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Complex
Numbers

9
...

In school, we soon find that in order to perform certain subtractions, we must extend
our concept of number to the integers, which include the natural numbers
...
Next we have to extend our understanding to the
real numbers, which include all the rationals and also include irrational numbers
...
We define the number i to be a number such that i2 =

-

1 The system of

...
Note that the
real numbers are included as those complex numbers with b = 0
...
However, the complex numbers
have a consistent set of rules of arithmetic, and the extension to complex numbers is
justified by the fact that they allow us to solve important mathematical and physical
problems that we could not solve using only real numbers
...
The set of all complex numbers is denoted by C
...
number'!
...
\ = X\ +

Y\i and z
...
efmed b�

Z1 + Z2= (x1 + X2) + (yl + Y2)i

Multiplication of complex numbers z1= x1 + y1 i and z2= x2 + y2i is defined by
Z1Z2= (x1 + Y1i)(x2 + Y2i)
2
i
= X1X2 + X1Y2i + X2Y1i + Y1Y2
= (x1x2- Y1Y2) + (X1Y2 + X2Y1)i

EXAMPLE 1

Perform the following operations
(a) (2 + 3i) + (S- 4i)

Solution: (2 + 3i) + (S- 4i)= 2 +S + (3- 4)i= 7- i
(b) (2 +3i)- cs - 4i)

Solution: (2 + 3i)- (S- 4i)= 2- S + (3- (-4))i= -3 + 7i
(c) (3- 2i)(-2 +Si)

Solution: (3- 2i)(-2 +Si)= [3(-2)- (-2)(S)] + [3(S) + (-2)(-2)]i= 4 +19i

EXERCISE 1

Calculate the following:
(a) (1- 4i) + (2 + Si)
(b) (2 + 2i)i
(c) (1 - 3i)(2 + i)
(d) (3- 2i)(3 +2i)

Remarks
1
...

2
...
We say that
the

imaginary part of z is y (not yi),

and we write Im (z)= y
...


3
...
" If y = 0, z = x is "purely
real
...


4
...
In physics and engineering, it is common to use j in
place of i since the letter i is often used to denote electric current
...
It is sometimes convenient to write x + iy instead of x + yi
...


The Complex Conjugate and Division
We have not yet discussed division of complex numbers
...


Definition

The

complex conjugate of the complex number z=x+yi is x - yi and is denoted

Complex Conjugate

z = x - yi

EXAMPLE2

2+Si= 2 - Si
-3 - 2i = -3+2i
x = x,

Theorem l

for any x

E

IR

Properties of the Complex Conjugate

For complex numbers zi = x+iy and z2 with x, y

E

IR we have

(1) Zi = Zi
(2) z1 is purely real if and only if ZI= zi

(3) zi is purely imaginary if and only if ZI = -z
(4) zi+z2=Zl+z2
(S) ZiZ2 = ZiZ2
(6) t: = Zl1
(7) zi+ZI= 2 Re(zi) = 2x1
(8) zi - ZI = i2 Im(zi)= i2yi
(9) ziZI=�+YT

EXERCISE2

Prove properties (1), (2), and (4) in Theorem 1
...

The

quotient of two complex numbers can now be displayed as a complex num­

ber in standard form by multiplying both the numerator and the denominator by the
complex conjugate of the denominator and simplifying
...


2+Si

(6-20)+(8+ 15)i
14 23
...

(a)

1+
�
l-t

(b)

�
l+t

(c)

4 -i
1+Si

Roots of Polynomial Equations
Complex conjugates are not only used to determine quotients of complex numbers, but
occur naturally as roots of polynomials with real coefficients
...


where a; E JR for

1 ::; i ::;

n
...


Proof: Suppose that z is a root, so that
a11i' +
...


EXAMPLE4

Find the roots of p(x) = x3 + 1
...
Therefore, by the Factor Theorem, (x+ 1) is a factor of p(x)
...
Then addition and multipli­

cation appear as follows:
Z1 +Z2

=

z1 z2

=

(x1,Y1) + (x2,Y2) = (x1+ Xz,y1+Y2)
(x1 ,Y1)(x2 ,Y2) = (x 1x2 - Y1Y2,X1Y2+ X2Y1)

In terms of this ordered pair notation, it is natural to represent complex numbers as
points in the plane, with the real part of z being the x-coordinate and the imaginary
part being the y-coordinate
...
See Figure 9
...
1
...

y
imaginary
axis

• � =

(x
...
1
...


(ex,cy)
...

=

c (x,y)

=

However, in the complex plane, we can also multiply by complex scalars
...


EXERCISE4

Plot the following complex numbers in the complex plane
...
If lzl * 0, let () be the angle measured counterclockwise
from the positive x-axis such that
x

=

r cos ()

and

y = r sin()

The angle () is unique up to a multiple of
shown in Figure

9
...
2, a polar form of z is

27r, and it is called an argument of z
...
1
...


Determine the modulus, an argument, and a polar form of z1 =

Solution: We have
lzil=

2-2i and z2=

-1 +

Y3i
...
Hence, a polar form

For z2, we have

1zz1 =
Since

l-1+ Y3il = )c-1)2 cY3)2 = 2
+

-1 = 2 cos() and Y3 = 2 sin(), we get () = ¥ + 27rk, k

E

Z
...
An important consequence of the definition is

2
z
l l = x2 + l = zZ

2
...
We will always use radians
...
Notice that every complex number z has infinitely many arguments and hence
infinitely many polar forms
...
It is tempting but incorrect to write()

=

arctan(y/x)
...
Also note that
y/x is not defined if x

=

0
...


r(cos() + i sin B)
...


The polar form is particularly convenient for multiplication and division because
of the trigonometric identities
cos(B1+B2)

=

cos Bi cos B2 - sin B1 sin B2

sin(B1+B2)

=

sin ()1 cos B2 +cos Bi sin B2

It follows that
z1z2

=

r1(cos Bi+i sin B1 )r2(cos B2 +i sin B2)

=

rir2((cos Bi cos B2 - sin Bi sin B2)+i(cos Bi sin B2 +sin Bi cos B2))

=

rir2( cos(B1+B2)+i sin(B1+B2))

In words, the modulus of a product is the product of the moduli of the factors, while
an argument of a product is the sum of the arguments
...
0, we have

The proof is left for you to complete in Problem D4
...
0
...


Describe Theorem 3 in words and use it to prove Corollary 4
...


( ( �)
( G; )

(1 - i)(- �+i) = Y2 cos = 2 Y2 cos
;<;

2+2i
1 + -
...
732+i(2
...
366+i(0
...
i) and

7r
12

( ))

+ isin -

2 - 2i
-1 +

-
...

3i

Powers and the Complex Exponential
From the rule for products, we find that
z2 = r2(cos 2e + isin 28)
Then
z3 = z2z = r2r(cos (W + B) + isin B(W+B))
= r3(cos 3e+isin 3B)

Theorem 5

[de Moivre's Formula]
Let z = r (cos e+isin B) with r

-:F

0
...
To prove that the theorem holds
for positive integers, we proceed by induction
...
Then
=

/+1 = lz =Ir[cos (ke+ ())+isin (k ()+ ())]
= /+1[cos((k+ l)())+isin((k+ 1 )()) ]

Therefore, the result is true for all non-negative integers n
...


EXAMPLE 7

Calculate

(2 + 2i)3
...
71828)
...


Definition

ei!I

Euler's Formula

Definition

E

=

cose + isine

For any complex number z = x + iy, we define

Remarks
1


...


2
...
There, one often uses the fact that

< +
e a bi)t

=

eateibt

=

e at(cos bt

+

isin bt)

Observe that Euler's Formula allows us to write every complex number z in the
form
where r = lzl and e is any argument of z
...

(a) (2 +2i)3

)

(

3
Solution: (2 +2i)3 = 2 "2
...
)3ei<3rr/4l = -1 6 + 1 6i
(b) (2i)3

(

)

Solution: (2i)3 = 2ei1112 3 = 23e i<31112l = -8i
(c)

cY3 + i)5

(

)

5 = 25ei511l6 = -16
...
/3
(
+ i)5 = 2ei1116

EXERCISE 9

Use polar form to calculate (1 - i)5 and ( -1 -
...


n-th Roots
Using de Moivre's Formula for n-th powers is the key to finding n-th roots
...
That is, we
need a number

w

such that w'1 = z
...
Then

w

n

= z implies that

Then R is the realn-th root of the positive real number r
...

Solution: We have 8 = 8e;co+2rrk), k E Z
...


0
If k = 0, we have the root w0 = 2e = 2
...
/3i
...
/3i
...


EXAMPLE9
(continued)

By increasing k further, we simply repeat the roots we have already found
...
The number 8 has three third
roots, w0, w1, and w2
...


Theorem 6

Let

z

be a non-zero complex number
...
'n - 1

Wk = rlf11ei(6+2rrk)/n,

EXAMPLE 10

z

Find the fourth roots of -81
...
Thus, the fourth roots are
(81)1/4ei(JT+2JTk)/4,

k= 0, 1,2,3

In Examples 9 and 10, we took roots of numbers that were purely real: we were
really solving x' - a= 0, where

a E

R By our earlier theorem, when the coefficients

of the equations are real, the roots that are not real occur in complex conjugate pairs
...


EXAMPLE 11

Find the third roots of Si and illustrate in an Argand diagram
...


PROBLEMS 9
...

(a) (2+Si)+ (3+2i)
(b) (2-7i)+ (-S+3i)

AS Express the following quotients in standard form
...

(a) (1 + 3i)(3- 2i)
(b) (-2- 4i)(3-i)
(c) (1 - 6i)(-4+ i)
(d) (-1- i)(l- i)

A3 Determine the complex conjugates of the following
numbers
...

(d)
4 l
-

z
A6 Use polar form to determine z1z2 and i if
Z2
(a) z1=1+ i, z2= 1 + Y3i
(b) Z1=--f3- i, Z2= 1 - i
(c) Z1 = 1 +2i, Z2=-2- 3i
(d) z1=-3+ i,z2=6- i
A 7 Use polar form to determine the following
...
/3i)4

A4 Determine the real and imaginary parts of the fol­
lowing
...

6 -[

2- Si

(b) z= (2+Si)(l- 3i)
-1
(d) z= -
...

1
1
(b) (-16i) 14
(a) (-1) 15
1
(c) c-Y3 - i)113
(d) (1 +4i) 13

l

Homework Problems
Bl Determine the following sums or differences
...

(a) (2+ i)(S- 3i)

(b) (-3- 2i)(S- 2i)
(c) (3- Si)(-1+6i)
(d) (-3- i)(3-i)
B3 Determine the complex conjugates of the following
numbers
...

(a) 4-7i

(b) (3+2i)(2- 3i)
s
( c)
4- i
1- 2i

...

l
(a)
3+4i
2
(b)
3- Si
1- 4i
(c)
3+Si
1+4i
(d)
4- Si
--

B6 Use polar form to determine z1z2 and �if
Z2
(a) Z1= l - -f3i, Z2= -1 + i
(b) Z1=- -
...

(a)

(1

+

VJi)4

(b) ( -2 2i)3
(c) ( -Y3 - i)4
-

(d) (-2 + 2 Y3i)5

BS Use polar form to determine all the indicated roots
...


D2 If z= r(cos e +i sin 8), what islzl? W hat is an argu­
ment ofz?

D3 Use Euler's Formula to show that
(a) eie= e-ie

� (e;e+e-ie)
;
sine= � (e e - e-ie)

(b) cos8=
(c)

i

D4 Prove Theorem 3

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