Browse Questions

# Prove that $\left ( 1+ \mathit{i\sqrt{3}} \right )^{\mathit{n}} + \left ( 1- \mathit{i\sqrt{3}} \right )^{\mathit{n}} = 2^{n+1} \cos \; \large\frac{n\pi}{3}$

m, n N.

This is the second part of the multi-part question Q4.

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 $1+\sqrt 3i=r(\cos \theta+i\sin \theta)$
Equating the real and imaginary parts separately
$r\cos\theta=1$
$r\sin\theta=\sqrt 3$
$r^2=4$
$r=2$
Step 2:
Also $\alpha=\tan^{-1}\sqrt 3=\large\frac{\pi}{3}$
Since $1+\sqrt 3i$ is represented by a point in quadrant 1.
$\theta=\alpha=\large\frac{\pi}{3}$
Therefore $1+\sqrt 3i=2(\cos \large\frac{\pi}{3}$$+i\sin\large\frac{\pi}{3}) 1-\sqrt 3i=1+\sqrt 3i \qquad\;\;\;\;\;=2(\cos \large\frac{\pi}{3}$$-i\sin\large\frac{\pi}{3})$
Step 3:
Now LHS=$(1+i\sqrt 3)^n+(1-i\sqrt 3)^n$
$\qquad\quad\;\;=[2(\cos\large\frac{\pi}{3}$$+i\sin\large\frac{\pi}{3})]^n+$$[2(\cos\large\frac{\pi}{3}$$-i\sin\large\frac{\pi}{3})]^n \qquad\quad\;\;=2^n[\cos\large\frac{n\pi}{3}$$+i\sin\large\frac{n\pi}{3}$$+\cos\large\frac{n\pi}{3}$$-i\sin\large\frac{n\pi}{3}$
$\qquad\quad\;\;=2^n.2\cos\large\frac{n\pi}{3}$
$\qquad\quad\;\;=2^{n+1}\cos\large\frac{n\pi}{3}$
$\qquad\quad\;\;$=RHS
Hence proved.