Originally Posted by

**Haven** Simple question but I can't seem to get it.

Show that the set of values of $\displaystyle log(i^{1/2})$ is $\displaystyle (n + 1/4)\pi\\i$ for any integer n.

But using the definition of a complex logarithm and the polar form of i. I get:

$\displaystyle i = e^{i\pi/2}$

So, $\displaystyle i^{1/2} = e^{i\pi/4}$

Since $\displaystyle log(z) = ln(r) + i(\theta + 2n\pi)$

So, $\displaystyle log(i^{1/2}) = ln(1) + i(\pi/4 + 2n\pi) = i(\pi/4 + 2n\pi)$

which is not the correct answer, is there something i'm missing?