Complex Numbers 5 Question 2
2. 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: 2. (c)
Solution:
$$ \text { Key Idea Use, } e^{i \theta}=\cos \theta+i \sin \theta $$
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, $\left(1+i z+z^{5}+i z^{8}\right)^{9}$
$=1+i e^{i \frac{\pi}{6}}+e^{i \frac{5 \pi}{6}}+i e^{i \frac{8 \pi}{6}}$
$=1+e^{i \frac{\pi}{2}} \cdot e^{i \frac{\pi}{6}}+e^{i \frac{5 \pi}{6}}+e^{i \frac{\pi}{2}} \cdot e^{i \frac{4 \pi}{3}}{ }^{9} \quad \because i=e^{i \frac{\pi}{2}}$
$=1+e^{i \frac{2 \pi}{3}}+e^{i \frac{5 \pi}{6}}+e^{i \frac{11 \pi}{6}}{ }^{9}$
$=1+\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}+\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}$ $+\cos \frac{11 \pi}{6}+i \sin \frac{11 \pi}{6}{ }^{9}$
$=1-\frac{1}{2}+\frac{i \sqrt{3}}{2}-\frac{\sqrt{3}}{2}+\frac{1}{2} i+\frac{\sqrt{3}}{2}-\frac{i}{2}^{9}$
$=\frac{1}{2}+\frac{\sqrt{3} i}{2}^{9}=\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}^{9}$
$=\cos 3 \pi+i \sin 3 \pi \quad[\because$ for any natural number ’ $n$ ’ $\left.(\cos \theta+i \sin \theta)^{n}=\cos (n \theta)+i \sin (n \theta)\right]$
$=-1$
