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MATH 211 Winter 2013
Lecture Notes(Adapted by permission of K. Seyffarth)
Appendix A
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Appendix A Complex Numbers
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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.
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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
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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
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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
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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
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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.
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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.
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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.
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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.
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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 .
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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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