Polar versus rectangular form.
A complex number written as a + bi is in rectangular form. There is
also the polar form.
Represent z = a + bi on the Argand plane as an arrow.
Then z is completely determined by knowing the angle θ it makes with
the positive x-axis, and its modulus or length.
Suppose the modulus of z = a + bi is r. Then by definition,
cos θ = a/r, and sin θ = b/r.
So a = r cos θ, b = r sin θ. Hence
z = a + bi
= r cos θ + r sin θ · i
= r(cos θ + i · sin θ).
Write cis θ = cos θ + i · sin θ.
We call z = rcis θ the polar form of z.
Here r = |z|, and θ is the argument of z, arg(z).
Note that |cis θ| = 1.
Example. Suppose z =
√
2cis π4 .
Express z in rectangular form z = a + bi for a, b real.
√
π
z =
2cis
4
√
π
π
2(cos + i · sin )
=
4
4
√
1
1
=
2( √ + √ i)
2
2
= 1 + i.
Example. Suppose |z| = 2 and arg(z) = 5π
6.
So z = 2cis 5π
6 . Give z in rectangular form.
Now π radians is 180 degrees.
5π · 180 = 150 degrees, or 180 − 30 degrees.
So 5π
radians
is
6
6
π.
Working in radians, 5π
=
π
−
6
6
Hence
π
π
cos(π + (− )) = − cos(− )
6
6
π
= − cos
6
√
3
= −
,
2
and likewise,
π
π
sin(π + (− )) = − sin(− )
6
6
π
= sin
6
1
=
.
2
√
5π
So cis 6 = − 23 + 1
2 i.
Hence
5π
6
√
1
3
= 2(−
+ i)
2
2
√
= − 3 + i.
z = 2cis
Example. Convert z = 1 + i to polar form z = r · cis θ.
Now r = |z| =
q
12 + 1 2 =
√
1+1=
√
2,
so write z as
z=
√
1
1
2( √ + √ i).
2
2
We must have
cis θ = cos θ + i · sin θ
1
1
= √ + √ i,
2
2
so
1
1
cos θ = √ , sin θ = √ .
2
2
So θ = π/4 (or 45 degrees).
So
z=
√
2cis
π
.
4
√
1
Example. Convert z = − 2 + 23 i to polar form.
Again, |z| =
q
1 + 3 = 1, so
4
4
√
1
cis θ = − 2 + 23 i, and so
√
3
1
cos θ = − , sin θ =
.
2
2
Now draw the situation! Must be in the second quadrant.
Let δ be the angle made by z as measured from the negative x-axis.
√
1 and sin δ = 3 .
Then cos δ = 2
2
So must have δ = π/3 (or 60 degrees).
So θ = π − π/3 = 2π/3 radians, (or 180 − 60 = 120 degrees).
So
2π
2π
z = 1 · cis
= cis
.
3
3
What happens when we multiply together two complex numbers in
polar form?
Let z1 = r1cis θ1, z2 = r2cis θ2. Then
z1z2 = (r1cis θ1)(r2cis θ2)
= r1r2cis θ1cis θ2.
On the worksheet, you are asked to prove that
cis θ1 · cis θ2 = cis(θ1 + θ2),
with the help of some trig identities. Hence
z1z2 = r1r2cis(θ1 + θ2).
This re-proves the fact proved earlier, that
|z1z2| = r1r2 = |z1| · |z2|.
We also have shown the following new fact:
arg(z1z2) = arg(z1) + arg(z2).
Example. Compute the product of 2 cis π/6 with 3 cis(−3π/2).
(2 cis π/6)(3 cis(−3π/2)) = 6 cis (π/6 − 3π/2)
π − 9π
= 6 cis (
)
6
−4π
).
= 6 cis (
3
Note that instead of cis θ, the notation eiθ is often used.
The above formula involving cis then has a more intuitive form:
eiθ1 eiθ2 = ei(θ1+θ2).
We do not pursue this further now.
Powers of complex numbers.
We can now prove (by induction!) De Moivre’s formula:
(r cis θ)n = rncis(nθ).
Example. Find (−1 +
√
3i)8, in rectangular form.
A key useful point here is that in general
cis(θ + 2π) = cos(θ + 2π) + i sin(θ + 2π)
= cos θ + i sin θ
= cis θ.
Let z = −1 +
Then |z| =
√
√
3i.
1+3=
√
4 = 2, so
√
3
1
i),
z = 2(− +
2
2
√
3,
so cos θ = − 1
,
sin
θ
=
2
2
so in second quadrant. (Do a drawing.)
We see that θ = π − π/3 = 2π/3.
So z = 2cis 2π
3 . Hence
(−1 +
√
2π 8
8
3i) = (2cis
)
3
2π
) by De Moivre
3
16π
8
2 cis
3
4π
28cis(4π +
)
3
4π
8
2 cis
3
π
28cis(π + )
3
= 28cis 8(
=
=
=
=
π
π
8
= 2 (cos(π + ) + i · sin(π + ))
3
π
π
− i · sin )
3
3
√
1
3
= 28(− −
i)
2
2
√
7
= 2 (−1 − 3i)
= 28(− cos
√
= −128 − 128 3i.
3