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# Prove that $\left ( 1+\mathit{i} \right )^{4n}$ and $\left ( 1+\mathit{i} \right )^{4n+2}$ are real and purely imaginary respectively.

m, n N This is the fourth 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 the real imaginary parts separately
$r\cos\theta=1$
$r\sin\theta=1$
$r^2=2$
$\Rightarrow r=\sqrt 2$
Now $\alpha=\tan^{-1}1=\large\frac{\pi}{4}$
$\theta=\alpha=\large\frac{\pi}{4}$
Since $1+i$ corresponds to a point in quadrant 1.
Therefore $1+i=\sqrt 2(\cos\large\frac{\pi}{4}$$+i\sin\large\frac{\pi}{4}) Step 3: (1+i)^{4n}=2^{\large\frac{4n}{2}}$$(\cos\large\frac{4n\pi}{4}$$+i\sin\large\frac{4n\pi}{4}) \qquad\;\;\quad=2^{2n}(\cos n\pi+i\sin n\pi) Now \sin n\pi=0 for n\in N Therefore (1+i)^{4n} is purely real. Step 4: (1+i)^{4n+2}=2^{\large\frac{4n+2}{2}}$$(\cos\large\frac{(4n+2)\pi}{4}$$+i\sin\large\frac{(4n+2)\pi}{4}) \qquad\qquad=2^{2n+1}(\cos(n\pi+\large\frac{\pi}{2})$$+i\sin(n\pi+\large\frac{\pi}{2}))$
Now $\cos(n\pi+\large\frac{\pi}{2})$$=0$ for $n\in N$
Therefore $(1+i)^{4n+2}$ is purely imaginary.