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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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1 Answer

  • 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
Squaring and adding we get
$\Rightarrow r=\sqrt 2$
Step 2:
Now $\alpha=\tan^{-1}1=\large\frac{\pi}{4}$
Since $1+i$ corresponds to a point in quadrant 1.
Therefore $1+i=\sqrt 2(\cos\large\frac{\pi}{4}$$+i\sin\large\frac{\pi}{4})$
Step 3:
$\qquad\;\;\quad=2^{2n}(\cos n\pi+i\sin n\pi)$
Now $\sin n\pi=0$ for $n\in N$
Therefore $(1+i)^{4n}$ is purely real.
Step 4:
Now $\cos(n\pi+\large\frac{\pi}{2})$$=0$ for $n\in N$
Therefore $(1+i)^{4n+2}$ is purely imaginary.
answered Jun 11, 2013 by sreemathi.v

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