Section 1: Complex Numbers
Examples and Problems
Example 1
[basic operations]
Perform each of the indicated operations:
(3  2 j )  (7  j )
(i)
(8  6 j )  (7  2 j )
(ii)
(2  3 j )( 4  2 j )
(iii)
(3  2 j ) /( 1  j )
(iv)
Solution
(3  2 j )  (7  j )  3  7  2 j  j  4  j
(i)
(8  6 j )  (7  2 j )  8  7  6 j  2 j  15  8 j
(ii)
(iii)
(2  3 j )(4  2 j )  2(4  2 j )  3 j (4  2 j )  8  4 j  12 j  6 j 2  8  4 j  12 j  6  14  8 j
(iv)
3  2 j 3  2 j 1 j  3  3 j  2 j  2 j 2  5  j
5 1




  j
2
1 j 1 j 1 j
2
2 2
1 j
Example 2
[modulus]
Let z1  2  j and z 2  3  2 j .
(i)
Evaluate 3z1  4 z 2
2
(ii)
(iii)
2 z 2  z1  5  i
Evaluate
2 z1  z 2  3  i
Show that the following relation holds for these complex numbers:
z1 z 2  z1 z 2
Solution
(i)
3z1  4 z 2  6  3 j  12  8 j   6  11 j  (6) 2  (11) 2  157
(ii)
2 z 2  z1  5  j
2 z1  z 2  3  j
2
64j 2 j 5 j

4 2j 3 2 j 3 j
or we could have used the fact that
Section 1 Examples & Problems - 1
2
34j

43j
2
3 4 j 43j

43j 43j
2
 25 j

25
za
za
, which usually makes thing easier.

zb
zb
2
1
(iii)
z1 z 2  2  j 3  2 j  2 2  12 3 2  (2) 2  5 13  65
z1 z 2  (2  j )(3  2 j )  8  j  8 2  (1) 2  65
Example 3
[conjugates]
Let z1  2  j and z 2  3  2 j . Show that z1  z 2  z1  z 2 . Graph z1 , z 2 , their conjugates and
z1  z 2 in the complex plane.
Solution
z1  z 2  (2  j )  (3  2 j )  5  j  5  j
z1  z 2  (2  j )  (3  2 j )  (2  j )  (3  2 j )  5  j
y
z2
z1  z 2
z1
z1
x
z1  z 2
z2
Example 4
[polar form and principal arguments]
Express the following in polar form and determine the principal value of the argument, Arg z
(and sketch in the complex plane)
22 3j
(i)
3j
(ii)
Express the following in the form x  jy
10(cos 0.4  j sin 0.4)
(iii)
Section 1 Examples & Problems - 2
Solution
(i)
The modulus is r  2  2 3 j  4  12  4
The argument is   tan 1
2 3

 60  rad
2
3
 2  2 3 j  r (cos   j sin  )
(ii)
 4(cos  / 3  j sin  / 3)
The principal value of the argument lies in the range       , so this argument,
 / 3 is the principal value, Arg z.
The modulus is r   3 j  9  3
3
3
 270 
rad
0
2
  3 j  r (cos   j sin  )
 3(cos 3 / 2  j sin 3 / 2)
The principal value of the argument lies in the range       , so
Arg z   / 2
The argument is   tan 1
y
4
270
60
x
3
(iii)
We have
10(cos 0.4  j sin 0.4)  r (cos   j sin 0.4)
z  x 2  y 2  10  x  100  y 2
tan 1

y
y
 0.4   0.4228
x
x
y
100  y 2
 0.4228  y  3.894  x  9.211
 z  9.211  3.894 j
Section 1 Examples & Problems - 3
Example 5
[polar form of the product, quotient and powers]
Let z1  2  j and z 2  3  2 j . Rewrite using the polar form and, using the trigonometric
identities
cos A  B   cos A cos B  sin A sin B
,
sin( A  B)  sin A cos B  cos A sin B
show that arg( z1 z 2 )  arg( z1 )  arg( z 2 ) (draw a sketch).
Solution
We have
z1 : r  2 2  12  5 , arg( z1 )  tan 1 (1 / 2)  0.4636
z 2 : r  32  (2) 2  13, arg( z 2 )  tan 1 (2 / 3)  0.5880
so that
z1  5 cos 0.4636  j sin 0.4636
z 2  13 cos (0.5880)  j sin (0.5880) 
Now
z1 z 2  5 13 cos 0.4636  j sin 0.4636cos (0.5880)  j sin (0.5880) 
 65cos 0.4636 cos (0.5880)  sin 0.4636 sin (0.5880)
 j cos 0.4636 sin (0.5880)  sin 0.4636 cos (0.5880)
The identities now give us
z1 z 2  65cos (0.4636  0.5880) j sin( 0.4636  0.5880)
The argument of this new complex number is evidently equal to the sum of the individual
arguments.
y
z1
x
z2
z1 z 2
From the graph, we can see that the sum of the two arguments ( 26.6  33.7 ) gives the argument
of the product (  7.1 ). Note also that the modulus of the new complex number is the product of
the moduli of the two complex numbers.
Section 1 Examples & Problems - 4
Problem 1
Evaluate the following (in the form x  j y )
(2  5 j )  (4  2 j )
(i)
[ ANSWER: 6  3 j ]
(2  j )  (2  j )
(ii)
[ ANSWER:  2 j ]
(1  5 j )( 4  3 j )
(iii)
[ ANSWER: 19  17 j ]
13 6
j]
(2  j ) /( 4  5 j )
(iv)
[ANSWER:  
41 41
(v)
 1 3
 
 2 2


j 

4
[ANSWER:
(vi)
(vii)
3 j 30  j 19
2 j 1
(1  j ) 8
(do not use the polar form)
7
6j]
4
[ANSWER: 1  j ]
[ANSWER: 16]
Problem 2
[modulus]
(i)
Evaluate  0.2 j
(ii)
Evaluate cos  j sin 
(iii)
Evaluate
(1  j ) 6
j 3 (1  4 j ) 2
[ANSWER: 0.2 ]
[ANSWER:1]
[ANSWER: 0.471]
Problem 3
[principal values]
Determine the principal value of (i)  2  2 j and (ii)   . Sketch both cases in the complex
plane.
[ANSWER: (i)  3 / 4 , (ii)  ]
Problem 4
[conjugates - this problem is good practice at working with abstract notation rather than with
numbers]
In the following, let z  x  j y , z1  x1  j y1 , etc.:
z
(i)
Evaluate Im  
z
(ii)
Show that the expression z   z implies that z is pure imaginary
(iii) Show that z1  z 2  z1  z 2
Section 1 Examples & Problems - 5
(iv)
(v)
z  z
Show that  1   1
 z2  z2
1
Show that Im z 
(z  z)
2j
Problem 5
[polar form]
Express the following in polar form (sketch in the complex plane)
2
(i)
[ANSWER: z  2(cos   j sin  ) ]
22j
(ii)
[ANSWER: z  8 / 41cos0.1107   j sin 0.1107  ]
54j
2j
(iii)
[ANSWER: z  2(cos(  / 2)  j sin(  / 2)) ]
Problem 6
[phasors]
Sketch a graph of the sinusoid X (t )  cos(t   ) , where 0     / 2 (time along the horizontal
axis). Also sketch the corresponding phasor on the unit circle in the complex plane,
corresponding to t  (5  4 ) / 4 .
Note: In the notes, slides 20-21, the imaginary part of the phasor represents the actual signal. The
imaginary part is the “height” of the phasor (in the y direction). Here, the real part of the phasor
will be the actual signal cos(t   ) , and the real part is measured along the real axis, so in this
case it is best to draw the complex plane above the signal, as indicated here:
y
x
t
Problem 7
[polar form of the product, quotient and powers]
By first putting the complex numbers into polar form, evaluate (a) j  (1  j ) , (b) (1  j )  j (in
polar form) and sketch the original and final complex numbers in the complex plane.
[ANSWER: (a) 2 (cos(3 / 4)  j sin( 3 / 4)) , (b) 2 cos5 / 4  j sin 5 / 4 ]
Section 1 Examples & Problems - 6