Question

If $n$ is an integer, show that $(1+i{)}^n+(1-i{)}^n=2^{\frac{n+2}{2}}\cos \left(\frac{n\pi }{4}\right)$

Answer 1

$1+i=\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$

$=\sqrt{2}\left(e^{i\frac{\pi }{4}}\right)$ ...$\left(i\right)$........$cos\left(\pi /4\right)=sin\left(\pi /4\right)=\frac{1}{\sqrt{2}}$

Similarly

$1-i=\sqrt{2}(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})$

$=\sqrt{2}\left(e^{-i\frac{\pi }{4}}\right)$ ...$\left(ii\right)$

Hence

${(1+i)}^n+{(1-i)}^n$

$={\left(\sqrt{2}\left(e^{i\frac{\pi }{4}}\right)\right)}^n+{\left(\sqrt{2}\left(e^{i\frac{-\pi }{4}}\right)\right)}^n$

$=2^{\frac{n}{2}}[\left(e^{i\frac{n\pi }{4}}\right)+\left(e^{i\frac{-n\pi }{4}}\right)]$

$=2^{\frac{n}{2}}[\cos \left(\frac{n\pi }{4}\right)+i\sin \left(\frac{n\pi }{4}\right)+(\cos \frac{n\pi }{4}-i\sin \frac{n\pi }{4})]$

$=2^{\frac{n}{2}}\left[2\cos \left(\frac{n\pi }{4}\right)\right]$

$=2^{\frac{n}{2}+1}\cos \left(\frac{n\pi }{4}\right)$

$=2^{\frac{n+2}{2}}\cos \left(\frac{n\pi }{4}\right)$