<s>[INST] Simplifying $(1+i)^n-(1-i)^n$ I'm trying to simplify this expression because there is a simpler way of writing this. But the furthest I've got is $$(1+i)^n-(1-i)^n=\sum_{k=0}^{n} \binom nk i^{k}-\sum_{k=0}^{n} \binom nk (-1)^{k}i^{k}.$$
 [/INST] In the case of you being not used to complex numbers, here is an answer with basic algebra.
Let $1+i = \alpha$, $1-i = \beta$, then $\alpha-\beta = 2i$, $\alpha\beta=2$, $\alpha/\beta = i$ and $\beta^2 = -2i$.
\begin{align*}
z_n:= \alpha^n - \beta^n &= (\alpha- \beta)(\alpha^{n-1} + \alpha^{n-2}\beta + \dots + \alpha \beta^{n-2} + \beta^{n-1}) \\
& = 2i \beta^{n-1}\left(1 + (\alpha/\beta) + \dots + (\alpha/\beta)^{n-1}\right) = 2i \beta^{n-1}(1 + i + \dots + i^{n-1})
\end{align*}
We separate cases of $n$ mod 4 since we have $1+i+i^2 + \dots + i^{n-1}$ term, which is periodic of period 4.
(1) $n$ is multiple of $4$: $1+i+i^2 + \dots + i^{n-1} =0$, so $z_n=0$.
(2) $n = 4k+1$ for some integer $k$: $1+i+i^2 + \dots + i^{n-1} = 1$, so $z_n =  2i \beta^{4k} = 2i (-2i)^{2k} = (-1)^k\cdot 2^{2k+1}i$.
(3) $n= 4k+2$ for some integer $k$: $1+i+i^2 + \dots + i^{n-1} = 1+ i = \alpha$, so $z_n = 2i \beta^{4k+1}\cdot \alpha  = 4i \beta^{4k} = (-1)^k\cdot 4^{k+1}i $.
(4) $n= 4k+3$ for some integer $k$: $1+i+i^2 + \dots + i^{n-1} = 1+ i -1= i$, so $z_n = 2i \beta^{4k+2}\cdot i  =
-2 (-2i)^{2k+1} =  (-1)^k \cdot 4^{k+1}i
$
So
$$ z_n = \begin{cases} 0 & n = 4k \\ 
(-1)^k \cdot 2^{2k+1} i & n = 4k+1 \\ 
(-1)^k \cdot 4^{k+1}i & n = 4k+2 \text{ or } 4k+3
\end{cases}$$
</s>