L01A - Complex Numbers

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COMPLEX NUMBERS:

DE MOIVRES THEOREM, POWERS,

& ROOTS

ADVANCED ENGINEERING MATHEMATICS

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Number systems

The natural numbers. These numbers are the positive (and

zero) whole numbers 0, 1, 2, 3, 4, 5, . If two such numbersare added or multiplied, the result is again a natural number.

The integers . These numbers are the positive and negative

whole numbers , 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, . If twosuch numbers are added, subtracted, or multiplied, the result

is again an integer.

The rational numbers. These numbers are the positive andnegative fractions p/q where p and q are integers and q 0. If

two such numbers are added, subtracted, multiplied, or

divided (except by 0), the result is again a rational number.

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The real numbers. These numbers are the positive and

negative infinite decimals(including terminating decimals that

can be considered as having an infinite sequence of zeros onthe end). If two such numbers are added, subtracted,

multiplied, or divided (except by 0), the result is again a real

number.

The complex numbers . These numbers are of the form x + iy

where x and y are real numbers and i = (1) . (For further

explanation, see the section Complex analysis.) If two such

numbers are added, subtracted, multiplied, or divided (except

by 0), the result is again a complex number.

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Complex Analysis

In the 18th century a far-reaching generalization of analysis was

discovered, centred on the so called imaginary number i = (1) . Inengineeringthis number is usually denoted byj.

The name imaginary arises because squares of real numbers are alwayspositive. In consequence, positive numbers have two distinct square

rootsone positive, one negative. Zerohas a single square rootnamely,zero. And negative numbershave no realsquare rootsat all.

The resulting objects (imaginary numbers) are NUMBERSin the sense thatarithmetic and algebra can be extended to them in a simple and naturalmanner; they are IMAGINARY in the sense that their relation to thephysical world is less direct than that of the real numbers. Numbersformed by combining real and imaginary components, such as 2 + 3i, aresaid to be COMPLEX (meaning composed of several parts rather thancomplicated).

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Formal definition of Complex Numbers

The modern approach is to define a complex numberx + iyas

a pair of real numbers (x, y) subject to certain algebraic

operations.

Thus one wishes to add or subtract, (a, b) (c, d), and to

multiply, (a, b) (c, d), or divide, (a, b)/(c, d), these quantities.

This is a formal way to set up a situation which, in effect,

ensures that one may operate with expressionsx+ iyusing all

the standard algebraic rules but recalling when necessary that

i2may be replaced by 1. For example,

(1 + 3i)2

= 12

+ 23i+ (3i)2

= 1 + 6i+ 9i2

= 1 + 6i9 = 8+ 6i

Real numbers can be described by a single number line, with

negative numbers to the left and positive numbers to the

right, the complex numbers require a number plane with two

axes, real and imaginary.ECE50 5


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Powers and Roots of Complex Numbers

Powers of Complex numbers

)2sin2(cos

)]sin()[cos(

)]sin(cos)sin(cos[)]sin(cos[

2

2

ir

irr

iririr

In the same way,

)3sin3(cos)]sin(cos[ 33 irir


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De MoivresTheorem

De MoivresTheoremIf r(cos + i sin) is a complex number, and nis any

real number, then

In compact form, this is written

).sin(cos)]sin(cos[ ninrir nn

).cis(]cis[ nrr nn


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Finding a Power of a Complex Number

Example Find and express the result inrectangular form.

Solution

8

)31( i

3128128

2

3

2

1256

)120sin120(cos256

)480sin480(cos256

)]608sin()608[cos(2

)]60sin60(cos2[)31(

8

88

i

i

i

i

i

ii

Convert to

trigonometric form.

480 : and 120 : are

coterminal.

cos120 : = -1/2;

sin120 : = 2/3

Rectangular form


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Roots of Complex Numbers

To find three complex cube roots of

8(cos 135: + i sin 135 :), for example, look for a

complex number, say r(cos + sin ), that will satisfy

nth Root

For a positive integer n, the complex number a+biis

the nthof the complex numberx + yiif

(a + bi)n=x + yi.

).135sin135(cos8)]sin(cos[ 3 iir


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By De Moivres Theorem,

becomes

Therefore, we must have r3= 8, or r= 2, and

).135sin135(cos8)]sin(cos[ 3 iir

).135sin135(cos8)3sin3(cos3 iir

integer.any,3

360135integerany,3601353

kk

kk



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Let ktake on integer values 0, 1, and 2.

It can be shown that for integers k= 3, 4, and 5, thesevalues have repeating solutions. Therefore, all of the cube roots

(three of them) can be found by letting k= 0, 1, and 2.

2853

720135,2

165

3

360135,1

453

0135,0

k

k

k



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When k = 0, the root is 2(cos 45: + i sin 45:).



nth Root Theorem

If nis any positive integer, r is a positive real number, and is

in degrees, then the nonzero complex number r(cos + i sin )has exactly ndistinctnth roots, given by

where

),sin(cos irn

1,,2,1,0,360

or360

nkn

k

nn

k



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Finding Complex Roots

Example Find the two square roots of 4i. Write theroots in rectangular form, and check your

results directly with a calculator.

Solution Firstwrite 4iin trigonometric form as

Here, r= 4 and = /2. The square roots have modulusand arguments as follows.

.2

sin2

cos44

ii

24

kk

42

2

2

2


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Since there are two roots, let k= 0 and 1.If k= 0, then

If k= 1, then

Using these values for , the

square roots are 2 cis and

2 cis which can be written

in rectangular form as

.4

04

.4

514

4

,45

.22and22 ii



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Example Find all fourth roots of Write the roots in

rectangular form.

Solution

If k= 0, then = 30: + 90:0 = 30:.If k= 1, then = 30: + 90:1 = 120:.

If k= 2, then = 30: + 90:2 = 210:.

If k= 3, then = 30: + 90:3 = 300:.

.388 i

kk

r

i

90304

360

4

120Arguments

216Modulus

120and16

120cis16388

4



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Using these angles on the previous slide, the fourth

roots are

2 cis 30:, 2 cis 120:, 2 cis 210:, and 2 cis 300:.

These four roots can be written in rectangular form as

.31,3,31,3 iiii



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Solving an Equation by Finding Complex Roots

Example Find all complex number solutions ofx51 = 0. Graph them as vectors in thecomplex plane.

Solution Write the equation as

To find the five complex number solutions, write 1 in

polar form as

The modulus of the fifth roots is

.1

015

5

x

x

).0sin0(cos1011 ii

.115


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Solving an Equation by Finding Complex Roots

The arguments are given by

Using these arguments, the fifth roots are

1(cos 0 : + i sin 0:), k = 01(cos 72 : + i sin 72:), k = 1

1(cos 144 : + i sin 144:), k = 2

1(cos 216 : + i sin 216:), k = 3

1(cos 288 : + i sin 288:), k = 4

.4or,3,2,1,0,720 kk


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References:

1. Advanced Engineering Mathematics by Kreyszig

2. A Graphical Approach to Algebra & Trigonometryby

Hornsby, et. al.

3. Encyclopedia Britannica: Ultimate Reference Suite

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L01A - Complex Numbers

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