Download Some important definitions

Let z = x + iy, where i is the square root of -1. Then
 The real part of z, written Re(z), is x;
 The imaginary part of z, written Im(z) is y;
 The complex conjugate of z, written z*, is
z* = x – iy.
Calculating with Complex Numbers:
Example: Suppose w = 2 – 3i and z = 3 – 4i.
Addition: Add the real and imaginary parts
separately, so w + z = 5 – 7i.
Subtraction: Subtract the real and imaginary parts
from each other, so
w – z = (2 – 3i) – (3 – 4i) = -1 + i
Multiplication: You multiply complex numbers by
using the rules for expanding two brackets AND
remembering that i2 = -1. So
wz = (2 – 3i)(3 – 4i) = 6 – 8i – 9i + 12i2
= 6 – 17i – 12 = -6 – 17i
Division: To divide complex numbers, you multiply
through by the complex conjugate of the
denominator:
w 2  3i 2  3i 3  4i



z 3  4i 3  4i 3  4i
6  8i  9i  12(1) 18  i
=

9  12i  12i  16(1)
25
Loci
Situation 1: The set of complex numbers satisfying
|z| = k is represented by a circle, centre O, radius k:
Im(z)
k
Argand Diagrams
Complex numbers can be represented on Argand
Diagrams.
The modulus of a complex number z is the length of
the line from 0 to z. So, if z = x + iy, then
| z | x  y .
The argument of a complex number is defined as
the angle it makes with the positive real axis. The
argument of a complex number z is usually written in
the range –π < arg(z) < π.
2
2
Imaginary axis
Solving Quadratic Eqns: e.g. 2z  2z  5  0
(2)  (2)  4  2  5
2  36
=
4
4
2  36  1 2  6i

 0.5  1.5i
=
4
4
The solutions are 0.5 + 1.5i and 0.5 – 1.5i.
Note: Complex solutions of a quadratic equation are
always complex conjugates of one another.
2
So z =
Im(z)
k
a
-2
-1
0

-1
Re(z)
Im
w
|w|
1
O
Situation 3: The set of complex numbers satisfying
the equation | z  a || z  b | is the perpendicular
bisector of the line joining a to b:
2
1
Re(z)
k
Situation 2: The set of complex numbers satisfying
| z  a | k is a circle, centre a, radius k.
Example: Suppose w = 2 + i and z = 3 – 2i
2
3
4

|z|
a
Real
z
| w | 22  12  5 and | z | 32  ( 2) 2  13 .
The argument of w is θ, where tan   12 .
So arg(w) = 0.464.
The argument of z is negative (as z is below the
axis). Since tan   23 , then  = 0.588.
Therefore, arg(z) = -0.588.
b
Real axis
-2
2
O
required locus
Situation 4: The complex numbers satisfying the
equation arg(z – a) =  is a half-line starting at a
making angle  with the positive real axis.
a
