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

Polar form plus De Moivre gives $2^{n+1}\cos\dfrac{n\pi}{2}$.

$1+i=\sqrt2\left(\cos\tfrac{\pi}{4}+i\sin\tfrac{\pi}{4}\right)$ and $1-i=\sqrt2\left(\cos\tfrac{\pi}{4}-i\sin\tfrac{\pi}{4}\right).$

By De Moivre's theorem,

$(1+i)^{2n}=(\sqrt2)^{2n}\left(\cos\tfrac{2n\pi}{4}+i\sin\tfrac{2n\pi}{4}\right)=2^{n}\left(\cos\tfrac{n\pi}{2}+i\sin\tfrac{n\pi}{2}\right),$

$(1-i)^{2n}=2^{n}\left(\cos\tfrac{n\pi}{2}-i\sin\tfrac{n\pi}{2}\right).$

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