Download Equal Complex Numbers

Lecture 1
Complex Numbers 1: Introduction
The aim of these notes is to expand our present number system so we can solve a larger
class of problems.
Motivating Problem
Consider the equation
x  y  10 and xy  25 .
The solution to this equation are easily found; it is x=5 and y=5. We now consider a very
similar equation
x  y  10 and xy  26
Unlike the first equation, this seems more difficult to solve. Since x  y  10 we must have
x  5  i and y  5  i where i is a number we will determine later. We put the conditions
x  5  i and y  5  i into xy  26 , this gives:
(5  i)(5  i)  26
Expanding now gives: 25  i 2  26 . Therefore, i 2  1 and hence i   1
This means if we want to solve the second equation we need to introduce a number i such
that i 2  1 .
This might seem rather strange, but it is similar to introducing negative numbers to solve
equations such as: x  5  3 , or fractions to solve an equation like 2x  1 . Of course, we
must ensure that our new number makes sense and doesn’t violate any of the previous rules
we have laid down for our number system!
(Note you might be tempted to think that we can just invent numbers to get ourselves out
of any sort of trouble like not being able to divide by zero. This certainly doesn’t work if we
want out invented numbers to be consistent with our existing number system. Suppose we
made the suggestion that 1/0=∞, where ∞ is our new symbol we have invented for use when
we divide by zero! This means 1=0×∞.
Now it is certainly true that 2×0=3×0. Multiplying both sides by ∞ to get (2×0) ×∞ = (3×0) ×∞.
We now have 2×(0 × ∞)=3×(0 × ∞). Using the fact that 1=0×∞ we get 2×1=3×1 and so 2=3!
Clearly, this is rubbish! So we are not free to add any number we want to our number
system.)
1
Lecture 1
Properties and uses of i
Consider the quadratic equation
x2  2x  2  0 .
(1)
Using the quadratic formula with a  1, b  2 and c  2
x

  2 
 22  4  1 2  2 
2 1
4
2
.

Since the discriminant b 2  4ac is negative  4 we cannot obtain roots to this equation
that are real numbers. The roots are can be simplified using our new number i. Returning
to equation (1)
x
2  4  1 2  4   1 2  2i


1 i .
2
2
2
The roots of equation 1 are therefore the complex numbers 1  i and 1  i .
We now check whether 1  i really does solve the original equation. We put 1  i into
equation (1)
(1  i) 2  2(1  i)  2
The expression (1+i)2 reminds us of expanding brackets, so we expand brackets in the usual
way.
(1  i) 2  2(1  i)  2  (1  2i  i 2 )  2  2i  2  0
since i 2  1 .
General Properties of Complex Numbers
We now explore the general properties of i. The number i is an example of a complex
number. The general form of a complex number is
z  x  yi
where x and y are real numbers. We call x the real part of z, written Re(z) and y is called
the imaginary part of z, written Im(z).
2
Lecture 1
Example 1
i)
ii)
iii)
Given z  3  5i , then Rez   3, Imz   5
If z  2i  0  2i , then Rez   0, Imz   2
If z  7  7  0i , then Rez   7, Imz   0 .
Equal Complex Numbers
Two complex numbers are defined x  yi and u  vi where x, y, u and v are real. They are
equal if and only if (iff) x  u and y  v. This might seem a rather trivial statement, but you
should compare equality of complex number with equal fractions.
Operations on Complex Numbers
For the complex numbers x  yi and u  vi we have the following basic operations
addition:
subtraction:
multiplication:
x  yi   u  vi  x  u    y  v i ,
x  yi   u  vi  x  u    y  v i ,
x  yi u  vi  xu  xvi  yui  yvi2
 xu  yv  xv  yu  i .
Example 2
i)
ii)
iii)
iv)
2  5i   3  i   2  3  5  1 i  5  6i
8  3i   6  4i   8  6  3  4 i  14  i
9  3i   6  2i   9  6   3  2 i  3  i
8  3i 6  2i   48  16i  18i  6i 2  54  2i .
So far we have not said how to divide complex numbers, before we can do this we need an
additional operation.
Complex Conjugate
Consider
2  3i 2  3i   4  6i  6i  9i 2  13 .
The product is a real number. 2  3i and 2  3i are complex conjugate numbers and 2  3i is
the conjugate of 2  3i .
Any pair of complex numbers of the form x  yi and x  yi are said to be conjugate, each is
the conjugate of the other. The conjugate of z  x  yi is z  x  yi .
3
Lecture 1
zz  x  yi x  yi   x 2  xyi  xyi  y 2i 2  x 2  y 2 .
Example 3
i)
ii)
8  i 8  i   82  12  65
 4  2i  4  2i    42  22  20 .
Division of a Complex Number
When dividing by a complex number we make the denominator real by multiplying by the
conjugate of the divisor.
Example 4
i)
ii)
iii)
iv)
1
, conjugate of 3  i is 3  i .
3i
1
1
3i
3i
3
i


 2 2   .
3  i 3  i 3  i 3  1 10 10
5i
5i
6  2i  10  30i
1 3i


 2
  .
2
6  2i 6  2i 6  2i
6 2
4 4
7  5i 7  5i 6  2i  42  14i  30i  10 4 11i
.


 
6  2i 6  2i 6  2i 
40
5 10
Square roots of complex numbers
We know how to find the square root of numbers such as 100, 4, 6 10 and so on. How do we
find the square root of a complex number? We make use of equality of complex numbers.
We find the square root of 3  4i . The square root of 3  4i will be a complex number, so we
write
3  4i  a  bi ,
where a and b are real and will be determined below.
Then
3  4i  a  bi   a 2  2abi  b 2i 2  a 2  b 2  2abi .
2
Equating real and imaginary parts (recall two complex numbers are equal if and only if their
real and imaginary parts are equal)
a 2  b2  3 (2), 2ab  4 (3) .
From (3)
4
Lecture 1
b
2
a
Substitute into (2)
2
a 
 3
 a 
4
a2  2  3
a
2
2
Multiply both sides by a 2
a 4  4  3a 2
a 4  3a 2  4  0
Factorising
a
2


 4 a2  1  0
a 2  4, a 2  1
a  2, a  i.
Discard a  i since a is real.
Substituting into (3)
a2
a  2
4b  1, b  1
 4b  1, b  1.
Hence the square root of 3  4i are the complex numbers  2  i  .
5
Lecture 1
Tutorial Questions
The tutorial questions consist of three sections: Warm-up questions, section A and section
B.



The warm-up questions are designed to allow you attempt some simple questions on
the lecture material and test your own learning. They also help prepare you for the
section A and B questions.
Section A questions show the level of learning required to answer the section A
questions on the test and exam.
Section B questions show the level of learning required to answer the section B
questions on the test and exam.
Note that not every set of tutorial questions will contain section A and B questions.
Warm-up Questions
Q1
Given that i   1 and i 2  1 simplify the following
a)
e)
Q2
i4
i 3
c)
g)
d)
h)
i5
i9
i7
i n , where n is a positive integer
x2  4x  5  0
x2  9  0
b)
d)
x2  2x  5  0
x4  1  0
Find the real and imaginary parts of the following complex numbers
a)
c)
e)
Q4
b)
f)
Solve the following equations
a)
c)
Q3
i3
i 1
z  2  6i
z  10
2  i 3  i 
b)
d)
f)
z  3  7i
z  16i
1  i 3
Simplify the following complex numbers
a)
c)
e)
g)
i)
k)
2  7i   5  3i 
3  5i 7  i 
3i7  4i 
2  i 2
i1  i 2  3i 
b)
d)
f)
2
1 i
l)
h)
j)
2  7i   5  3i 
3  3i   4  i 
3  4i 3  4i 
1  i 3
74  3i 
3i
4  3i
6
Lecture 1
m)
o)
4i
4i
3i
i
n)
1 i
1 i
Section A Questions
Q5
Solve the following equations for x and y
a)
b)
c)
Q6
x  yi  3  i 2  3i 
3  4i  x  yi 1  i 
x  yi
5i
2i
Find the square roots of the following complex numbers
a)
15  8i
b)
21 20i
c)
2i
Q7
For the complex numbers z  x  yi and w  u  vi solve the simultaneous equations
3z  2w  38 and 2z  iw  7
Q8
For z  x  yi solve the following
z 2  16  0
z 2  5iz  6  0
a)
b)
Q9
Let z=(3+4i)/(2-3i). Find the complex number w which satisfies the equation zw=1.
Q10
Find the values of a and b (with a>0) which satisfy (a+bi)2=5+12i.
7
Lecture 1
Solutions
Should you get stuck on a tutorial question then the scheduled weekly tutorials will offer
you support and advice with them. Note you are expected to have tried the questions
before the weekly tutorials.
1
a
e
h
i
b
1
c
i
d
i
f
i
g
i
If n 4 without remainder, i n  1
if n 4 gives remainder 1, i n  i
if n 4 gives remainder 2, i n  1
if n 4 gives remainder 3, i n  i
2
a
c
x  2i
x  3i
3
a
c
e
Rez   2, Imz   6
Rez   10, Imz   0
Rez   7, Imz   1
a
c
e
g
i
7  10i
26  32i
12  21i
3  4i
5i
b
d
f
h
j
k
1 i
l
m
4 16i

17 17
1 3i
n
i
b
 5  2i 
4
o
c
9  7i
7 i

2 2
11 3i
6
a
c
 4  i 
 1  i 
7
z
8
a
5
a
b
b
d
x  1 2i
x  1,  i
Rez   3, Imz   7
Rez   0, Imz   16
Re  z   2, Im  z   2
b
d
f
 3  4i
 1 4i
25
 2  2i
28  21i
9 13i

25 25
34 22i
44 33i

, w

5
5
5
5
z  4i
b
i
z  2i or z  3i
8
Lecture 1
The following are sample worked solutions to questions 9 and 10.
9
If zw=1 then w=1/z, so
2  3i (2  3i)(3  4i) 6  17i  12  6  17i  6  17
w





i
3  4i (3  4i)(3  4i)
25
25
25
25
10
(a  bi ) 2  5  12i  a 2  2abi  b 2  5  12i
Equating real and imaginary parts gives :
2ab  12  b 
6
a
36
 5  a 4  5a 2  36  0  (a 2  9)( a 2  4)  0
2
a
Since a is real and positive, a  3 and therefore b  2.
a2  b2  5  a2 
9