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I have tried to look when does :$\cos²(z)+\sin²(\bar{z})=1$ with $z$ is a complex variable and $\bar{z}$ is it's conjugate ,and I have got that $z=i$ is the best example for that where :$\cos²(i)+\sin²(-i)=1$ , really my question about the nature of the identity :$ \cos^n(i)+\sin^n(-i)$ for $n>2$ , now my question here is :

Question: Could be this : $ \cos^n(i)+\sin^n(-i)$ integer for $n>2$?

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    You could try proving it directly by the identity $\cos(z) = \frac{e^{iz}+e^{-iz}}{2}$ and $\sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$. – 2017-02-25

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Clearly $n$ should be even, because $\sin(-i)=-\sin i$ is purely imaginary.

If $n$ is a multiple of $4$, we have $$ \cos^ni+\sin^ni= \frac{(1+e^2)^n}{2^ne^n}+\frac{(1-e^2)^n}{2^ne^n} $$ If this is an integer $N$, then we get $$ (1+e^2)^n+(1-e^2)^n=2^ne^nN $$ which is impossible, because we'd obtain a polynomial of degree $2n$ with integer coefficients having $e$ as root.

So the only possibility is to have $n\equiv 2\pmod{4}$, when the equality would be $$ \frac{(1+e^2)^n}{2^ne^n}-\frac{(1-e^2)^n}{2^ne^n}=N $$ With the binomial theorem, $$ \sum_{k=0}^n\binom{n}{k}e^{2k}- \sum_{k=0}^n\binom{n}{k}(-1)^ke^{2k}=2^nNe^n $$ so $$ 2\sum_{\substack{0\le k\le n\\k\text{ odd}}}\binom{n}{k}e^{2k}=2^nNe^n $$ If $n>2$, this would again give a polynomial with integer coefficients having $e$ as root.

So the only case is indeed $n=2$.