Browse Questions

The least positive integer $n$ so that $\large(\frac{1+i}{1-i})^n$$=1$ is?

$\begin{array}{1 1} 1 \\ 2 \\ 3 \\ 4 \end{array}$

Toolbox:
• $i^2=-1$
• If $i^n=1$ then $n=$multiple of 4
$\large\frac{1+i}{1-i}=\frac{1+i}{1-i}\times\frac{1+i}{1+i}=\frac{2i}{2}=i$
$\large(\frac{1+i}{1-i})^n=1$
$\Rightarrow\:i^n=1$
$\Rightarrow$ The least positive n=4