What is the product of all the values of $(1-i)^{\frac{2}{5}}$ is equal to
What is the product of all the values of $(1-i)^{\frac{2}{5}}$ is equal to
$-2 i$
$2 i$
$-2$
$2$
Solution
Let $z=(1-i)^{2 / 5}$
$\begin{aligned} & z=\left[(1-i)^2\right]^{1 / 5} \\ & z=(1-1-2 i)^{1 / 5} \\ & z=(-2 i)^{1 / 5} \Rightarrow z^5=-2 i \Rightarrow z^5+2 i=0\end{aligned}$
The product of all value of $z$ is $-\frac{2 i}{1}=-2 i$.