Complex Part2

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  • 8/6/2019 Complex Part2

    1/13

    ===== The square root of a complex number =========================

    To find the two square roots of the

    complex number a ib do t

    e followi g:

    (i) Let t

    e sq are root be x iy ie

    a ib x iy ! (ii) Sq are bot

    sides, t

    e apply t

    e eq ality property to get two

    sim

    lta

    eo

    s eq

    atio

    s i

    x a

    d y, solve t

    em.EXAMPLE: Fi d t

    e two sq are roots of t

    e complex mber 5-12i

    SOLUION:

    2 2

    2 2

    2 2

    2 2

    ( i ) let 5-12i x iy

    ( ii ) Sq

    are bot

    e sides: 5-12i=x y 2 xyi

    x y 5...........( 1)

    2 xy 12......( 2 )

    ( 1 ) ( 2 )

    x y 13........( 3 )

    Solvi

    g ( 1 )&( 3 ) : x 3 & y 25 12i ( 3 2 y ) A

    s

    !

    @ !!

    !

    ! s ! s@ ! s

    Ot

    er sol tio :

    2 2 2 2

    2 2

    From eq

    atio

    s (1),(2)

    e q

    a

    tities x y ;2 xy; x y form t

    e t

    ree sides of a

    rig

    t a

    gled tria

    gle i

    w

    ic

    x y is t

    e

    ypoti

    eo

    s. So we ca

    form a rt. a

    gled tria

    glew

    ose legs are 5;12 a

    d get t

    2 2

    2 2

    e

    ypoti

    eo

    s

    eq

    als 13, a

    d write

    x y 5..................( 1 )

    2xy 12.............( 2 )x y 13...............( 3 )

    Solvi

    g( 1 ) &( 3 ) x 3 ; y 2 5 12i ( 3 2i ).

    !

    ! !

    ! s ! ! s m

    Ot

    er sol tio :

    2 2 2 2 2

    1T

    i

    k abo t t

    o

    mbers

    ose prod ct 6 ( t

    e imagi

    ary part )2

    a

    d t

    e differe

    ce of t

    eir sq ares is 5 (t

    e real part); t

    e

    mbers

    ill

    be 3 a d 2

    5-12i

    3 12i 2 i 9 12i 4i ( 3 2i )

    5 12i ( 3 2i )

    s.

    @ ! !

    @ !

    Now, try t

    ese problems:

    3 4i ; 7 24i ; 8 15i

  • 8/6/2019 Complex Part2

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    EXAMPLE: Solve t

    e followi

    g eq atio

    s i

    C(i) !z z2 1 0 (ii) x ( i )x !22 5 6 0

    SOLU ION

    1 1 4 1 1 1 3i 1 3( i )z i

    2 2 2 2

    s v v s s

    2

    2 2 2 2

    1 2

    ( 5 i ) ( 5 i ) 4 2 6 ( 5 i ) 24 10i ( ii ) x

    4 424 10i l mi l m 24 ;l m 26

    l 1 ; m 5

    24 10i ( 1 5i )

    ( 5 i ) ( 1 5i ) 3 3i x x ; x 1 i

    4 2

    s v v s

    s s

    s s

    EXAMPLE: Solve t!

    e followi" g eq# atio" i" C2

    ( 1 i )z ( 1 3i )z 2( 2 3i ) 0 SOLU ION:

    2

    2

    1 2

    Divi de t$

    e eq%

    atio&

    by (1+i) a&

    d simplify

    z ( 2 i )z ( 1 5i ) 0

    ( 2 i ) ( 2 i ) 4( 1 5i ) ( 2 i ) 7 24i z

    2 2

    7 24i ( 4 3i )

    ( 2 i ) ( 4 3i ) ( 2 i ) ( 4 3i )z 3 2i ; z 1 i

    2 2

    s s

    s

    @

    EXAMPLE:

    Form t!

    e q#

    adratic eq#

    atio" wit!

    real coefficie" ts, if o" e of its roots is 3+i

    SOLU ION:

    2

    2

    3 i is a root 3-i is t $

    e ot$

    er roots

    %

    m of roots = 3+i + 3-i=6

    prod%

    ct of roots=(3+i)(3-i)=10

    t$

    e eq%

    atio&

    is

    x ( s%

    m )x prod %

    ct 0

    i .e . x 6 x 10 0

    @

    @

    Q

    EXAMPLE:

    1 2i 1 2i If x ; y ,then find 5x ' 3y .

    1 i 1 i

    ! !

    SOLU(

    ION:1 2i 1 2i 1 i 1 3i x

    1 i 1 i 1 i 21 2i 1 2i 1 i 1 3i

    y1 i 1 i 1 i 2

    2 3 25 x 3 y 4 3i ( i )

    2 2

    ! ! v !

    ! ! v !

    ! ! s

  • 8/6/2019 Complex Part2

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    Exercise 2

    2 2

    4 2 2

    2

    2

    1 If x ) then find the solution set of

    a. x 25 0 b.x 4 x 13 0

    c.x 7 x 12 0 d .x 6 x 9 2i 0

    e.x ( 5 i )x ( 8 i ) 0

    f .( 2 i )x ( 9 7 i )x 5( 3 2i ) 0

    20

    ind the square roots of the follo1

    ing complex numbersa.z 5

    ! !

    ! !

    !

    !

    !

    3-2

    1 1

    2 2

    2 2

    12i b.z 3 4i

    c.z 7 24i d .z 1 i

    13 65i e.z i f .z

    5 i

    3 If x 3 4i , then find x

    4 If x 2 21 3 20i, then find the value of x 29x .

    5 If x and y are real values, find these values if

    y iy 6

    ! ! !

    ! !

    !

    2 2 2

    2

    2 2

    *

    i 2 x ix

    ( x i ) ( 2 y i ) 4( 3 1 )i 2 y x

    6 If l and m are the roots of the equation x ( 4 6i )x ( 10 20i ) 0 ,then find

    then find the equation1

    hose roots are l and m .

    74

    olve the simultaneously the follo1

    ing equ

    ! !

    !

    ations, 1 here x and y are real

    2 5i and x y i

    x y ! !

  • 8/6/2019 Complex Part2

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    Geometric Represe5 tatio5 Of Complex N6

    mbers

    Arga5

    d Fig6 res

    7

    8

    e Fre5 c8

    mat8

    ematicia5 Arga5 d establis8

    ed t8

    e geometric represe5 tatio5

    of t8

    e complex5 6 mber x + y i as a5 ordered pair (x , y) R2,8

    e called t8

    e x-axis as

    t8

    e real axis a5

    d t8

    e y-axis as t8

    e imagi5

    ary axis. So, t8

    e Cartesia5

    pla5

    e is5

    amed

    as Arga

    5

    d pla

    5

    e a

    5

    d t

    8

    e fig6

    res t

    8

    at represe

    5

    t t

    8

    e complex

    5

    6

    mbers or a

    5

    yoperatio5 performed o5 t8

    em as Arga5 d fig6 res.

    For t8

    is, t8

    e complex5 6 mbers will be represe5 ted i5 Arga5 d pla5 e by a poi5 t

    (x , y), ( sometimes t8

    ey call it vector), as yo6

    see i5 t8

    e fig6

    re:

    9

    @

    e complexAB

    mberz1 = 3 + 4i is represeA ted by poiA t A(3,4), z2 = -1 + 2i is

    represeA ted by t@

    e poiA t B(-1,2) aA dz3 = -2 3i is represeA ted by t

    @

    e poiA t C(-2,-3),

    aA d so oA .

    RepreseA tatioA of SB

    m

    Ifz1=(x1,y1) aA dz2=(x2,y2) t@

    eA

    z1 + z2 = (x1 + x2 , y1 + y2). From t@

    e

    figB re we fiA d t@

    at t@

    e poiA ts (x1,y1) ;

    (x2,y2); aA d (x1+x2 ,y1+y2) are t@

    ree vertices

    of t@

    e parallelogram OACB, t@

    at is t@

    e s B m of

    two complexAB

    mbers is t@

    e foB

    rt@

    vertex C. A(x1,y1)

    B(x2,y2)

    C(x1+x2,y1+y2)

  • 8/6/2019 Complex Part2

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    TheC

    D mbers izaC

    d iz

    RepreseC t theC

    Dmberz= 1+2i, the

    C

    represeC t the two C D mbers izaC d iz. What do

    yoD

    C otice.

    1

    2

    1

    1

    2

    z1 2i is t

    he poi

    E

    t A (1,2)iz=-2+i is t he poi

    E

    t A (-2,1)

    -iz=2-i is t he poiE

    t A (2,-1)

    UsiE

    g the slope, we fiE

    d that OA OA

    aE

    d OA OA . This meaE

    s that we rotate OA

    aE

    ticlock 90 . OE

    the otherhaE

    d OA OA

    OA

    !

    B

    !

    r B

    2OA ,This mea

    F

    s that we rotate OA

    clockwise90

    !

    r

    ModG lG s- ArgG meH t-TrigoH ometric Form

    of a ComplexHG

    mber

    We kH ew that the complexH

    G mberz = x + iy caH be represeH ted iH the arga

    H d

    plaH

    e by the poiH

    t A(x , y). This poiH

    t caH

    also fG lly determiH

    ed as sooH

    as we kH

    ow

    the distaH ce OA aH d the polar a

    H gle betweeH OA aH d the (+)ve x-axis (Ur), meas G red

    aH ticlockwise.

    The distaH ce OA is called the modG lG s ofz, a

    H d is deH oted by r

    2 2r x y@ ! The aH gle U is called the ArgG meH t ofz, aH d is deH oted by Arg(z) where

    ytaH

    x!U

    SiH

    ce if we kH

    ow taH

    U, theH

    U willhave maH

    y valG es ( 2FU Ts ), it is agreed

    that [0 ,2 [ aH

    d is called the priH

    cipal ArgG meH

    t ofzU T .

    the priH

    cipal ArgG meH

    t [0,2 [ U T@

    From the figG re we fiH

    d that

    xcos x r cos

    ry

    siH y r si H

    r

    U U

    U U

    ! !

    ! !

    z r(cos i si H )U U!

    A(1,2)

    A1(-2,1)

    (2,-1)A2

    U

    A

    x

    y

    O

  • 8/6/2019 Complex Part2

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    The previo I s form is called the TrigoP

    ometric Form of the complexP

    I mber.

    YoI

    mI

    stP otice that the previoI

    s form is the priP cipal form, which mI

    st be, wheP

    weP

    eed to fiP

    d its modI lI s aP

    d ArgI meP

    t, aP

    d yo I o I ght to modify aP

    y other form

    before yoI

    determiP e the modI

    lI

    s aP d the Arg.

    SpecialTrigoP

    ometric forms

    z 1 z cos 0 i si Q

    0

    z i z cos i si Q

    2 2z 1 z cos i si

    Q

    3 3z i z cos i si

    Q

    2 2

    T T

    T T

    T T

    ! R ! r r

    ! R !

    ! !

    ! !

    EXAMPLE:

    FiS

    d the modT lT s aS

    d the priS

    cipal Arg. of the followiS

    g complexS

    T mbers

    theS write the trigoS ometric form of eachST

    mber.

    ( i )2 2i ; 3 3i ; 1 3i ; 3 i

    ( ii )5 ; 2i ; 4 ; 4i

    SOLUTION:

    2 2

    2 2

    2 2

    ( i ) z 2 2i r 2 2 2 2

    2ta

    U

    1 ( x 0 , y 0 )2 4

    z 2 2(cos i si U

    )4 4

    z 3 3i r 3 3 3 2

    3 3ta

    U

    1 ( x 0, y 0 )

    3 43 3z 3 2(cos i si

    U

    )4 4

    z 1 3i r ( 1 ) ( 3 ) 2

    3 4ta

    U

    3 ( x 0, y 0 )1 3

    4 4z 2(cos i si

    U

    )3 3

    z 3

    TU U

    T T

    TU U

    T