Complex numbers $\left(\frac{1+i}{1-i}\right)^k = 1$ what is $k$? The smallest possible integer $k$ for which $\left(\frac{1+i}{1-i}\right)^k = 1$ is?
I tried solving this, but my answer doesn't match the given answer. Correct me if I'm wrong at some place
My solution:
\begin{align}
\left(\frac{1+i}{1-i}.\frac{1+i}{1+i}\right)^k&= 1\\
\left(\frac{1+2i+i^2}{1-i^2}\right)^k&= 1\\
\left(\frac{1+2i-1}{1-(-1)}\right)^k&= 1\\
\left(\frac{2i}{2}\right)^k&= 1\\
i^k&= 1\\
i^4&= 1\\
\end{align}
EDIT: The question is part of the multiple choice section and the answer is 2.
Other options include: 4, 8, 16

$(\frac {1+i}{1-i})^k=1$
$(1+i)^k = (1-i)^k$
$\sum_{j=0}^k {k\choose j}i^j = \sum_{j=0}^k {k\choose j}i^j*(-1)^j$
As $(-1)^{even} = 1$ and $(-1)^{odd} = -1$
$\sum_{j=0;j odd}{k\choose j}i^j =0$
As $i^{4k + 1} = i$ and $i^{4k - 1} = -i$
we need to find the smallest $k$ where $\sum_{h=0}^{4h+1 \le k}{k\choose 4h+1} =\sum_{h=1}^{4h-1 \le k}{k\choose 4h-1}$.
${k \choose 4h -1} = {k\choose k -4h + 1}$ so if $k = 4m$ and $g = m-h$ we will have ${k\choose 4h -1} = {k \choose k - 4h + 1} = {k\choose 4(m-h) + 1} = {k\choose 4g + 1}$ so any $k = 4m$ will be a solution so $k =4$ will be a solution.
It's easy to show directly that $k = 1,2,3$ are not:
$1 + i \ne 1 -i$
$(1+i)^2 =1 + 2i -1 \ne 1 - 2i -1 =(1-i)^2$
$(1+i)^3 = 1 + 3i + 3i^2 + i^3 = -2 + 2i \ne 2-2i = 1 - 3i + 3i^2 -i^3 = (1+i)^3$
However $(1 + i)^4 = 1 + 4i + 6i^2 + 4i^3 + i^4 = 1 - 4i^3 + 6i^2 - 4i + i^4 = 1 - 4i + 6i^2 - 4i^3 + i^4 = (1 - i)^4$.
