Question: Let the square root be computed using the branch $L_{\pi/2}$ of the logarithm. Where is $(z^2-i)^\frac{1}{2}$ not analytic?
My solution:
The function is not analytic when $\sqrt{z^2-i}=yi, y \in \mathbb{R}/{\mathbb{R^-}}$ (along the non-negative imaginary axis). Hence, solving for the $z$ of the equation is sufficient.
\begin{equation} \begin{split} z^2-i &= -y^2\\ z^2 &= -y^2+i\\ z &= \pm \sqrt{\underbrace{-y^2+i}_\beta}\\ \end{split} \end{equation}
Now, let's convert $\beta$ to the form of $|\beta|e^{i\theta}$.
$\theta = \arctan{\frac{1}{-y^2}}$ and $|\beta| = y^4 + 1$.
Hence, $z = \pm\sqrt{\beta} = \pm\sqrt{(y^4 + 1)e^{i\arctan{\frac{1}{-y^2}}}}$
Now, my solution is going south since $-y^2$ became a denominator and can be $0$.
Can someone please point out where I am doing wrong?