Q.If $n$ is an integer then show that $(1+i)^{2n}+(1-i)^{2n}=2^{n+1}\cos\frac{n\pi}{2}$.

Write $1+i$ and $1-i$ in polar form, apply De Moivre's theorem, and add the two conjugate results.

$1+i = \sqrt2\,\text{cis}\frac\pi4, \qquad 1-i=\sqrt2\,\text{cis}\left(-\frac\pi4\right).$

By De Moivre's theorem,

$(1+i)^{2n} = (\sqrt2)^{2n}\,\text{cis}\!\left(2n\cdot\frac\pi4\right) = 2^n\,\text{cis}\frac{n\pi}2,$

$(1-i)^{2n} = (\sqrt2)^{2n}\,\text{cis}\!\left(-2n\cdot\frac\pi4\right) = 2^n\,\text{cis}\!\left(-\frac{n\pi}2\right).$

Adding, and using $\text{cis}\,\theta+\text{cis}(-\theta)=2\cos\theta$:

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