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

which isclearly true,and the theorem holds. Whenn = 2,wefind
(cosθ +jsinθ) 2 = (cosθ +jsinθ)(cosθ +jsinθ)
=cos 2 θ+jsinθcosθ+jcosθsinθ+j 2 sin 2 θ
=cos 2 θ −sin 2 θ +j(2sinθcosθ)
9.9 De Moivre’s theorem 345
Recallingthetrigonometricidentitiescos2θ = cos 2 θ − sin 2 θ andsin2θ = 2sin θ cos θ,
wecan write the previous expression as
cos2θ +jsin2θ
Therefore,
(cosθ +jsinθ) 2 =cos2θ +jsin2θ
and DeMoivre’s theorem has been verified whenn = 2.
The theorem also holds whennis a rational number, that isn = p/q where pandqare
integers.Thus wehave
(cosθ +jsinθ) p/q =cos p q θ+jsinp q θ
Inthisform itcan be used toobtain roots of complex numbers. For example,
√
3
cosθ +jsinθ =(cosθ +jsinθ) 1/3 =cos 1 3 θ+jsin1 3 θ
Insuchacasetheexpressionobtainedisonlyoneofthepossibleroots.Additionalroots
can be found as illustratedinExample 9.18.
De Moivre’s theorem is particularly important for the solution of certain types of
equation.
Example9.18 Findallcomplex numberszwhich satisfy
z 3 =1 (9.9)
Solution The solution of this equation is equivalent to finding the solutions ofz = 1 1/3 ; that is,
findingthecuberootsof1.Sinceweareallowingztobecomplex,thatisz ∈ C,wecan
write
z=r(cosθ+jsinθ)
Then,usingDeMoivre’s theorem,
z 3 =r 3 (cosθ +jsinθ) 3
=r 3 (cos3θ +jsin3θ)
Wenextconverttheexpressiononther.h.s.ofEquation(9.9)intopolarform.Figure9.17
shows the number1 = 1 +0jon anArgand diagram.
From the Argand diagram we see that its modulus is 1 and its argument is 0, or possibly
±2π, ±4π,..., thatis2nπ wheren ∈ Z.Consequently, wecan write
1 =1(cos2nπ +jsin2nπ) n ∈ Z