Browse Questions

# Find all the value of the following : $(-\sqrt 3-i)^{\large\frac{2}{3}}$

This is the third part of the multi-part question Q1

Toolbox:
• From De moivre's theorem we have
• (i) $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta,n\in Q$
• (ii) $(\cos\theta+i\sin\theta)^{-n}=\cos n\theta-i\sin n\theta$
• (iii) $(\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta$
• (iv) $(\sin \theta+i\cos \theta)^n=[\cos(\large\frac{\pi}{2}$$-\theta)+i\sin(\large\frac{\pi}{2}$$-\theta)]^n=\cos n(\large\frac{\pi}{2}$$-\theta)+i\sin n(\large\frac{\pi}{2}$$-\theta)$
• $e^{i\theta}=\cos\theta+i\sin\theta$
• $e^{-i\theta}=\cos\theta-i\sin\theta$,also written as $\cos\theta$ and $\cos(-\theta)$
Step 1:
Let $-\sqrt 3-i=r(\cos\theta+i\sin\theta)$
Equating the real and imaginary parts separately
$r\cos\theta=-\sqrt 3$
$r\sin\theta=-1$
$r^2=4$
$r=2$
$\alpha=\tan^{-1}\bigg|\large\frac{-1}{\sqrt 3}\bigg|=\large\frac{\pi}{6}$
Step 2:
Now $-\sqrt{3}-i$ is represented by a point in quadrant 3.
Therefore $\theta=-\pi+\alpha$
We know $\alpha=\large\frac{\pi}{6}$
$\theta=-\pi+\large\frac{\pi}{6}=\large\frac{-5\pi}{6}$
Step 3:
$(-\sqrt 3-i)^{\large\frac{2}{3}}=[2(\cos(\large\frac{-5\pi}{6})+$$i\sin(\large\frac{-5\pi}{6})]^{\large\frac{2}{3}} \qquad\qquad\;\;\;=2^{\large\frac{2}{3}}[\cos(2k\pi-\large\frac{5\pi}{6})$$+i\sin(2k\pi-\large\frac{5\pi}{6})]^{\large\frac{2}{3}}$$\;\;\;k\in z \qquad\qquad\;\;\;=2^{\large\frac{2}{3}}[\cos(\large\frac{\pi}{6}$$(12k-5))$$+i\sin(\large\frac{\pi}{6}$$(12k-5))]$$\;\;\;k\in z \qquad\qquad\;\;\;=2^{\large\frac{2}{3}}[\cos(\large\frac{2\pi}{18}$$(12k-5))$$+i\sin(\large\frac{2\pi}{18}$$(12k-5))]$$\;\;\;k=0,1,2 \qquad\qquad\;\;\;=2^{\large\frac{2}{3}}[\cos(\large\frac{\pi}{9}$$(12k-5))$$+i\sin(\large\frac{\pi}{9}$$(12k-5))]$$\;\;\;k=0,1,2 Step 4: The roots are 2^{\large\frac{2}{3}}[\cos(\large\frac{-5\pi}{9})+$$i\sin(\large\frac{-5\pi}{9})$$,2^{\large\frac{2}{3}}[\cos(\large\frac{7\pi}{9})$$+i\sin(\large\frac{7\pi}{9}),$$2^{\large\frac{2}{3}}$$[\cos(\large\frac{19\pi}{9})+$$i\sin(\large\frac{19\pi}{9}). \Rightarrow 2^{\large\frac{2}{3}}[\cos(\large\frac{19\pi}{9}-$$2\pi)+i\sin(\large\frac{19\pi}{9}-$$2\pi)] \Rightarrow 2^{\large\frac{2}{3}}[\cos(\large\frac{\pi}{9})$$+i\sin(\large\frac{\pi}{9})]$