Complex Numbers 4 Question 1
1. Let $z=\frac{\sqrt{3}}{2}+\frac{i}{2}^{5}+\frac{\sqrt{3}}{2}-\frac{i}{2}^{5}$. If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$, then
(2019 Main, 10 Jan II)
(a) $R(z)>0$ and $I(z)>0$
(b) $I(z)=0$
(c) $R(z)<0$ and $I(z)>0$
(d) $R(z)=-3$
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Answer:
Correct Answer: 1. (b)
Solution:
- Given, $z=\frac{\sqrt{3}}{2}+\frac{i}{2}^{5}+\frac{\sqrt{3}}{2}-\frac{i}{2}^{5}$
$\because$ Euler’s form of
$$ \begin{aligned} & \quad \begin{array}{l} \frac{\sqrt{3}}{2}+\frac{i}{2}=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}=e^{i(\pi / 6)} \\ \text { and } \frac{\sqrt{3}}{2}-\frac{i}{2}=\cos \frac{-\pi}{6}+i \sin -\frac{\pi}{6}=e^{-i \pi / 6} \\ \text { So, } z=\left(e^{i \pi / 6}\right)^{5}+\left(e^{-i \pi / 6}\right)^{5}=e^{i \frac{5 \pi}{6}}+e^{-i \frac{5 \pi}{6}} \\ =\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}+\cos \frac{5 \pi}{6}-i \sin \frac{5 \pi}{6} \\ \\ \qquad=2 \cos \frac{5 \pi}{6} \end{array} \quad\left[\because e^{i \theta}=\cos \theta+i \sin \theta\right] \end{aligned} $$
$\therefore \quad I(z)=0$ and $R(z)=-2 \cos \frac{\pi}{6}=-\sqrt{3}<0$
$$ \because \cos \frac{5 \pi}{6}=\cos \pi-\frac{\pi}{6}=-\cos \frac{\pi}{6} $$
