<s>[INST] 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.
 [/INST] Calculate! 
$(1+i)^0=1$, so $(1-i)^0=1$, so $(1+i)^0+(1-i)^0=2$.
$(1+i)^1=1+i$, so $(1-i)^1=1-i$, so $(1+i)^1+(1-i)^1=2$ .
$(1+i)^2=2i$, so $(1-i)^2=-2i$, so $(1+i)^2+(1-i)^2=0$.
$(1+i)^3=-2+2i$, so $(1-i)^3=-2-2i$,  so $(1+i)^3+(1-i)^3=-4$. 
$(1+i)^4=-4$, so $(1-i)^4=-4$, so $(1+i)^4+(1-i)^4=-8$.
Now the game starts all over again. The pattern of the first four entries continues forever, except that every time $n$ is incremented by $4$,  we multiply by $-4$, for the simple reason that $(1+i)^4=(1-i)^4=-4$. 
Let $n=4k+r$, where $r=0$, $1$, $2$, or $3$.
Then
$(1+i)^n+(1-i)^n=2(-4)^k=(-1)^k 2^{2k+1}$ if $\;r=0\;$  or $\;r=1$.
$(1+i)^n+(1-i)^n=0$ if $\;r=2$.
$(1+i)^n+(1-i)^n=(-4)^{k+1}=(-1)^{k+1}2^{2k+2}$ if $\;r=3$.
Comment: More briefly, since $(1+i)^4=(1-i)^4=-4$, we have
$$(1+i)^{4k+r}=(-4)^k (1+i)^r\qquad\text{and}\qquad (1-i)^{4k+r}=(-4)^k (1-i)^r,$$
and therefore
$$(1+i)^{4k+r}+(1-i)^{4k+r}=(-4)^k\left((1+i)^r+ (1-i)^r  \right).$$
Note that the "cases" expression for $(1+i)^n+(1-i)^n$ can be made into a single expression in various ways. 
</s>