Function $f(n)=({1+i\over\sqrt2})^n+({1-i\over\sqrt2})^n$ problem

Question

If $n\in\mathbb N$, $$f(n)=\bigg({1+i\over\sqrt2}\bigg)^n+\bigg({1-i\over\sqrt2}\bigg)^n$$ where $i^2=-1$, how much is $f(2017)+f(2013)?$

What I did here is, I expressed:

$(1+i)^2=1+2i-i^2=1+2i-1=2i\\(1-i)^2=1-2i+i^2=1-2i-1=-2i$

So:

$f(2017)=\bigg({1+i\over\sqrt2}\bigg)^n+\bigg({1-i\over\sqrt2}\bigg)^n={(2^{1008}i^{1008})(1+i)\over\sqrt2^{2017}}+{(-2^{1008}i^{1008})(1-i)\over\sqrt2^{2017}}=...$

From here I continued solving this until for both $f(2017)$ and $f(2013)$ I got

$f(2017)={2^{1009}i\over\sqrt2^{2017}}\\f(2013)={-2^{1007}\over\sqrt{2}^{2013}}$

I thought this method would result in much "prettier" answers, but I can't do anything with these 2 to sum them up, so I'm sure that this method is not the required one for solving this.

Answer 1

Hint: $\dfrac{1\pm i}{\sqrt{2}}$ are primitive 8-th root of unity.

Answer 2

Take $\sqrt2^{2013}=2^{1006}\sqrt2$ and $\sqrt2^{2017}=2^{1008}\sqrt2$. Then subtract exponents and use $\frac{2}{\sqrt2}=\sqrt2$ to get $f(2017)=i\sqrt2$ and $f(2013)=-\sqrt2$.

Answer 3

So, by using De Moivre's formula I've gotten the answer to this:

$f(n)=({1+i\over\sqrt2})^n+({1-i\over\sqrt2})^n=\\({1\over\sqrt2}+{i\over\sqrt2})^n+\big({1\over\sqrt2}+(-{i\over\sqrt2})\big)^n=\\({\sqrt2\over2}+{i\sqrt2\over2})^n+({\sqrt2\over2}+{-i\sqrt2\over2})^n=\\(\cos{\pi\over4}+i\sin{\pi\over4})^n+(\cos{\pi\over4}+i\sin\big(-{\pi\over4})\big)^n=\\\cos n{\pi\over4}+i\sin n{\pi\over4}+\cos n{\pi\over4}+i\sin (-n{\pi\over4})=\\2\cos n{\pi\over4}$

So:

$f(n)=2\cos n{\pi\over4}\\f(2017)=2\cos {2017\pi\over4}=2\cos(504\pi+{\pi\over4})=2\cos{\pi\over4}=\sqrt2\\f(2013)=2\cos {2013\pi\over4}=2\cos(503\pi+{\pi\over4})=2\cos{5\pi\over4}=-\sqrt2\\f(2017)+f(2013)=\sqrt2-\sqrt2=0$