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 $$(1+i)^n-(1-i)^n=\sum_{k=0}^{n} \binom nk i^{k}-\sum_{k=0}^{n} \binom nk (-1)^{k}i^{k}.$$
 A: 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
$$ z_n = \begin{cases} 0 & n = 4k \\ 
(-1)^k \cdot 2^{2k+1} i & n = 4k+1 \\ 
(-1)^k \cdot 4^{k+1}i & n = 4k+2 \text{ or } 4k+3
\end{cases}$$
A: Hint: Let $z = a + bi$ for real numbers $a,b$. Note that

*

*$(\bar z)^n = \overline{z^n}$, where $\overline{a + bi} = a- bi$,

*$z - \bar z = 2 b$.

From there, you should use either Demoivre's or Euler's formula, depending on what you are familiar with.
A: Personally I find polar coordinates make powers easier,  so $1+i= {\sqrt 2} e^{i\frac \pi 4}$ and $1-i= {\sqrt 2} e^{-i\frac \pi 4}$,  so raising both to the $n$ gets us to $$( {\sqrt 2})^n(e^{ni\frac \pi 4}-e^{-n{i\frac \pi 4}})=$$
$$( {\sqrt 2} )^n (\cos(\frac {n\pi}4)+i\sin (\frac {n\pi}4)-\cos(\frac {-n\pi}4)-i\sin (\frac {n\pi}4)) $$
Now you can use the even/odd nature of cosine and sine to simplify this
A: $$(1+i)^n - (1-i)^n =$$
$$\sum_{k=0}^{n}\binom{n}{k}i^k - \sum_{k=0}^{n}\binom{n}{k}(-1)^ki^k =\sum_{k=0}^{n}\binom{n}{k} \left( i^k - (-1)^ki^k \right)=\sum_{k=0}^{n}\binom{n}{k}i^k\left( 1-(-1)^k \right)$$

*

*If k is even $(k=2n)$ $\Rightarrow$ $\left(1-(-i)^k\right)=0$


*If k is odd $(k=2n+1)$$\Rightarrow$ $i^k(1-(-1)^k)=2i(-1)^n=2i(-1)^{\frac{k-1}{2}}$.
Hence, $(1+i)^n - (1-i)^n$ = $$2i\sum_{k=1}^{n}\binom{n}{k}(-1)^\frac{k-1}{2}$$
A: In a more compact way, there is a simpler way of writing the expression:
$(1+i)^{n}-(1-i)^{n}=2^{(1+\frac{n}{2})}i\sin(\frac{n\pi}{4})$
