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Q)

Prove that $\left ( 1+\mathit{i} \right )^{\mathit{n}} + \left ( 1-\mathit{i} \right )^{\mathit{n}} = 2^{\large\frac{n+2}{2}} \cos \; \large\frac{n\pi}{4}$

m, n N.

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

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