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.
 A: $(1)$ 
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 $\displaystyle z^4+\frac1{z^4}=\left(z^2+\frac1{z^2}\right)^2-2\cdot z^2\cdot\frac1{z^2}$
A: Inspired by Lab-bhattacharjee:
We have $z^2-iz-1 = 0$,  multiplying by $z+i$ gives $z^3-i = 0$, or $z^3 = i$.
$2014 = 3 \cdot (4 \cdot 167+3)+1$, and so $z^{3 \cdot (4 \cdot 167+3)} = i^{4 \cdot 167+3} = -i$.
Hence $z^{2014}+ {1 \over 2^{2014}} = z^{2013+1}+ {1 \over 2^{2013+1}} = -iz+i{1\over z}= (-i)(z - {1 \over z}) = (-i) i = 1$.
A: Another approach:
Try to solve the equation: $z-\frac{1}{z}=i \Rightarrow z^2-iz-1=0$ we have:
$z_1 =\frac{\sqrt{3}}{2}+\frac{i}{2}=cos(\frac{\pi}{6})+isin(\frac{\pi}{6})$
$z_2 =\frac{-\sqrt{3}}{2}+\frac{i}{2}=cos(\frac{5\pi}{6})+isin(\frac{5\pi}{6})$
Applying the De' Moivre formula we have:
$z_1^{2014}+\frac{1}{z_1^{2014}}=(cos(\frac{\pi}{6})+isin(\frac{\pi}{6}))^{2014}+(cos(\frac{\pi}{6})-isin(\frac{\pi}{6}))^{2014}=2cos(\frac{2014\pi}{6})=1$
Similarly, we have: $z_2^{2014}+\frac{1}{z_2^{2014}}=2cos(\frac{2014.5\pi}{6})=1$
Consequently, $z^{2014}+\frac{1}{z^{2014}}=1$
