6. Zadatak

Računamo

$$f(n+2016)=\left(\frac{1+i}{\sqrt2}\right)^{n+2016}+\left(\frac{1-i}{\sqrt2}\right)^{n+2016}$$ $$=\left(\frac{1+i}{\sqrt2}\right)^{2016}\left(\frac{1+i}{\sqrt2}\right)^n+\left(\frac{1-i}{\sqrt2}\right)^{2016}\left(\frac{1-i}{\sqrt2}\right)^n.$$

Kako je

$$\left(\frac{1+i}{\sqrt2}\right)^{2016}=\left[\left(\frac{1+i}{\sqrt2}\right)^2\right]^{1008}=\left(\frac{2i}{2}\right)^{1008}=i^{1008}=1,$$

i slično $\left(\frac{1-i}{\sqrt2}\right)^{2016}=1$, slijedi

$$f(n+2016)=\left(\frac{1+i}{\sqrt2}\right)^n+\left(\frac{1-i}{\sqrt2}\right)^n=f(n).$$

Analogno računamo

$$f(n-2016)=\left(\frac{1+i}{\sqrt2}\right)^{n-2016}+\left(\frac{1-i}{\sqrt2}\right)^{n-2016}$$ $$=\left(\frac{1+i}{\sqrt2}\right)^{-2016}\left(\frac{1+i}{\sqrt2}\right)^n+\left(\frac{1-i}{\sqrt2}\right)^{-2016}\left(\frac{1-i}{\sqrt2}\right)^n$$ $$=\left(\left(\frac{2i}{2}\right)^{1008}\right)^{-1}\left(\frac{1+i}{\sqrt2}\right)^n+\left(\left(\frac{-2i}{2}\right)^{1008}\right)^{-1}\left(\frac{1-i}{\sqrt2}\right)^n$$ $$=\left(\frac{1+i}{\sqrt2}\right)^n+\left(\frac{1-i}{\sqrt2}\right)^n=f(n).$$

Zaključujemo

$$f(n+2016)-f(n-2016)=f(n)-f(n)=0.$$