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# Complex Numbers Introduction to Complex Numbers and some Complex Number Laws.

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University of Waterloo

Mathematics

MATH 135

Wentang Kuo

Fall

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MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010
WENTANG KUO
Chapter 8 Complex Numbers
We recall the Quadradic formula: if a;b;c 2 R where a 6= 0, then the solution of ax +bx+c = 0
is p
▯b ▯ b ▯ 4ac
x = :
2a
The solutions are if b ▯ 4ac ▯ 0. What if b ▯ 4ac < 0?
8.2 Complex numbers
De▯nition
2
Let the symbol i be a square root of ▯1, i.e.,i = ▯1.
A complex number z is a number of the form z = x + iy where x;y 2 R (Cartesian form).
We call x the real part of z, denoted by Re(z) and y the imaginary part of z, denoted it by Im(z).
Note that x 2 R if and only if y = 0.
The set of all complex numbers is denoted by C.
The addition and multiplication on C are de▯ned as follows: let z = x + iy and w = u + iv. Then
z + w = (x + iy) + (u + iv) = (x + u) + i(y + v)
and
z ▯ w = (x + iy) ▯ (u + iv)
2
= xu + iyu + xiv + i yv
= (xu ▯ yv) + i(yu + xv):
Example Let z = 2 + i and w = 4 ▯ i. Find Im(z + w) and Re(z w).
We have
z + w = (2 + 4) + i(1 + (▯1)) = 6 + 0 ▯ i = 6:
Thus, Im(z + w) = 0. Also,
z w = (2 + i) (4 ▯ i) = (2 + 4i + i )(4 ▯ i) = (3 + 4i)(4 ▯ i) = 16 + 13i:
Thus, Re(z w) = 16.
p
Example Find z 2 C such that z = 1 + 2 2i.
Write z = x + iy with x;y 2 R. Then
2 2 2 2 2 2 2 p
z = (x + iy) = x + 2xyi + i y = (x ▯ y ) + 2xyi = 1 + 2 2i:
p p p
Thus, we have x ▯ y = 1 and 2xy = 2 2, i.e., xy = 2. Write y = x. We have
p
▯ 2▯2 2
x ▯ y = x ▯2 = x ▯ = 1:
x x2
Date: November 15, 2010.
1 2 WENTANG KUO
Multiplying by x , we get
x ▯ 2 = x i.e.x ▯ x ▯ 2 = 0i.e.(x ▯ 2)(x + 1) = 0:
p p p
Hence, x = 2 or ▯1. Since x 2 R, x ▯ 0. Hence, x = 2, i.e2=▯ 2 = ▯1. Also, y =
p p
Hence, z 2 + i or z = ▯ 2 ▯ i.
8.3 The complex plane & 8.4 Properties of complex numbers
The complex plane
A way to visualize C and its operations.
Idea
Let x-axis be the real axis and y-axis the imaginary axis. We associate z = x + iy to the point
(x;y). imaginary
y sz = x + iy = (x;y)
▯▯1
▯ ▯▯
▯ ▯▯ -
x real
Example Let z = x + iy and w = u + iv. Then z + w = (x + u) + i(y + v).
6
y + v ▯ z + w
▯ ▯▯>▯
▯▯ ▯ ▯ ▯
v ws ▯ ▯ ▯
▯ ▯▯ ▯
▯ ▯ ▯
y ▯ ▯ sz
▯ ▯ ▯▯1
▯ ▯▯ ▯▯
▯ ▯▯ -
u x x + u
Addition can be visualized as the parallelogram law for adding vectors.
Given z = x + iy 2 C, we have
(x + iy) ▯ (x ▯ iy) = x + y 2 R:
(Thus, by multiplying (x + iy) by (x ▯ iy), we get back a real number)
De▯nition
The complex conjugate of z = x + iy is
6 ▯ = x ▯ iy:
y sz = x + iy
▯▯1
▯ ▯▯
▯▯
P P -
PP P x
P PP
▯y qs▯ = x ▯ iy MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010 3
Conjugation is just the re
ection with respect to the real axis.
De▯nition
The modulus of z = x + iy is p
jzj = x + y :
Note that the distance from (0;0) to (x;y) is
p
x + y = jzj:
6
p
r = x + y 21sz = x + iy
▯ ▯ ▯ y
▯▯ ▯
▯ ▯ -
x
Thus, jzj is the distance from 0 to z.
Remark (1) We have
z ▯ = jzj :
(2) Note that if y = 0, i.e., z = x 2 R, then
p
jzj = x = jxj;
i.e., for real numbers, modulus is just the absolute value.
Example Find the inverse of z = x + iy 2 C, where x and y are not both 0.
We have
▯1 1 1 1 x ▯ iy x ▯ iy ▯
z = = = ▯ = 2 2 = 2:
z x + iy x + iy x ▯ iy x + y jzj
Example Find z;w 2 C such that
z + w = 7 | (1)
2iz + 7w = 9 | (2):
From (1), we have z = 7 ▯ w. Put it into (2). We get
2i(7 ▯ w) + 7w = 9; i.e.,(7 ▯ 2i)w = 9 ▯ 14i:
Thus,
9 ▯ 14i 9 ▯ 14i 7 + 2i 63 ▯ 98i + 18i ▯ 28i (63 + 28) ▯ 80i 91 ▯ 80i
w = 7 ▯ 2i= 7 ▯ 2i▯ 7 + 2i 7 + 22 = 7 + 2 2 = 53 :
Also,
91 ▯ 80i 371 ▯ (91 ▯ 80i) 280 + 80i
z = 7 ▯ w = 7 ▯ = = :
53 53 53
Properties of Conjugation 8.42
If z;w 2 C, we have
(1) z + w =▯+ w▯.
(2) z ▯ w ▯▯ ▯.
▯
(3)▯ = z.
(4) z ▯ = jzj . 4 WENTANG KUO
(5) z ▯ = 2Re(z).
(6) z ▯ = 2iIm(z).
Proof. (1) and (2) are easy and we have shown (4). Let z = x▯ = x ▯ iy. z
(3) We have
▯ = x ▯ iy = x + iy = z:
(5) We have
z + ▯ = (x + iy) + (x ▯ iy) = 2x = 2Re(z):
(6) We have
z ▯▯ = (x + iy) ▯ (x ▯ iy) = 2iy = 2iIm(z):
Properties of Modulus 8.44
(1) jzj = 0 if and only if z = 0.
(2)▯j = jzj.
(3) jzwj = jzjjwj.
(4) jz + wj ▯ jzj +(triangle Inequality).
Proof. Let z = x + iy.
(1)
p
jzj = 0 () x + y = 0 () x + y = 0 () x = 0 = y; i.e., z = 0:
(2)
p p
j▯j = jx ▯ iyj(x + (▯y) = x + y = jzj:
(3) Since jzwj and jzjjwj are non-negative real number, it su▯ces to prove jzwj = (jzjjwj) . We
have
jzwj = (zw)(zw)(by Proposition 8.42(4))
= (zw)▯▯) (by Proposition 8.42(2))
= (▯)(w▯)
2 2
= jzj jwj(by Proposition 8.42(4))
2
= (jzjjwj) :
It follows that jzwj = jzjjwj.
(4)
6
z + w
▯ ▯▯
▯ ▯▯ ▯ ▯
w▯ ▯ ▯ ▯ ▯
▯ ▯
▯▯jz+w▯ ▯ jwj
▯ ▯ ▯ ▯z
▯ ▯ ▯ ▯1
▯ ▯ ▯▯ ▯
▯▯ jzj
▯ - MATH 135 CLASSNOTE FOR MAKEUP CLASS ON NOVEMBER 16, 2010 5
We have the sum of two sides of a triangle ▯ the third side.

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