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

9.9 De Moivre’s theorem 347
y
2
5p —4
p –4
x
2
Figure9.19
Solution ofz 2 = 4j.
by De Moivre’s theorem. Furthermore, 4j has modulus 4 and argument π/2 + 2nπ,
n ∈ Z, thatis
4j = 4{cos(π/2 +2nπ) +jsin(π/2 +2nπ)}
n ∈ Z
Therefore,
r 2 (cos2θ +jsin2θ) = 4{cos(π/2 +2nπ) +jsin(π/2 +2nπ)}
Comparing both sides of this equation, we see that
and
r 2 =4 andso r=2
2θ=π/2+2nπ andso θ=π/4+nπ
Whenn = 0wefind θ = π/4,andwhenn = 1wefind θ = 5π/4.Usinglargervaluesof
n simply repeats solutions already obtained. These solutions are shown in Figure 9.19.
We note that in this example the solutions are equally spaced at intervals of 2π/2 = π.
InCartesian form,
z 1
= 2 √
2
+j 2 √
2
= √ 2(1+j) and z 2
=− 2 √
2
−j 2 √
2
= − √ 2(1 +j)
Ingeneral, thenrootsofz n =a +jbare equallyspacedatangles2π/n.
Once the technique for solving equations like those in Examples 9.18 and 9.19 has
beenmastered,engineersfinditsimplertoworkwiththeabbreviatedformr̸ θ.Example
9.19 reworked inthisfashion becomes
Letz=r̸ θ, then z 2 =r 2̸ 2θ
Furthermore, 4j = 4̸ π/2 +2nπ, and hence ifz 2 = 4j, we have
( ) π
2 +2nπ
r 2̸
fromwhich
2θ=4̸
r 2 =4 and 2θ= π 2 +2nπ
as before.Rework Example 9.18 foryourself usingthisapproach.