m211f2012 Slides AppA Handout

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MATH 211 Winter 2013

Lecture Notes(Adapted by permission of K. Seyffarth)

Appendix A

Appendix A Page 1/1


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Appendix A Complex Numbers

Appendix A Page 2/1


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Why complex numbers?Counting numbers: 1, 2, 3, 4, 5, . . .Integers: 0, 1, 2, 3, 4 . . . but also 1, 2, 3 . . ..To solve 3x + 2 = 0, integers arent enough, so we have rationalnumbers (fractions), i.e.,

If 3x + 2 = 0, then x = 23

.

We still cant solve x

2

2 = 0 because there are no rational numbersx with the property that x2 2 = 0, so we have irrational numbers,i.e.,

If x2 2 = 0, then x =

2.

The set of real numbers, R, consists of all rational and irrational

numbers (note that integers are rational numbers). However, we stillcant solve

x2 + 1 = 0

because this requires x2 = 1, but any real number x has theproperty that x

2

0.Appendix A Page 3/1


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Definitions

The imaginary unit, denoted i, is defined to be a number with theproperty that i2 = 1.A pure imaginary number has the form bi where b R, b= 0, and iis the imaginary unit.

A complex number is any number z of the form

z = a + bi

where a, b R and i is the imaginary unit. a is called the real part of z. b is called the imaginary part of z. If b = 0, then z is a real number.

Appendix A Page 4/1


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Operations with Complex Numbers

Definitions

Let z = a + bi and w = c + di be complex numbers.

Equality. z = w if and a = c and b = d.

Addition and Subtraction.

z + w = (a + bi) + (c + di) = (a + c) + (b+ d)i

z w = (a + bi) (c + di) = (a c) + (b d)i

Multiplication.

zw = (a + bi)(c + di) = (ac bd) + (ad + bc)i

Appendix A Page 5/1


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Examples

(3 + 6i) + (5 i) = (2 + 5i)(4

7i)

(6

2i) =

2

5i.

(2 3i)(3 + 4i) = 6 + 8i + 9i + 12 = 6 + 17i.

Appendix A Page 6/1


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Example

Find all complex number z so that z2 = 3 + 4i.Let z = a + bi. Then

z2 = (a + bi)2 = (a2 b2) + 2abi = 3 + 4i,so

a2 b2 = 3 and 2ab = 4.Since 2ab = 4, a = 2

b. Substituting this into the first equation gives us

a2 b2 = 32

b

2 b2 = 3

4b2

b2 = 34 b4 = 3b2

b4 3b2 4 = 0.

Appendix A Page 7/1


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Example (continued)

Now, b

4

3b2

4 = 0 can be factored into(b2 4)(b2 + 1) = 0

(b 2)(b+ 2)(b2 + 1) = 0.

Since b R, and b2

+ 1 has no real roots, b = 2 or b = 2.Since a = 2

b, it follows that

when b = 2, a = 1, and z = a + bi = 1 + 2i;

when b =

2, a =

1, and z = a + bi =

1

2i.

Therefore, if z2 = 3 + 4i, then z = 1 + 2i or z = 1 2i.

Appendix A Page 8/1


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Definitions

Let z = a + bi and w = c + di be complex numbers.

The conjugate of z is the complex number

z = a bi.

Division. Suppose that c, d are not both zero. Then

a + bi

c + di=

a + bi

c + di c di

c di=

(ac + bd) + (bc ad)ic2 + d2

= ac + bdc2 + d2

+ bc adc2 + d2

i.

Appendix A Page 9/1


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Examples

1

i=

1

i ii =

ii2 = i.

1

3 + 4i=

1

3 + 4i 3 4i3 4i

=3 4i

33 + 42=

3

25 4

25i.

1 2i2 + 5i =

1 2i2 + 5i

2 5i2 5i =

(2 10) + (4 5)i22 + 52

= 1229

129

i.

Appendix A Page 10/1


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Definition

The absolute value or modulus of a complex number z = a + bi is

|z| =

a2 + b2.

Note that this is consistent with the definition of the absolute value of a

real number.

Examples

| 3 + 4i| = 32 + 42 = 25 = 5.

|3

2i

|=

32 + 22 =

13.

|i| = 12 = 1.



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Properties of the Conjugate and Absolute Value (p. 507)

Let z and w be complex numbers.

C1. z

w = z

w.

C2. (zw) = z w.

C3.

z

w

= z

w.

C4. (z) = z.

C5. z is real if and only if z = z.C6. z z = |z|2.C7. 1

z= z|z|2 .

C8. |z| 0 for all complex numbers zC9. |z| = 0 if and only if z = 0.

C10. |zw| = |z| |w|.C11.

z

w

= |z||w| .

C12. Triangle Inequality

|z + w

| |z

|+

|w

|.



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The Complex Plane

Represent z = a + bi as a point (a, b) in the plane, where the x-axis is the

real axis and the y-axis is the imaginary axis.

0 x

y

(a, b)a

b

Real numbers: a + 0i lie on the x-axis.

Pure imaginary numbers: 0 + bi (b

= 0) lie on

the y-axis.

|z| = a2 + b2 is the distance from z to the origin.z is the reflection of z in the x-axis.



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If z = a + bi and w = c + di, then

|z

w

|= (a c)

2 + (b

d)2.

0 x

yz = a + bi

w = c + di

b

d

b d

c a

a c

This is used to derive the triangle inequality: |z + w| |z| + |w|.



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Triangle Inequality

0 x

yz + w

z

|z + w|

|w|

|z|

|z + w| |z| + |w|.



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Addition

If z = a + bi and w = c + di, then z + w = (a + c) + (b+ d)i.Geometrically, we have:

0 x

y

z

wz + w

c a a + c

b

d

b + d

0, z, w, and z + w are the vertices of a parallelogram.



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Representing Complex Numbers in Polar Form

Suppose z = a + bi, and let r = |z| = a2 + b2. Then r is the distancefrom z to the origin. Denote by the angle that the line through 0 and z

makes with the positive x-axis.

0 x

y

z = a + bi

a

br

Then is an angle defined by cos = ar

and sin = br

, so

z = rcos + rsin i = r(cos + isin ).

is called the argument of z, and is denoted arg z.



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Definitions

The principal argument of z = r(cos + isin ) is the angle suchthat <

( is measured in radians).

If z is a complex number with |z| = r and arg z = , then we writez = rei = r(cos + isin ).

Note that since arg z is not unique, rei is a polar form of z, not the polarform of z.



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Examples

Convert each of the following complex numbers to polar form.

1 3i

2 1 i3

3 i

4

3 + 3iSolutions.

1 3e(/2)i

2

2e(3/4)i =

2e(5/4)i

3 2e(/6)i

4 2

3e(/3)i



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Problems involving multiplication of complex numbers can often besimplified by using polar forms of the complex numbers.

Theorem (Appendix A, Theorem 1 Multiplication Rule)If z1 = r1e

i1 and z2 = r2ei2 are complex numbers, then

z1z2 = r1r2ei(1+2).

Theorem (Appendix A, Theorem 2 De Moivres Theorem)

If is any angle, then(ei)n = ein

for all integers n.(This is an obvious consequence of Theorem 1 when n 0, but also holdswhen n < 0.)



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Example

Express (1 i)6(3 + i)3 in the form a + bi.

Solution.Let z = 1 i = 2e(/4)i and w = 3 + i = 2e(/6)i. Then we want tocompute z6w3.

z6w3 = (

2e(/4)i)6(2e(/6)i)3

= (23e(6/4)i)(23e(3/6)i)= (8e(3/2)i)(8e(/2)i)= 64ei

= 64ei

= 64(cos + isin )

= 64.



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Example

Express

12

3

2 i17

in the form a + bi.

Solution.

Let z = 12

32 i = e

(/3)i.Then

z17

=

e(/3)i17

= e(17/3)i

= e(/3)i

= cos

3

+ isin

3=

1

2+

3

2i.


Roots of U it


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Roots of Unity

Example

Find all complex number z so that z3

= 1, i.e., find the cube roots ofunity. Express each root in the form a + bi.

Solution.

Let z = rei. Since 1 = 1ei0 in polar form, we want to solve

rei

3= 1ei0,

i.e.,r3ei3 = 1ei0.

Thus r3 = 1 and 3 = 0 + 2k = 2k for k = 0, 1, 2, . . .Since r3 = 1 and r is real, r = 1.


E l ( ti d)


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Example (continued)

Now 3 = 2k for k = 0, 1, 2, . . ., so = 23 k for k = 0, 1, 2, . . .k ei

3 2 e2i = e0i = 12 43 e(4/3)i = e(2/3)i1 23 e(2/3)i = e(4/3)i0 0 e0i = 1

1 2

3

e(2/3)i

2 43 e(4/3)i

3 2 e2i = e0i = 1

The three cube roots of unity are

e0i = 1

e(2/3)i = cos 23 + isin23 = 12 +

3

2 i

e(4/3)i = cos 43 + isin43 = 12

3

2 i


th


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Theorem (Appendix A, Theorem 3 nth Roots of Unity)

For n 1, the (complex) solutions to zn = 1 are

z = e(2k/n)i

for k = 0, 1, 2, . . . , n 1.Example

The sixth roots of unity, i.e., the solutions to z6 = 1, are

z = e(2k/6)i = e(k/3)i for k = 0, 1, 2, 3, 4, 5.k z

0 z = e0i = 1

1 z = e(/3)i = 12 +

32 i

2 z = e(2/3)i

= 12 + 32 i3 z = ei = 14 z = e(4/3)i = 12

3

2 i

5 z = e(5/3)i = 12

32 i


Example


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Example

Find all complex numbers z such that z4 = 2(

3i 1), and express eachin the form a + bi.

Solution. First, convert 2(3i 1) = 2 + 23i to polar form:

|z4| =

(2)2 + (2

3)2 =

16 = 4.

If = arg(z4

), thencos =

24

=12

sin =2

3

4=

3

2

Thus, = 23 , and

z4 = 4e(2/3)i.



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Example (continued)

. . . z4 = 4e(2/3)i.

Let z = rei. Then z4 = r4ei4, so r4 = 4 and 4 = 2

3

+ 2k,k = 0, 1, 2, 3.

Since r4 = 4, r2 = 2. But r is real, and so r2 = 2, implying r = 2.However r 0, and therefore r = 2.Since 4 = 2

3 + 2k, k = 0, 1, 2, 3,

=2

12+

2k

4

=

6

+k

2=

(3k + 1)

6

for k = 0, 1, 2, 3.



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Example (continued)

r =

2 and = 3k+16 , k = 0, 1, 2, 3.k = 0 : z =

2e(/6)i =

2( (

3

2 +12 i) =

6

2 +

22 i

k = 1 : z =

2e(2/3)i =

2(12 +

32 i) =

2

2 +

62 i

k = 2 : z =

2e(7/6)i =

2(

3

2

12 i) =

6

2

2

2 i

k = 3 : z = 2e(5/3)i = 2( 12 32 i) = 22 62 i

Therefore, the fourth roots of 2(

3i 1) are:

6

2 +

2

2 i,

2

2 +

6

2 i,

6

2

2

2 i,

2

2

6

2 i.



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Two additional examples are posted on Blackboard under

Supplementary Notes

Roots of Complex Numbers


Real Quadratics


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Real Quadratics

Definition

A real quadratic is an expression of the form ax2 + bx+ c wherea, b, c R and a = 0.To find the roots of a real quadratic, we can either factor by inspection, oruse the quadratic formula:

x =b b2 4ac

2a.

The expression b2

4ac in the quadratic formula is called the

discriminant, and

if b2 4ac 0, then the roots of the quadratic are real;if b2 4ac < 0, then the quadratic has no real roots.


Real Quadratics


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Real QuadraticsIn the case b2 4ac < 0, the quadratic is called irreducible, and

b2 4ac = (1)(4ac b2) = i4ac b2.Therefore, the roots of an irreducible quadratic are

b i4ac b2

2a

= b2a +

4acb2

2a i

b

2a

4ac

b2

2a i

The two roots are complex conjugates of each other, and are denoted

u = b

2a +

4ac

b2

2a i

and

u = b2a

4ac b22a

i



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Example

The quadratic x2

14x + 58 has roots

x =14 196 4 58

2

=14 196 232

2

=14 36

2

=14 6i

2= 7

3i,

so the roots are 7 + 3i and 7 3i.


Conversely given u = a + bi with b = 0 there is an irreducible quadratic


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Conversely, given u = a + bi with b= 0, there is an irreducible quadratichaving roots u and u.

Example

Find an irreducible quadratic with u = 5 2i as a root. What is the otherroot?

Solution.

(x u)(x u) = (x (5 2i))(x (5 + 2i))= x2 (5 2i)x (5 + 2i)x + (5 2i)(5 + 2i)= x2 10x + 29.

Therefore, x2 10x + 29 is an irreducible quadratic with roots 5 2i and5 + 2i.Notice that 10 = (u+ u) and 29 = uu = |u|2.



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ExampleFind an irreducible quadratic with root u = 3 + 4i, and find the otherroot.Solution.

(x u)(x u) = (x (3 + 4i))(x (3 4i))= x2 + 6x + 25.

Thus x2 + 6x + 25 has roots 3 + 4i and 3 4i.


Quadratics with Complex Coefficients


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Q p

Example

Find the roots of the quadratic x

2

(3 2i)x + (5 i) = 0.Solution. Using the quadratic formula

x =3 2i

((3 2i))2 4(5 i)

2

Now,

((3 2i))2 4(5 i) = 5 12i 20 + 4i = 15 8i,

so

x = 3 2i 15 8i2

To find 15 8i, solve z2 = 15 8i for z.


Example (continued)


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p ( )

Let z = a + bi and z2 = 15 8i. Then

(a2

b2) + 2abi =

15

8i,

so a2 b2 = 15 and 2ab = 8.Solving for a and b gives us z = 1 4i, 1 + 4i, i.e., z = (1 4i).Therefore,

x = 3 2i (1 4i)2

and32i+(14i)

2

= 46i2

= 2

3i,

32i(14i)2 =

2+2i2 = 1 + i.

Thus the roots of x2 (3 2i)x + (5 i) are 2 3i and 1 + i.Appendix A Page 36/1


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Problem

Find the roots of x2 3ix + (3 + i) = 0.

Solution. 1 + i , 1 + 2i .


Example


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p

Verify that u1 = (4 i) is a root of

x2

(2

3i)x

(10 + 6i)

and find the other root, u2.

Solution. First,

u21

(2

3i)u1

(10 + 6i) = (4

i)2

(2

3i)(4

i)

(10 + 6i)

= (15 8i) (5 14i) (10 + 6i)= 0,

so u1 = (4 i) is a root.

Recall that if u1 and u2 are the roots of the quadratic, then

u1 + u2 = (2 3i) and u1u2 = (10 + 6i).

Solve for u2 using either one of these equations.



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Example (continued)

Since u1 = 4 i and u1 + u2 = 2 3i,u2 = 2 3i u1 = 2 3i (4 i) = 2 2i.

Therefore, the other root is u2 = 2 2i.

You can easily verify your answer by computing u1u2:

u1u2 = (4 i)(2 2i) = 10 6i = (10 + 6i).


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