<s>[INST] 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
 [/INST] 
The result $k=4$ is correct. The given multiple choice answer $k=2$ is wrong, since
  \begin{align*}
\left(\frac{1+i}{1-i}\right)^2=i^2=-1\ne 1
\end{align*}

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