082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

346 Chapter 9 Complex numbers
z 2
1
y
y
1 x
2p —3
4p —3
1 z 1
1
z 3
x
Figure9.17
The complex numberz = 1 +0j.
Figure9.18
Solutionsofz 3 = 1.
Using the polar form, Equation (9.9) becomes
r 3 (cos3θ +jsin3θ) = 1(cos2nπ +jsin2nπ)
Comparing both sides of thisequation we see that
and
r 3 =1 thatis r=1sincer∈R
3θ = 2nπ thatis θ = 2nπ/3 n ∈ Z
Apparently θ can take infinitely many values, but, as we shall see, the corresponding
complex numbers aresimplyrepetitions. Whenn = 0,we find θ = 0,so that
z=z 1
=1(cos0+jsin0)=1
isthe first solution.Whenn = 1 wefind θ = 2π/3,so that
(
z=z 2
=1 cos 2π )
3 +jsin2π = − 1 √
3
3 2 +j 2
isthe second solution.Whenn = 2 we find θ = 4π/3,so that
(
z=z 3
=1 cos 4π )
3 +jsin4π = − 1 √
3
3 2 −j 2
is the third solution. If we continue searching for solutions using larger values ofnwe
find that we only repeat solutions already obtained. It is often useful to plot solutions
on an Argand diagram and this is easily done directly from the polar form, as shown in
Figure 9.18. We note thatthe solutions are equally spaced atangles of 2π/3.
Example9.19 Findthe complex numberszwhichsatisfyz 2 = 4j.
Solution Sincez ∈ Cwewritez =r(cos θ +jsin θ). Therefore,
z 2 =r 2 (cosθ +jsinθ) 2
=r 2 (cos2θ +jsin2θ)