CN1.2: POLAR FORM OF A
COMPLEX NUMBER
Rectangular and Polar Form
When a complex number is expressed in the form
z = x + yi
it is said to be in rectangular form.
But a point P with Cartesian coordinates (x,y) can also be represented by the polar coordinates (r, θ ) where r is
the distance of the point P from the origin and
θ
is the angle that

OP makes with the positive x-axis
y
P (x,y) or (r, θ )
r
x
x
NB: x = rcos θ
and
y = rsin θ
x2 + y 2
and x2 + y2 = r2 or r =
To express a complex number z in polar form:
z = x + yi
= rcos θ
= r (cos θ
+ rsin θ i
+ sin θ i )
which we abbreviate to z = rcis
θ
So, the polar form of the complex number z is
z = r cis
where r =
CN1.2 – Complex Numbers: Polar Form
θ
 y
x 2 + y 2 and θ = tan-1  
x
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June 2012
Modulus of z
The modulus of z,
z is the distance of the point z from the origin.
z = x + yi =
mod z =
x2 + y 2 = r
The argument of z, arg z, is the angle measured from the positive direction of the
x-axis to

OP
If arg z =
θ
θ
then sin
=
y
z
and cos
θ
=
x
z
and tan
θ
=
y
x
An infinite number of arguments of z exist
eg If z = i then arg z =
π
2
+ 2π n, n ∈ Z .
argument and Argument of z
We define the Argument of z:
Arg z =
θ , where −π ≤ θ ≤ π
Examples
1. Express in polar form z = 1 – i
[NB: z is in the 4th quadrant]
x = 1, y = -1
r=
z =
x2 + y 2 = 1 + 1 =
y −1
= -1
=
x 1
−π
θ = tan-1(-1) =
4
∴z =
rcisθ
 −π 
= 2 cis 

 4 
tan
θ
2
x
●
=
2. Express 2 cis
2 cis
y
z
[since z is in the 4th quadrant]
 4π 

 in the form x + yi
 3 
  4π
 4π 

 = 2 cos 
 3 
  3
= 2
 1
× − 
 2
= -1 -

 4π  
 + sin 
 i

 3  

3
+ 2 × −
 2  i


3i
See Exercise 1
CN1.2 – Complex Numbers: Polar Form
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June 2012
Operations on Complex Numbers in Polar Form
Addition and Subtraction
Complex numbers in polar form are best converted to the form x + yi before addition or subtraction
Multiplication and Division
If
z1 = r1cisθ1 and z2 = r2 cisθ 2 then it can be shown using trigonometric identities that
z1 r1
z1 z2 = r1r2 cis (θ1 + θ 2 ) and
=
cis (θ1 − θ 2 )
z2 r2
Examples:
1. If z1 = 2cis
5π
find z1 z2 in polar form, −π ≤ θ ≤ π
6
4
π 
5π 
z1 z2 = 2cis ×  −3cis

4 
6 
 π 5π 
= -6cis  +

4 6 
 13π 
= -6cis 

 12 
 −11π 
since −π ≤ θ ≤ π
= -6cis 

 12 
π
and z2 = -3cis
2. If u = 1 + 3i and v = 2 – i find
u
in polar form with −π ≤ θ ≤ π
v
Two approaches are possible:
u = 1 + 3i
ie. x = 1, y = 3
1 + 3 = 10
3
θ = tan-1   = 1.25R
1
∴ u = 10 cis1.25,
r=
v= 2–i
2
ie x = 2, y = -1
22 + ( −1) =
2
r=
θ
2
5
 −1 
R
 = -0.46
 2 
= tan-1 
∴v=
u
Then
=
v
5 cis0.46R
R
10cis1.25
5cis 0.46 R
10
cis(1.25 + 0.46)R
5
=
2 cis 1.71R
=
OR
1 + 3i
u
=
2−i
v
1 + 3i 2 + i
=
×
2−i 2+i
−1 + 7i
=
4 +1
−1 + 7i
=
5
1 7
= − + i
5 5
1
7
∴ x= − , y=
5
5
x 2 + y 2 = (−0.2) 2 + 1.42 =
1.4
tan θ =
= -7, θ = 1.71R
−0.2
u
∴ = 2 cis(1.71)R
v
r=
2
See Exercise 2
CN1.2 – Complex Numbers: Polar Form
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June 2012
Exercises
Exercise 1
1. Find the polar form (in radians) of the following complex numbers:
(b) z = - 3 + i
(d) z = -2 – 4i
(a) z = -1 + i
(c) z = - 3i
2. Express each of the following complex numbers in rectangular form
π
(a) 3cis
(c) 8cis
7cisπ
(b)
4
π
(d) 10cis0.41R
2
3. If z = 2 + i and w = 1 – 4i find each of the following in polar form using radians where appropriate:
(a)
(b)
z
(e) Arg(zw)
(c) Arg z
w
(d)
w
(f) zw
Exercise 2
1. Simplify
(a)
4cis
π
× 3cis
3
3cis
π
(b)
4
5π
6
12cis
π
6
2. If u = 6cis
3π
u
 π
and v = 4cis  −  express
in polar form
4
v
 4
3. If z = 1-
3 i, find z and express both z and z in polar form using radians.
Answers
Exercise 1
1. (a)
2cis
3π
(b)
2cis
5π
4
(c)
3cis
6
−π
20cis ( −2.03)
(d)
R
2
3
2. (a)
3
+
2
i
(b) -
7
2
(c) 8i
(d) 9.2 + 4i
3. (a)
5
(d)
17
(b)
(c) 0.46R
17
(e) -0.86R
(f) 9.22cis(-0.86)R
Exercise 2
1. (a) 12cis
7π
(b)
12
2. (a)
3
1
4
cis
2π
3
cisπ
2
3.
 π

 3
z = 2cis  −
π
z = 2cis  
3
CN1.2 – Complex Numbers: Polar Form
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June 2012