Is there a formula for $(1+i)^n+(1-i)^n$? I'm wondering if there is a formula for the value of $(1+i)^n+(1-i)^n$?
I calculated the first terms starting with $n=1$ to be, in order, $2$, $0$, $-4$, $-8$, $-8$, $0$, $16$, $\dots$
So it seems to be some sequence of positive and negative powers of $2$ with $0$s thrown in. Is there a more explicit formulation of what $(1+i)^n+(1-i)^n$ is, based on $n$? 
With the binomial theorem, I get it equal to
$$
\sum_{k=0}^n\binom{n}{k}i^{n-k}(1+(-1)^{n-k}).
$$
Can this be made nicer?
Thanks.

This is not a particularly elegant solution, but an alternative route is to simply note that $(1 \pm i)^2 = \pm 2i$.  I will show a lot of steps, but this method involves only very easy calculations.  If $n$ is even, then
$$
(1\pm i)^n = \left((1+i)^2\right)^{ \frac{n}{2}}= (\pm 2i)^{n/2} = \left\{ \begin{array}{ccl}
2^{n/2} & & n = 8k
\\
\pm i \cdot 2^{n/2}& &  n = 8k +2
\\
i \cdot 2^{n/2} & & n = 8k + 4
\\
\mp i 2^{n/2} & & n = 8k + 6
\end{array} \right.
$$
Hence,
$$(1+i)^n + (1-i)^n = 
\left\{
\begin{array}{ccl}
2^{\frac{n}{2}+1} & & n = 8k
\\
0 & & n = 8k+2 \text{ or } n = 8k+6
\\
i \cdot 2^{\frac{n}{2}+1} & & n = 8k + 4
\end{array}
\right.
$$
This is all you need since if $n$ is odd, then $n-1$ is even.  For example, when $n = 8k + 1$, we have
$$\begin{align}
(1+i)^n + (1 - i)^n &= (1+i)^{n-1}(1+i) + (1 -i)^{n-1}(1-i)
\\
&= 2^{\frac{n-1}{2}}(1+i) + 2^{\frac{n-1}{2}} (1-i)
\\
&= 2^{\frac{n-1}{2}} + i 2^{\frac{n-1}{2}} + 2 ^{\frac{n-1}{2}} - i 2^{\frac{n-1}{2}}
\\
&= 2\cdot 2^{\frac{n-1}{2}}
\\
&= 2^{\frac{n+1}{2}}.
\end{align}$$
The cases $n = 8k+3$, $n=8k+5$, and $n = 8k + 7$ follow similarly.
