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Complex Numbers in Polar Form; DeMoivre’s Theorem

Dr .Hayk MelikyanDepartmen of Mathematics and CS

melikyan@nccu.edu

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We know that a real number can be represented as a point on a number line. By contrast, a complex number z = a + bi is represented as a point (a, b) in a coordinate plane, shown below. The horizontal axis of the coordinate plane is called the real axis. The vertical axis is called the imaginary axis. The coordinate system is called the complex plane. Every complex number corresponds to a point in the complex plane and every point in the complex plane corresponds to a complex number.

Imaginary axis

Real axis

a

b

z = a + bi

The Complex Plane

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Plot in the complex plane: a. z = 3 + 4i b. z = -1 – 2i c. z = -3 d. z = -4i

-5 -4 -3 -2 1 2 3 4 5

5

4

3

2

1

-3

-4

-5

• The complex number z = -1 – 2i corresponds to the point (-1, -2) in the rectangular coordinate system. Plot the complex number by moving one unit to the left on the real axis and two units down parallel to the imaginary axis.

Solution

• We plot the complex number z = 3 + 4i the same way we plot (3, 4) in the rectangular coordinate system. We move three units to the right on the real axis and four units up parallel to the imaginary axis.

z = 3 + 4i

z = -1 – 2i

Text Example

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Plot in the complex plane: a. z = 3 + 4i b. z = -1 – 2i c. z = -3 d. z = -4i

Solution• Because z = -3 = -3 + 0i, this complex

number corresponds to the point (-3, 0). We plot –3 by moving three units to the left on the real axis.

-5 -4 -3 -2 1 2 3 4 5

5

4

3

2

1

-3

-4

-5

z = 3 + 4i

z = -1 – 2i

z = -3

• Because z = -4i = 0 – 4i, this complex number corresponds to the point (0, -4). We plot the complex number by moving three units down on the imaginary axis.

z = -4i

Text Example cont.

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The Absolute Value of a Complex Number

22 babiaz

The absolute value of the complex number a + bi is

Determine the absolute value of z=2-4iSolution:

5220

164)4(2 22

22

babiaz

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Polar Form of a Complex Number

The complex number a + bi is written in polar form as

z = r (cos + i sin )

Where

a = r cos , b = r sin ,

tan =b/a

The value of r is called the modulus (plural: moduli) of the

complex number z, and the angle is called the

argument of the

complex number z, with 0 < < 2

r a2 b2

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Plot z = -2 – 2i in the complex plane. Then write z in polar form.

Solution The complex number z = -2 – 2i, graphed below, is in rectangular form a + bi, with a = -2 and b = -2. By definition, the polar form of z is r(cos + i sin ). We need to determine the value for r and the value for , included in the figure below.

2

-2

-2

è

r

Imaginaryaxis

Realaxis2

z = -2 – 2i

Text Example

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Since tan 4 = 1, we know that lies in quadrant III. Thus,

SolutionText Example cont.

r a2 b2 ( 2)2 ( 2)2 4 4 8 2 2

tan b

a

2

21

4

44

4

54

z r(cos i sin ) 2 2(cos54

i sin54

)

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Product of Two Complex Numbers in Polar Form

Let z1 = r1 (cos 1+ i sin 1) and

z2 = r2 (cos 2 + i sin 2)

be two complex numbers in polar form. Their product, z1z2, is

z1z2 = r1 r2 (cos ( 1 + 2) + i sin ( 1 + 2))

To multiply two complex numbers, multiply moduli and add arguments.

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Find the product of the complex numbers. Leave the answer in polar form.z1 = 4(cos 50º + i sin 50º) z2 = 7(cos 100º + i sin 100º)

Solution

Form the product of the given numbers.

= (4 · 7)[cos (50º + 100º) + i sin (50º + 100º)]

= 28(cos 150º + i sin 150º)

Multiply moduli and add arguments.

Simplify.

Text Example

= [4(cos 50º + i sin 50º)][7(cos 100º + i sin 100º)]z1z2

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Quotient of Two Complex Numbers in Polar Form

Let z1 = r1 (cos 1 + i sin 1) and

z2 = r2 (cos 2 + i sin 2) be two complex numbers in polar form. Their quotient, z1/z2, is

To divide two complex numbers, divide moduli and subtract arguments.

)]sin()[cos( 21212

1

2

1 ir

r

z

z

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DeMoivre’s Theorem

• Let z = r (cos + i sin ) be a complex numbers in polar form. If n is a positive integer, z to the nth power, zn, is

)sin(cos

)]sin(cos[

ninr

irzn

nn

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Find [2 (cos 10º + i sin 10º)]6. Write the answer in rectangular form a + bi.

Solution By DeMoivre’s Theorem,

= 26[cos (6 · 10º) + i sin (6 · 10º)]Raise the modulus to the 6th power and multiply the argument by 6.

= 64(cos 60º + i sin 60º) Simplify.

Text Example

Write the answer in rectangular form.

Multiply and express the answer in a + bi form.

641

2 i

3

2

32 32 3i

[2 (cos 10º + i sin 10º)]6

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1,...,3,2,1,0

360sin

360cos

nkwhere

n

ki

n

krz n

k

DeMoivre’s Theorem for Finding Complex Roots Let =r(cos+isin) be a complex number in polar form. If

0, has n distinct complex nth roots given by the formula

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Example

Find all the complex fourth roots of 81(cos60º+isin60º)

)15sin15(cos3

4

0*36060sin

4

0*36060cos81

360sin

360cos

4

i

i

n

ki

n

krz n

k

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Objectives (sec 1.4 from math 1100 staff)

• Add and subtract complex numbers.

• Multiply complex numbers.

• Divide complex numbers.

• Perform operations with square roots of negative numbers.

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Complex Numbers and Imaginary Numbers

The set of all numbers in the form a + bi

with real numbers a and b, and i, the imaginary unit, is called the set of complex numbers.

The imaginary unit i is defined as 21, where 1.i i

The standard form of a complex number isa + bi.

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Operations on Complex Numbers

The form of a complex number a + bi is like the binomiala + bx. To add, subtract, and multiply complex numbers, we use the same methods that we use for binomials.

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Example: Adding and Subtracting Complex Numbers

Perform the indicated operations, writing the resultin standard form:

a. (5 2 ) (3 3 )i i

(5 3) ( 2 3)i 8 i

b. (2 6 ) (12 )i i

2 6 12i i (2 12) (6 1)i

10 7i

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Example: Multiplying Complex Numbers

Find the product:

a. 7 (2 9 )i i214 63i i

14 63( 1)i

14 63i

63 14i

b. (5 4 )(6 7 )i i

5(6) 5( 7 ) 4 (6) 4 ( 7 )i i i i 230 35 24 28i i i

30 11 28( 1)i

30 11 28i

58 11i

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Conjugate of a Complex Number

For the complex number a + bi, we define its complex conjugate to be a – bi.

The product of a complex number and its conjugateis a real number.( )( )a bi a bi

( ) ( ) ( ) ( )a a a bi bi a bi bi 2 2 2a abi abi b i 2 2 ( 1)a b 2 2a b

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Complex Number Division

The goal of complex number division is to obtain a realnumber in the denominator. We multiply the numeratorand denominator of a complex number quotient by theconjugate of the denominator to obtain this real number.

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Example: Using Complex Conjugates to Divide Complex Numbers

Divide and express the result in standard form:5 44

ii

5 4 44 4

i ii i

2

2

20 5 16 416 4 4

i i ii i i

20 21 ( 4)16 ( 1)

i

16 2117

i

16 21

17 17i

In standard form, the result is

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Principal Square Root of a Negative Number

For any positive real number b, the principal square root ofthe negative number – b is defined by

b i b

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Example: Operations Involving Square Roots of Negative Numbers

Perform the indicated operations and write the result instandard form:

a. 27 48

3 3 4 3i i

7 3i

2b. ( 2 3)

22 3

2 3 2 3i i 24 2 3 2 3 3i i i

4 4 3 3( 1)i

1 4 3i