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$$ Show\ that,\ (1+i)^{\mathrm{n}}+(1-i)^{\mathrm{n}}=2^{\frac{\mathrm{n}+2}{2}} \cos \frac{\mathrm{n} \pi}{4} \text {. } $$
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written 2.1 years ago by
prajapatijaimin
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• modified 2.1 years ago |
Solution:
Let, $1+i=r(\cos \theta+i \sin \theta)=r \cos \theta+i \sin \theta$ Equating real \& imaginary parts on both sides,
$$ r \cos \theta=1 \quad \& \quad r \sin \theta=1\\ $$
Now,
$ (r \cos \theta)^{2}+(r \sin \theta)^{2}=(1)^{2}+(1)^{2}\\ $
Now,
$ (\operatorname{rcos} \theta)^{2}+(\operatorname{rsin} \theta)^{2}=(1)^{2}+(1)^{2} \\ $
$ \Rightarrow \mathrm{r}^{2} \cos ^{2} \theta+\mathrm{r}^{2} …
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