‘๐‘–๐‘  (๐œ‹

6)]5= ๐‘๐‘–๐‘  (

5๐œ‹

6) = cos (

5๐œ‹

6) + ๐‘– sin (

5๐œ‹

6)

Similarly, (โˆš3

2โˆ’

๐‘–

2)5

= cos (5๐œ‹

6) โˆ’ ๐‘– sin (

5๐œ‹

6)

(โˆš3

2+

๐‘–

2)5

+ (โˆš3

2โˆ’

๐‘–

2)5

= cos (5๐œ‹

6) + ๐‘– sin (

5๐œ‹

6) + cos (

5๐œ‹

6) โˆ’ ๐‘– sin (

5๐œ‹

6)

= 2 cos (5๐œ‹

6) = 2 cos (

5ร—180

6) = 2 cos 150 = โˆ’2 cos 30 = โˆ’2 ร—

โˆš3

2= โˆ’โˆš3

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ‘๐Ÿ Simplify (๐‘–)(1 + ๐‘–)18 (๐‘–๐‘–)(โˆ’โˆš3 + 3๐‘–)31

3.Find the value of (1+sin

๐œ‹

10+๐‘– cos

๐œ‹

10

1+sin๐œ‹

10โˆ’๐‘– cos

๐œ‹

10

)10

Let ๐‘ง = sin๐œ‹

10+ ๐‘– cos

๐œ‹

10 โŸน

1

๐‘ง= sin

๐œ‹

10โˆ’ ๐‘– cos

๐œ‹

10

(1+sin

๐œ‹

10+๐‘– cos

๐œ‹

10

1+sin๐œ‹

10โˆ’๐‘– cos

๐œ‹

10

)10

= (1+๐‘ง

1+1

๐‘ง

)

10

= (1+๐‘ง

1+๐‘งร— ๐‘ง)

10= ๐‘ง10

๐‘ง10 = (sin๐œ‹

10+ ๐‘– cos

๐œ‹

10)10= [๐‘– (cos

๐œ‹

10โˆ’ ๐‘– sin

๐œ‹

10)]10

= ๐‘–10 [cos (10 ร—๐œ‹

10) โˆ’ ๐‘– sin (10 ร—

๐œ‹

10)]

= โˆ’1[cos๐œ‹ โˆ’ ๐‘– sin ๐œ‹] = โˆ’1(โˆ’1 โˆ’ 0) = 1

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ— Simplify (sin๐œ‹

6+ ๐‘– cos

๐œ‹

6)18

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ‘๐ŸŽ Simplify (1+cos2๐œƒ+๐‘– sin 2๐œƒ

1+cos2๐œƒโˆ’๐‘– sin 2๐œƒ)30

4.If 2 cos๐›ผ = ๐‘ฅ +1

๐‘ฅ and 2 cos๐›ฝ = ๐‘ฆ +

1

๐‘ฆ , show that

(๐‘–)๐‘ฅ

๐‘ฆ+๐‘ฆ

๐‘ฅ= 2 cos(๐›ผ โˆ’ ๐›ฝ) (๐‘–๐‘–)๐‘ฅ๐‘ฆ โˆ’

1

๐‘ฅ๐‘ฆ= 2๐‘– sin(๐›ผ + ๐›ฝ)

(๐‘–๐‘–๐‘–)๐‘ฅ๐‘š

๐‘ฆ๐‘›โˆ’

๐‘ฆ

๐‘ฅ๐‘š

๐‘›= 2๐‘– sin(๐‘š๐›ผ โˆ’ ๐‘›๐›ฝ) (๐‘–๐‘ฃ)๐‘ฅ๐‘š๐‘ฆ๐‘› +

1

๐‘ฅ๐‘š๐‘ฆ๐‘›= 2 cos(๐‘š๐›ผ + ๐‘›๐›ฝ)

๐‘ฅ +1

๐‘ฅ= 2 cos๐›ผ โŸน ๐‘ฅ = ๐‘ฅ = cos๐›ผ + ๐‘– sin ๐›ผ

๐‘ฆ +1

๐‘ฆ= 2 cos๐›ฝ โŸน ๐‘ฆ = cos๐›ฝ + ๐‘– sin ๐›ฝ

(๐‘–) ๐‘ฅ

๐‘ฆ=

cos๐›ผ+๐‘– sin๐›ผ

cos๐›ฝ+๐‘– sin๐›ฝ= cos(๐›ผ โˆ’ ๐›ฝ) + ๐‘– sin(๐›ผ โˆ’ ๐›ฝ)

๐‘ฆ

๐‘ฅ= (

๐‘ฅ

๐‘ฆ)โˆ’1= cos(๐›ผ โˆ’ ๐›ฝ) โˆ’ ๐‘– sin(๐›ผ โˆ’ ๐›ฝ)

๐‘ฅ

๐‘ฆ+๐‘ฆ

๐‘ฅ= 2 cos(๐›ผ โˆ’ ๐›ฝ)

(๐‘–๐‘–) ๐‘ฅ๐‘ฆ = (cos๐›ผ + ๐‘– sin ๐›ผ)(cos๐›ฝ + ๐‘– sin ๐›ฝ) = cos(๐›ผ + ๐›ฝ) + ๐‘– sin(๐›ผ + ๐›ฝ) 1

๐‘ฅ๐‘ฆ= (๐‘ฅ๐‘ฆ)โˆ’1 = cos(๐›ผ + ๐›ฝ) โˆ’ ๐‘– sin(๐›ผ + ๐›ฝ)

๐‘ฅ๐‘ฆ โˆ’1

๐‘ฅ๐‘ฆ= 2๐‘– sin(๐›ผ + ๐›ฝ)

(๐‘–๐‘–๐‘–) ๐‘ฅ๐‘š = (cos๐›ผ + ๐‘– sin ๐›ผ)๐‘š = cos๐‘š๐›ผ + ๐‘– sin๐‘š๐›ผ

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Padasalai TrbTnpsc

Page 15: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 15

๐‘ฆ๐‘› = (cos๐›ฝ + ๐‘– sin ๐›ฝ)๐‘› = cos๐‘›๐›ฝ + ๐‘– sin ๐‘›๐›ฝ ๐‘ฅ๐‘š

๐‘ฆ๐‘›=

cos๐‘š๐›ผ+๐‘– sin๐‘š๐›ผ

cos๐‘›๐›ฝ+๐‘– sin๐‘›๐›ฝ= cos(๐‘š๐›ผ โˆ’ ๐‘›๐›ฝ) + ๐‘– sin(๐‘š๐›ผ โˆ’ ๐‘›๐›ฝ)

๐‘ฆ

๐‘ฅ๐‘š

๐‘›= (

๐‘ฅ๐‘š

๐‘ฆ๐‘›)โˆ’1

= cos(๐‘š๐›ผ โˆ’ ๐‘›๐›ฝ) โˆ’ ๐‘– sin(๐‘š๐›ผ โˆ’ ๐‘›๐›ฝ)

๐‘ฅ๐‘š

๐‘ฆ๐‘›โˆ’

๐‘ฆ

๐‘ฅ๐‘š

๐‘›= 2๐‘– sin(๐‘š๐›ผ โˆ’ ๐‘›๐›ฝ)

(๐‘–๐‘ฃ)๐‘ฅ๐‘š = (cos ๐›ผ + ๐‘– sin ๐›ผ)๐‘š = cos๐‘š๐›ผ + ๐‘– sin๐‘š๐›ผ ๐‘ฆ๐‘› = (cos๐›ฝ + ๐‘– sin ๐›ฝ)๐‘› = cos๐‘›๐›ฝ + ๐‘– sin ๐‘›๐›ฝ ๐‘ฅ๐‘š๐‘ฆ๐‘› = (cos๐‘š๐›ผ + ๐‘– sin๐‘š๐›ผ)(cos๐‘›๐›ฝ + ๐‘– sin ๐‘›๐›ฝ)

= cos(๐‘š๐›ผ + ๐‘›๐›ฝ) + ๐‘– sin(๐‘š๐›ผ + ๐‘›๐›ฝ) 1

๐‘ฅ๐‘š๐‘ฆ๐‘›= cos(๐‘š๐›ผ + ๐‘›๐›ฝ) โˆ’ ๐‘– sin(๐‘š๐›ผ + ๐‘›๐›ฝ)

๐‘ฅ๐‘š๐‘ฆ๐‘› +1

๐‘ฅ๐‘š๐‘ฆ๐‘›= 2 cos(๐‘š๐›ผ + ๐‘›๐›ฝ)

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ๐Ÿ– If ๐‘ง = cos ๐œƒ + ๐‘– sin ๐œƒ, show that ๐‘ง๐‘› +1

๐‘ง๐‘›= 2 cos ๐‘›๐œƒ and

๐‘ง๐‘› โˆ’1

๐‘ง๐‘›= 2 sin ๐‘›๐œƒ.

5.Solve the equation ๐‘ง3 + 27 = 0

๐‘ง3 + 27 = 0 โŸน ๐‘ง3 = โˆ’27โŸน ๐‘ง = (โˆ’27)13โ„ = (27 ร— โˆ’1)

13โ„

๐‘ง = 3(โˆ’1)13โ„ = 3[๐‘๐‘–๐‘ (๐œ‹)]

13โ„ = 3[๐‘๐‘–๐‘ (2๐‘˜๐œ‹ + ๐œ‹)]

13โ„ , ๐‘˜ = 0,1,2

= 3๐‘๐‘–๐‘ (2๐‘˜ + 1)๐œ‹

3, ๐‘˜ = 0,1,2

= 3๐‘๐‘–๐‘  (๐œ‹

3) , 3๐‘๐‘–๐‘ (๐œ‹), 3๐‘๐‘–๐‘  (

5๐œ‹

3)

6. If ๐œ” โ‰  1 is a cube root of unity, show that the roots of the equation (๐‘ง โˆ’ 1)3 +8 = 0 are โˆ’1, 1 โˆ’ 2๐œ”, 1 โˆ’ 2๐œ”2. (๐‘ง โˆ’ 1)3 + 8 = 0 โŸน (๐‘ง โˆ’ 1)3 = โˆ’8

โŸน (๐‘ง โˆ’ 1)3 = (โˆ’2)3 โŸน(๐‘งโˆ’1)3

(โˆ’2)3= 1

โŸน (๐‘งโˆ’1

โˆ’2)3= 1

โŸน๐‘งโˆ’1

โˆ’2= (1)

13โ„

โŸน ๐‘งโˆ’ 1 = โˆ’2(1)13โ„ โŸน ๐‘ง โˆ’ 1 = โˆ’2[1 (๐‘œ๐‘Ÿ) ๐œ” (๐‘œ๐‘Ÿ) ๐œ”2]

โŸน {๐‘ง โˆ’ 1 = โˆ’2(1) โŸน ๐‘ง โˆ’ 1 = โˆ’2 โŸน ๐‘ง = โˆ’1

๐‘ง โˆ’ 1 = โˆ’2๐œ” โŸน ๐‘ง = 1 โˆ’ 2๐œ”๐‘ง โˆ’ 1 = โˆ’2๐œ”2 โŸน ๐‘ง = 1 โˆ’ 2๐œ”2

Thus the roots of the equation (๐‘ง โˆ’ 1)3 + 8 = 0 are โˆ’1, 1 โˆ’ 2๐œ”, 1 โˆ’ 2๐œ”2.

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ‘๐Ÿ’ Solve the equation ๐‘ง3 + 8๐‘– = 0, where ๐‘ง โˆˆ โ„‚.

Note: ๐‘ง3 + 8๐‘– = 0 โŸน ๐‘ง = (8๐‘–)13โ„ โŸน ๐‘ง = 2(๐‘–)

13โ„

๐‘“๐‘–๐‘›๐‘‘ ๐‘๐‘œ๐‘™๐‘Ž๐‘Ÿ ๐‘“๐‘œ๐‘Ÿ๐‘š ๐‘“๐‘œ๐‘Ÿ ๐‘– ๐‘Ž๐‘›๐‘‘ ๐‘ข๐‘ ๐‘’ ๐‘‘๐‘’ ๐‘€๐‘œ๐‘–๐‘ฃ๐‘Ÿ๐‘’โ€ฒ๐‘กโ„Ž๐‘’๐‘œ๐‘Ÿ๐‘’๐‘š

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ‘๐Ÿ“ Find all cube roots of โˆš3 + ๐‘–.

Note: (โˆš3 + ๐‘–)13โ„ ; ๐‘“๐‘–๐‘›๐‘‘ ๐‘๐‘œ๐‘™๐‘Ž๐‘Ÿ ๐‘“๐‘œ๐‘Ÿ๐‘š ๐‘“๐‘œ๐‘Ÿ โˆš3 + ๐‘– ๐‘Ž๐‘›๐‘‘ ๐‘ข๐‘ ๐‘’ ๐‘‘๐‘’ ๐‘€๐‘œ๐‘–๐‘ฃ๐‘Ÿ๐‘’โ€ฒ๐‘กโ„Ž๐‘’๐‘œ๐‘Ÿ๐‘’๐‘š

7.Find the value of โˆ‘ (cos2๐‘˜๐œ‹

9+ ๐‘– sin

2๐‘˜๐œ‹

9)8

๐‘˜=1

Let ๐‘ฅ = (1)19โ„ โŸน ๐‘ฅ = [๐‘๐‘–๐‘ (0)]

19โ„ = [๐‘๐‘–๐‘ (2๐‘˜๐œ‹ + 0)]

19โ„ , ๐‘˜ = 0,1,2,โ€ฆ ,8

โŸน ๐‘ฅ = ๐‘๐‘–๐‘  (2๐‘˜๐œ‹

9) , ๐‘˜ = 0,1,2,โ€ฆ ,8

Thus the roots are ๐‘๐‘–๐‘ (0), ๐‘๐‘–๐‘  (2๐œ‹

9) , ๐‘๐‘–๐‘  (

4๐œ‹

9) , ๐‘๐‘–๐‘  (

6๐œ‹

9) , ๐‘๐‘–๐‘  (

8๐œ‹

9) , ๐‘๐‘–๐‘  (

10๐œ‹

9),

๐‘๐‘–๐‘  (12๐œ‹

9) , ๐‘๐‘–๐‘  (

14๐œ‹

9) , ๐‘๐‘–๐‘  (

16๐œ‹

9) .

W.K.T, sum of all the roots Is equal to zero.

๐‘๐‘–๐‘ (0) + ๐‘๐‘–๐‘  (2๐œ‹

9) + ๐‘๐‘–๐‘  (

4๐œ‹

9) + ๐‘๐‘–๐‘  (

6๐œ‹

9) + ๐‘๐‘–๐‘  (

8๐œ‹

9) + ๐‘๐‘–๐‘  (

10๐œ‹

9) +

๐‘๐‘–๐‘  (12๐œ‹

9) + ๐‘๐‘–๐‘  (

14๐œ‹

9) + ๐‘๐‘–๐‘  (

16๐œ‹

9) = 0

โŸน 1+ ๐‘๐‘–๐‘  (2๐œ‹

9) + ๐‘๐‘–๐‘  (

4๐œ‹

9) + ๐‘๐‘–๐‘  (

6๐œ‹

9) + ๐‘๐‘–๐‘  (

8๐œ‹

9) + ๐‘๐‘–๐‘  (

10๐œ‹

9) +

๐‘๐‘–๐‘  (12๐œ‹

9) + ๐‘๐‘–๐‘  (

14๐œ‹

9) + ๐‘๐‘–๐‘  (

16๐œ‹

9) = 0

โŸน ๐‘๐‘–๐‘  (2๐œ‹

9) + ๐‘๐‘–๐‘  (

4๐œ‹

9) + ๐‘๐‘–๐‘  (

6๐œ‹

9) + ๐‘๐‘–๐‘  (

8๐œ‹

9) + ๐‘๐‘–๐‘  (

10๐œ‹

9) +

๐‘๐‘–๐‘  (12๐œ‹

9) + ๐‘๐‘–๐‘  (

14๐œ‹

9) + ๐‘๐‘–๐‘  (

16๐œ‹

9) = โˆ’1

โŸนโˆ‘ ๐‘๐‘–๐‘  (2๐‘˜๐œ‹

9)8

๐‘˜=1 = โˆ’1

โŸนโˆ‘ (cos2๐‘˜๐œ‹

9+ ๐‘– sin

2๐‘˜๐œ‹

9)8

๐‘˜=1 = โˆ’1

8. If ๐œ” โ‰  1 is a cube root of unity, show that (๐‘–)(1 โˆ’ ๐œ” + ๐œ”2)6 + (1 + ๐œ” โˆ’ ๐œ”2)6 = 128

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Padasalai TrbTnpsc

Page 16: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 16

(๐‘–๐‘–)(1 + ๐œ”)(1 + ๐œ”2)(1 + ๐œ”4)(1 + ๐œ”8)โ€ฆ . (1 + ๐œ”211) = 1

W.K.T, ๐œ”3 = 1 ; 1 + ๐œ” + ๐œ”2 = 0

(๐‘–)(1 โˆ’ ๐œ” + ๐œ”2)6 + (1 + ๐œ” โˆ’ ๐œ”2)6 = (โˆ’๐œ” โˆ’ ๐œ”)6 + (โˆ’๐œ”2 โˆ’๐œ”2)6 = (โˆ’2๐œ”)6 + (โˆ’2๐œ”2)6 = 26๐œ”6 + 26๐œ”12 = 64(๐œ”3)2 + 64(๐œ”3)4 = 64 + 64 = 128

(๐‘–๐‘–)(1 + ๐œ”)(1 + ๐œ”2)(1 + ๐œ”4)(1 + ๐œ”8)โ€ฆ . (1 + ๐œ”211)

= (1 + ๐œ”)(1 + ๐œ”2)(1 + ๐œ”22)(1 + ๐œ”2

3)โ€ฆ . (1 + ๐œ”2

11)

Clearly the above expression contains 12 terms, = (1 + ๐œ”)(1 + ๐œ”2)(1 + ๐œ”4)(1 + ๐œ”8)โ€ฆ . .12 ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘  = [(1 + ๐œ”)(1 + ๐œ”2)][(1 + ๐œ”)(1 + ๐œ”2)]โ€ฆ .6 ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘  = [(1 + ๐œ”)(1 + ๐œ”2)]6 = (1 + ๐œ” + ๐œ”2 +๐œ”3)6 = (0 + 1)6 = 1 9.If ๐‘ง = 2 โˆ’ 2๐‘–, find the rotation of ๐‘ง by ๐œƒ radians in the counter clockwise

direction about the origin when (๐‘–)๐œƒ =๐œ‹

3 (๐‘–๐‘–)๐œƒ =

2๐œ‹

3 (๐‘–๐‘–๐‘–)๐œƒ =

3๐œ‹

2

๐‘ง = 2 โˆ’ 2๐‘– = ๐‘Ÿ๐‘๐‘–๐‘ (๐œƒ) โ†’ (1)

๐‘Ÿ = โˆš๐‘ฅ2 + ๐‘ฆ2 = โˆš(2)2 + (โˆ’2)2 = โˆš4 + 4 = โˆš4 ร— 2 = 2โˆš2

๐›ผ = ๐‘ก๐‘Ž๐‘›โˆ’1 |๐‘ฆ

๐‘ฅ| = ๐‘ก๐‘Ž๐‘›โˆ’1 |

โˆ’2

2| =

๐œ‹

4

Principal argument lies in the 4th quadrant ๐œƒ = โˆ’๐›ผ = โˆ’๐œ‹

4

(1) โŸน ๐‘ง = 2 โˆ’ 2๐‘– = 2โˆš2๐‘๐‘–๐‘  (โˆ’๐œ‹

4)

(๐‘–)๐‘คโ„Ž๐‘’๐‘› ๐œƒ =๐œ‹

3โŸน ๐‘ง = 2โˆš2๐‘๐‘–๐‘  (โˆ’

๐œ‹

4) ๐‘๐‘–๐‘  (

๐œ‹

3)

= 2โˆš2๐‘๐‘–๐‘  (โˆ’๐œ‹

4+๐œ‹

3)

= 2โˆš2๐‘๐‘–๐‘  (โˆ’3๐œ‹+4๐œ‹

12)

= 2โˆš2๐‘๐‘–๐‘  (๐œ‹

12)

(๐‘–๐‘–)๐‘คโ„Ž๐‘’๐‘› ๐œƒ =2๐œ‹

3โŸน ๐‘ง = 2โˆš2๐‘๐‘–๐‘  (โˆ’

๐œ‹

4) ๐‘๐‘–๐‘  (

2๐œ‹

3)

= 2โˆš2๐‘๐‘–๐‘  (โˆ’๐œ‹

4+2๐œ‹

3)

= 2โˆš2๐‘๐‘–๐‘  (โˆ’3๐œ‹+8๐œ‹

12)

= 2โˆš2๐‘๐‘–๐‘  (5๐œ‹

12)

(๐‘–๐‘–)๐‘คโ„Ž๐‘’๐‘› ๐œƒ =3๐œ‹

2โŸน ๐‘ง = 2โˆš2๐‘๐‘–๐‘  (โˆ’

๐œ‹

4) ๐‘๐‘–๐‘  (

3๐œ‹

2)

= 2โˆš2๐‘๐‘–๐‘  (โˆ’๐œ‹

4+3๐œ‹

2)

= 2โˆš2๐‘๐‘–๐‘  (โˆ’๐œ‹+6๐œ‹

4)

= 2โˆš2๐‘๐‘–๐‘  (5๐œ‹

4)

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ‘๐Ÿ” Suppose ๐‘ง1, ๐‘ง2 and ๐‘ง3 are the vertices of an equilateral triangle

inscribed in the circle |๐‘ง| = 2. If ๐‘ง1 = 1 + ๐‘–โˆš3, then find ๐‘ง2 and ๐‘ง3. |๐‘ง| = 2 represents the circle with centre (0,0) and radius 2. Let ๐ด, ๐ต ๐‘Ž๐‘›๐‘‘ ๐ถ be the vertices of the given triangle. Since the vertices ๐‘ง1, ๐‘ง2 and ๐‘ง3 form an equilateral triangle inscribed in the circle |๐‘ง| = 2, the sides of this triangle

๐ด๐ต, ๐ต๐ถ ๐‘Ž๐‘›๐‘‘ ๐ถ๐ด subtend 2๐œ‹

3 radian at the origin.

We obtain ๐‘ง2 and ๐‘ง3 by the rotation of ๐‘ง1 by 2๐œ‹

3 and

4๐œ‹

3 respectively.

Given: ๐‘ง1 = 1 + ๐‘–โˆš3

๐‘ง2 = ๐‘ง1๐‘๐‘–๐‘  (2๐œ‹

3) = (1 + ๐‘–โˆš3) [cos (

2๐œ‹

3) + ๐‘– sin (

2๐œ‹

3)]

= (1 + ๐‘–โˆš3)[cos(180 โˆ’ 60) + ๐‘– sin(180 โˆ’ 60)]

= (1 + ๐‘–โˆš3)(โˆ’cos 60 + ๐‘– sin 60)

= (1 + ๐‘–โˆš3) (โˆ’1

2+ ๐‘–

โˆš3

2)

= โˆ’2

๐‘ง3 = ๐‘ง2๐‘๐‘–๐‘  (2๐œ‹

3) = โˆ’2 [cos (

2๐œ‹

3) + ๐‘– sin (

2๐œ‹

3)]

= โˆ’2[cos(180 โˆ’ 60) + ๐‘– sin(180 โˆ’ 60)] = โˆ’2(โˆ’cos 60 + ๐‘– sin 60)

= โˆ’2(โˆ’1

2+ ๐‘–

โˆš3

2)

= 1 โˆ’ ๐‘–โˆš3

10.Prove that the values of โˆšโˆ’14

are ยฑ1

โˆš2(1 ยฑ ๐‘–)

Let ๐‘ฅ = โˆšโˆ’14

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Padasalai TrbTnpsc

Page 17: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 17

โŸน ๐‘ฅ = (โˆ’1)14โ„ โŸน ๐‘ฅ4 = โˆ’1 โŸน ๐‘ฅ4 + 1 = 0

โŸน (๐‘ฅ4 + 2๐‘ฅ2 + 1) โˆ’ (2) = 0

โŸน (๐‘ฅ2 + 1)2 โˆ’ (โˆš2๐‘ฅ)2= 0

โŸน (๐‘ฅ2 + 1 + โˆš2๐‘ฅ)(๐‘ฅ2 + 1 โˆ’ โˆš2๐‘ฅ) = 0

โŸน (๐‘ฅ2 + 1 + โˆš2๐‘ฅ) = 0 ; (๐‘ฅ2 + 1 โˆ’ โˆš2๐‘ฅ) = 0

๐‘ฅ =โˆ’๐‘ ยฑ โˆš๐‘2 โˆ’ 4๐‘Ž๐‘

2๐‘Ž

โŸน ๐‘ฅ =โˆ’โˆš2ยฑโˆš(โˆš2

2)โˆ’4(1)(1)

2(1) ;

โˆ’(โˆ’โˆš2)ยฑโˆš(โˆ’โˆš22)โˆ’4(1)(1)

2(1)

โŸน ๐‘ฅ =โˆ’โˆš2ยฑโˆš2โˆ’4

2 ;

โˆš2ยฑโˆš2โˆ’4

2

โŸน ๐‘ฅ =โˆ’โˆš2ยฑโˆšโˆ’2

2 ;

โˆš2ยฑโˆšโˆ’2

2

โŸน ๐‘ฅ =โˆ’โˆš2ยฑโˆš2๐‘–2

2 ;

โˆš2ยฑโˆš2๐‘–2

2

โŸน ๐‘ฅ =โˆ’โˆš2ยฑโˆš2๐‘–

2 ;

โˆš2ยฑโˆšโˆš2๐‘–

2

โŸน ๐‘ฅ =โˆ’โˆš2

2(1 ยฑ ๐‘–) ;

โˆš2

2(1 ยฑ ๐‘–)

โŸน ๐‘ฅ = ยฑ1

โˆš2(1 ยฑ ๐‘–)

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ‘๐Ÿ Find the cube roots of unity.

Let ๐‘ง3 = 1 โŸน ๐‘ง = (1)13โ„ = [๐‘๐‘–๐‘ (0)]

13โ„

โŸน ๐‘ง = [๐‘๐‘–๐‘ (2๐‘˜๐œ‹ + 0)]13โ„ , ๐‘˜ = 0,1,2

โŸน ๐‘ง = ๐‘๐‘–๐‘  (2๐‘˜๐œ‹

3) , ๐‘˜ = 0,1,2

โŸน ๐‘ง = ๐‘๐‘–๐‘ (0), ๐‘๐‘–๐‘  (2๐œ‹

3) , ๐‘๐‘–๐‘  (

4๐œ‹

3)

โŸน ๐‘ง = ๐‘๐‘–๐‘ (0), ๐‘๐‘–๐‘  (๐œ‹ โˆ’๐œ‹

3) , ๐‘๐‘–๐‘  (๐œ‹ +

๐œ‹

3)

โŸน ๐‘ง = cos 0 + ๐‘– sin 0 , cos (๐œ‹ โˆ’๐œ‹

3) + ๐‘– sin (๐œ‹ โˆ’

๐œ‹

3) , cos (๐œ‹ +

๐œ‹

3) + ๐‘– sin (๐œ‹ +

๐œ‹

3)

โŸน ๐‘ง = 1,โˆ’ cos๐œ‹

3+ ๐‘– sin

๐œ‹

3, โˆ’ cos

๐œ‹

3โˆ’ ๐‘– sin

๐œ‹

3

โŸน ๐‘ง = 1,โˆ’1

2+ ๐‘–

โˆš3

2, โˆ’

1

2+ ๐‘–

โˆš3

2โŸน ๐‘ง = 1,

โˆ’1+๐‘–โˆš3

2,โˆ’1โˆ’๐‘–โˆš3

2

โˆด The cube roots of unity 1,๐œ”, ๐œ”2 are1,โˆ’1+๐‘–โˆš3

2,โˆ’1โˆ’๐‘–โˆš3

2

๐„๐—๐€๐Œ๐๐‹๐„ ๐Ÿ. ๐Ÿ‘๐Ÿ‘ Find the fourth roots of unity.

Let ๐‘ง4 = 1 โŸน ๐‘ง = (1)14โ„ = [๐‘๐‘–๐‘ (0)]

14โ„

โŸน ๐‘ง = [๐‘๐‘–๐‘ (2๐‘˜๐œ‹ + 0)]14โ„ , ๐‘˜ = 0,1,2,3

โŸน ๐‘ง = ๐‘๐‘–๐‘  (2๐‘˜๐œ‹

4) , ๐‘˜ = 0,1,2,3

โŸน ๐‘ง = ๐‘๐‘–๐‘ (0), ๐‘๐‘–๐‘  (2๐œ‹

4) , ๐‘๐‘–๐‘  (

4๐œ‹

4) , ๐‘๐‘–๐‘  (

6๐œ‹

4)

โŸน ๐‘ง = ๐‘๐‘–๐‘ (0), ๐‘๐‘–๐‘  (๐œ‹

2) , ๐‘๐‘–๐‘ (๐œ‹), ๐‘๐‘–๐‘  (

3๐œ‹

2)

โŸน ๐‘ง = cos 0 + ๐‘– sin 0 , cos (๐œ‹

2) + ๐‘– sin (

๐œ‹

2) , cos(๐œ‹) + ๐‘– sin(๐œ‹) , cos (

3๐œ‹

2) + ๐‘– sin (

3๐œ‹

2)

โŸน ๐‘ง = 1, ๐‘–, โˆ’1,โˆ’๐‘– โˆด The fourth roots of unity 1,๐œ”,๐œ”2, ๐œ”3 are 1, ๐‘–, โˆ’1,โˆ’๐‘–.

1.Prove that the multiplicative inverse of a non- zero complex number ๐’› = ๐’™ + ๐’Š๐’š

is ๐’›โˆ’๐Ÿ = (๐’™

๐’™๐Ÿ+๐’š๐Ÿ) + ๐’Š (โˆ’

๐’š

๐’™๐Ÿ+๐’š๐Ÿ)

PROOF: Let ๐‘งโˆ’1 = ๐‘ข + ๐‘–๐‘ฃ be the inverse of ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ W.K.T, ๐‘ง๐‘งโˆ’1 = 1 โŸน (๐‘ฅ + ๐‘–๐‘ฆ)(๐‘ข + ๐‘–๐‘ฃ) = 1 โŸน (๐‘ฅ๐‘ข โˆ’ ๐‘ฆ๐‘ฃ) + ๐‘–(๐‘ฅ๐‘ฃ + ๐‘ฆ๐‘ข) = 1 + ๐‘–0 Equating the Real and Imaginary parts on both sides, ๐‘ฅ๐‘ข โˆ’ ๐‘ฆ๐‘ฃ = 1 โ†’ (1) ; ๐‘ฅ๐‘ฃ + ๐‘ฆ๐‘ข = 0 โ†’ (2)

โˆ†= |๐‘ฅ โˆ’๐‘ฆ๐‘ฆ ๐‘ฅ | = ๐‘ฅ

2 + ๐‘ฆ2; โˆ†๐‘ข= |1 โˆ’๐‘ฆ0 ๐‘ฅ

| = ๐‘ฅ; โˆ†๐‘ฃ= |๐‘ฅ 1๐‘ฆ 0

| = โˆ’๐‘ฆ

๐‘ข =โˆ†๐‘ข

โˆ†=

๐‘ฅ

๐‘ฅ2+๐‘ฆ2 ; ๐‘ฃ =

โˆ†๐‘ฃ

โˆ†=

โˆ’๐‘ฆ

๐‘ฅ2+๐‘ฆ2

๐‘งโˆ’1 = ๐‘ข + ๐‘–๐‘ฃ โŸน ๐‘งโˆ’1 = (๐‘ฅ

๐‘ฅ2+๐‘ฆ2) + ๐‘– (โˆ’

๐‘ฆ

๐‘ฅ2+๐‘ฆ2)

2.For any two complex numbers ๐’›๐Ÿ ๐’‚๐’๐’… ๐’›๐Ÿ prove that ๐ณ๐Ÿ + ๐ณ๐Ÿ = ๐ณ๐Ÿ + ๐ณ๐Ÿ PROOF: Let ๐‘ง1 = ๐‘Ž + ๐‘–๐‘ and ๐‘ง2 = ๐‘ + ๐‘–๐‘‘ , ๐‘Ž, ๐‘, ๐‘, ๐‘‘ โˆˆ ๐‘…

z1 + z2 = (๐‘Ž + ๐‘–๐‘) + (๐‘ + ๐‘–๐‘‘)

= (๐‘Ž + ๐‘) + ๐‘–(๐‘ + ๐‘‘) = (๐‘Ž + ๐‘) โˆ’ ๐‘–(๐‘ + ๐‘‘) = (๐‘Ž โˆ’ ๐‘–๐‘) + (๐‘ โˆ’ ๐‘–๐‘‘)

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Padasalai TrbTnpsc

Page 18: 2.COMPLEX NUMBERS 2000 Padasalai TrbTnpsc ร—2001 =๐‘– 2 ...ย ยท 12.06.2019ย ยท v.gnanamurugan, m.sc.,m.ed., p.g.t, ghss, s.s.kottai, sivagangai dt โ€“ 94874 43870, 82489 56766 page

V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT โ€“ 94874 43870, 82489 56766 Page 18

= z1 + z2 3.Prove that ๐ณ๐Ÿ. ๐ณ๐Ÿ = ๐ณ๐Ÿ . ๐ณ๐Ÿ ๐’˜๐’‰๐’†๐’“๐’† ๐’‚, ๐’ƒ, ๐’„, ๐’… โˆˆ ๐‘น. PROOF: Let ๐‘ง1 = ๐‘Ž + ๐‘–๐‘ and ๐‘ง2 = ๐‘ + ๐‘–๐‘‘ , ๐‘Ž, ๐‘, ๐‘, ๐‘‘ โˆˆ ๐‘…

z1. z2 = (๐‘Ž + ๐‘–๐‘)(๐‘ + ๐‘–๐‘‘)

= (๐‘Ž๐‘ โˆ’ ๐‘๐‘‘) + ๐‘–(๐‘Ž๐‘‘ + ๐‘๐‘) = (๐‘Ž๐‘ โˆ’ ๐‘๐‘‘) โˆ’ ๐‘–(๐‘Ž๐‘‘ + ๐‘๐‘) โ†’ (1)

z1. z2 = ๐‘Ž + ๐‘–๐‘ . ๐‘ + ๐‘–๐‘‘ = (๐‘Ž โˆ’ ๐‘–๐‘)(๐‘ โˆ’ ๐‘–๐‘‘) = (๐‘Ž๐‘ โˆ’ ๐‘๐‘‘) โˆ’ ๐‘–(๐‘Ž๐‘‘ + ๐‘๐‘) โ†’ (2)

From (1) and (2), z1. z2 = z1. z2 4.Prove that ๐’› is purely imaginary if and only if ๐’› = โˆ’๏ฟฝ๏ฟฝ PROOF: Let ๐‘ง = ๐‘ฅ + ๐‘–๐‘ฆ โŸน ๐‘ง = ๐‘ฅ โˆ’ ๐‘–๐‘ฆ ๐‘ง = โˆ’๐‘ง โŸบ ๐‘ฅ + ๐‘–๐‘ฆ = โˆ’(๐‘ฅ โˆ’ ๐‘–๐‘ฆ)

โŸบ ๐‘ฅ + ๐‘–๐‘ฆ = โˆ’๐‘ฅ + ๐‘–๐‘ฆ โŸบ 2๐‘ฅ = 0 โŸบ ๐‘ฅ = 0

โŸบ ๐‘ง ๐‘–๐‘  ๐‘๐‘ข๐‘Ÿ๐‘’๐‘™๐‘ฆ ๐‘–๐‘š๐‘Ž๐‘”๐‘–๐‘›๐‘Ž๐‘Ÿ๐‘ฆ 5.State and prove triangle inequality of two complex numbers (or) For any two complex numbers ๐’›๐Ÿ ๐’‚๐’๐’… ๐’›๐Ÿ prove that |๐’›๐Ÿ + ๐’›๐Ÿ| โ‰ค |๐’›๐Ÿ| + |๐’›๐Ÿ| PROOF:

|๐‘ง1 + ๐‘ง2|2 = (๐‘ง1 + ๐‘ง2)(๐‘ง1 + ๐‘ง2) (โˆต ๐‘ง๐‘ง = |๐‘ง|2) = (๐‘ง1 + ๐‘ง2)(๐‘ง1 + ๐‘ง2) (โˆต z1 + z2 = z1 + z2) = ๐‘ง1๐‘ง1 + (๐‘ง1๐‘ง2 + ๐‘ง1๐‘ง2) + ๐‘ง2๐‘ง2 = ๐‘ง1๐‘ง1 + (๐‘ง1๐‘ง2 + ๐‘ง1๐‘ง2 ) + ๐‘ง2๐‘ง2 (โˆต ๐‘ง = ๐‘ง) = |๐‘ง1|

2 + 2๐‘…๐‘’(๐‘ง1๐‘ง2) + |๐‘ง2|2 (โˆต ๐‘ง + ๐‘ง = 2๐‘…๐‘’(๐‘ง))

โ‰ค |๐‘ง1|2 + 2|๐‘ง1๐‘ง2| + |๐‘ง2|

2 (โˆต ๐‘…๐‘’(๐‘ง) โ‰ค |๐‘ง|) = |๐‘ง1|

2 + 2|๐‘ง1||๐‘ง2| + |๐‘ง2|2 (โˆต z1. z2 = z1. z2 &|๐‘ง| = |๐‘ง|)

= (|๐‘ง1| + |๐‘ง2|)2

Taking square root on both sides, |๐‘ง1 + ๐‘ง2|

2 โ‰ค |z1| + |z2| 6.For any two complex numbers ๐’›๐Ÿ ๐’‚๐’๐’… ๐’›๐Ÿ prove that |๐ณ๐Ÿ๐ณ๐Ÿ| = |๐ณ๐Ÿ||๐ณ๐Ÿ|

PROOF:

|z1z2|2 = (z1z2)(z1z2) (โˆต ๐‘ง๐‘ง = |๐‘ง|2) = (z1)(z2)(z1)(z2) (โˆต z1. z2 = z1. z2 ) = (z1๐‘ง1)(z2๐‘ง2) = |๐‘ง1|

2|๐‘ง2|2

Taking square root on both sides, |z1z2| = |z1||z2|

7.If ๐’› = ๐’“(๐œ๐จ๐ฌ๐œฝ + ๐’Š ๐ฌ๐ข๐ง๐œฝ) then prove that ๐’›โˆ’๐Ÿ =๐Ÿ

๐’“(๐œ๐จ๐ฌ๐œฝ โˆ’ ๐’Š ๐ฌ๐ข๐ง๐œฝ)

PROOF: ๐‘ง = ๐‘Ÿ(cos ๐œƒ + ๐‘– sin ๐œƒ)

๐‘งโˆ’1 =1

๐‘ง=

1

๐‘Ÿ(cos ๐œƒ+๐‘– sin๐œƒ)

= 1

๐‘Ÿ

1

cos ๐œƒ+๐‘– sin๐œƒร—cos๐œƒโˆ’๐‘– sin๐œƒ

cos๐œƒโˆ’๐‘– sin๐œƒ

=1

๐‘Ÿ.cos๐œƒโˆ’๐‘– sin๐œƒ

๐‘๐‘œ๐‘ 2๐œƒ+๐‘ ๐‘–๐‘›2๐œƒ

=1

๐‘Ÿ(cos๐œƒ โˆ’ ๐‘– sin ๐œƒ)

8. If ๐’›๐Ÿ = ๐’“๐Ÿ(๐œ๐จ๐ฌ๐œฝ๐Ÿ + ๐’Š ๐ฌ๐ข๐ง๐œฝ๐Ÿ), ๐’›๐Ÿ = ๐’“๐Ÿ(๐œ๐จ๐ฌ๐œฝ๐Ÿ + ๐’Š ๐ฌ๐ข๐ง๐œฝ๐Ÿ) ๐’•๐’‰๐’†๐’ ๐’›๐Ÿ๐’›๐Ÿ = ๐’“๐Ÿ๐’“๐Ÿ[๐œ๐จ๐ฌ(๐œฝ๐Ÿ + ๐œฝ๐Ÿ) + ๐’Š ๐ฌ๐ข๐ง(๐œฝ๐Ÿ + ๐œฝ๐Ÿ)] PROOF: z1z2 = [๐‘Ÿ1(cos ๐œƒ1 + ๐‘– sin ๐œƒ1)][๐‘Ÿ2(cos๐œƒ2 + ๐‘– sin ๐œƒ2)]

= ๐‘Ÿ1๐‘Ÿ2[(cos๐œƒ1 cos ๐œƒ2 โˆ’ sin ๐œƒ1 sin ๐œƒ2) + ๐‘–(sin ๐œƒ1 cos ๐œƒ2 + cos ๐œƒ1 sin ๐œƒ2)] = ๐‘Ÿ1๐‘Ÿ2[cos(๐œƒ1 + ๐œƒ2) + ๐‘– sin(๐œƒ1 + ๐œƒ2)]

9. If ๐’›๐Ÿ = ๐’“๐Ÿ(๐œ๐จ๐ฌ๐œฝ๐Ÿ + ๐’Š ๐ฌ๐ข๐ง๐œฝ๐Ÿ), ๐’›๐Ÿ = ๐’“๐Ÿ(๐œ๐จ๐ฌ๐œฝ๐Ÿ + ๐’Š ๐ฌ๐ข๐ง๐œฝ๐Ÿ) ๐’•๐’‰๐’†๐’ ๐’›๐Ÿ

๐’›๐Ÿ=

๐’“๐Ÿ

๐’“๐Ÿ[๐œ๐จ๐ฌ(๐œฝ๐Ÿ โˆ’ ๐œฝ๐Ÿ) + ๐’Š ๐ฌ๐ข๐ง(๐œฝ๐Ÿ โˆ’ ๐œฝ๐Ÿ)]

PROOF: ๐’›๐Ÿ

๐’›๐Ÿ=

๐‘Ÿ1(cos๐œƒ1+๐‘– sin๐œƒ1)

๐‘Ÿ2(cos๐œƒ2+๐‘– sin๐œƒ2)

=๐‘Ÿ1

๐‘Ÿ2.cos๐œƒ1+๐‘– sin๐œƒ1

cos๐œƒ2+๐‘– sin๐œƒ2ร—cos๐œƒ2โˆ’๐‘– sin๐œƒ2

cos๐œƒ2โˆ’๐‘– sin๐œƒ2

=๐‘Ÿ1

๐‘Ÿ2.(cos๐œƒ1 cos๐œƒ2+sin๐œƒ1 sin๐œƒ2)+๐’Š(sin๐œƒ1 cos๐œƒ2โˆ’cos๐œƒ1 sin๐œƒ2)

cos2(ฮธ2)+sin2(ฮธ2)

=๐‘Ÿ1

๐‘Ÿ2[cos(๐œƒ1 โˆ’ ๐œƒ2) + ๐‘– sin(๐œƒ1 โˆ’ ๐œƒ2)]

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