Complex Numbers Ques 102
If $z=\frac{\sqrt{3}}{2}+\frac{i}{2}(i=\sqrt{-1})$, then $\left(1+i z+z^{5}+i z^{8}\right)^{9}$ is equal to
(2019 Main, 8 April II)
(a) $1$
(b) $(-1+2 i)^{9}$
(c) -$1$
(d) $0$
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Answer:
Correct Answer: 102.(c)
Solution:
Formula:
- Key Idea: Use, $e^{i \theta}=\cos \theta+i \sin \theta $
Given, $z=\frac{\sqrt{3}}{2}+(\frac{1}{2}) i=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}=e^{i \frac{\pi}{6}}$
so, $\left(1+i z+z^{5}+i z^{8}\right)^{9}$
$=(1+i e^{i \frac{\pi}{6}}+e^{i \frac{5 \pi}{6}}+i e^{i \frac{8 \pi}{6}})^9$
$=(1+e^{i \frac{\pi}{2}} \cdot e^{i \frac{\pi}{6}}+e^{i \frac{5 \pi}{6}}+e^{i \frac{\pi}{2}} \cdot e^{i \frac{4 \pi}{3}}){ }^{9} \quad [\because i=e^{i \frac{\pi}{2}}]$
$=(1+e^{i \frac{2 \pi}{3}}+e^{i \frac{5 \pi}{6}}+e^{i \frac{11 \pi}{6}}){ }^{9}$
$=[1+ (\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3})+(\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6})$ $+(\cos \frac{11 \pi}{6}+i \sin \frac{11 \pi}{6})]{ }^{9}$
$=(1-\frac{1}{2}+\frac{i \sqrt{3}}{2}-\frac{\sqrt{3}}{2}+\frac{1}{2} i+\frac{\sqrt{3}}{2}-\frac{i}{2})^{9}$
$=(\frac{1}{2}+\frac{\sqrt{3} i}{2})^{9}=(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3})^{9}$
$=\cos 3 \pi+i \sin 3 \pi \quad[\because$ for any natural number ’ $n$ ’ $\left.(\cos \theta+i \sin \theta)^{n}=\cos (n \theta)+i \sin (n \theta)\right]$
$=-1$
BETA
