<s>[INST] Exponentials in complex numbers If $\displaystyle z-\frac1z=i$, then find $\displaystyle z^{2014}+\frac{1}{z^{2014}}$.
The answer should be in terms of $1, -1,\;i\;or\;-i$. I am not able to understand how to simplify the given expression so that its $2014^{th}$ power can be found out easily.
 [/INST] $(1)$ 
We have $\displaystyle z^2+\frac1{z^2}=1$
$\displaystyle\implies z^4-z^2+1=0$
Using $\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2),$
$z^6+1=(z^2+1)(z^4-z^2+1)=0\implies z^6=-1$
$(2A)$ We have $\displaystyle z^2=iz+1$
$\displaystyle \implies z^3=z\cdot z^2=z(iz+1)$
$\displaystyle=iz^2+z$
$\displaystyle=i(iz+1)+z=i+z(i^2+1)=i$
$(2B)$ As $\displaystyle z^2-iz-1=0, z^2-iz+i^2=0$
$\displaystyle\implies z^3+i^3=(z+i)(z^2-iz+i^2)=0\implies z^3=-i^3=i$
$\displaystyle\implies z^6=-1$
Now, $2014=335\cdot6+4$
Finally $\displaystyle z^4+\frac1{z^4}=\left(z^2+\frac1{z^2}\right)^2-2\cdot z^2\cdot\frac1{z^2}$
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