So in polar form the product becomes(√
3 −j)( 1 −√
3 j)= 2 (
−π
6)
2 (
−π
3)= 4 (
−π
6−π
3)= 4 (
−π
2)
=− 4 jon conversion back to Cartesian form. You can now check directly that
(√
3 −j)( 1 −√
3 j)=√
3 −j− 3 j+√
3 j^2
=− 4 jYou may feel that this is a rather roundabout approach to a simple
multiplication – converting to polar form to multiply. However, in practice one already has
the polar form – for example when using phasors to represent currents in alternating current
theory. Also, the real power of conversion to polar form comes later, when considering
powers and roots.
Exercises on 12.4
- Evaluate all possible products, excluding powers, of the three complex numbers below
and compare with the results obtained ina+jbform (see Q2 Exercises on 12.3).
z 1 =^ (π/ 2 )z 2 = 3 (
3 π
4)
z 3 = 1 (
−π
3)- Show that ifz=r^ (θ),then
z^2 =r^2 ( 2 θ), z^3 =r^3 ( 3 θ)What do you think is the result forznwherenis a positive integer? You will see more
of this in Section 12.7.Answers
1.z 1 z 2 = 3
(
−3 π
4)
=−3
√
2−3
√
2jz 1 z 3 =^(π
6)
=√
3
2+1
2jz 2 z 3 = 3 (
5 π
12)
=3 (√
3 − 1 )
2√
2+3 (√
3 + 1 )
2√
2jz 1 z 2 z 3 = 3 (
11 π
12)
=−3 (√
3 + 1 )
2√
2+3 (√
3 − 1 )
2√
2j2.zn=rn^ (nθ)