## branch

Suppose a branch of $z^{-1/4}(z^{2}+1)$ assumes negative real values for $y = 0, \ x > 0$. There is a branch cut along the line $x = y, \ y \geq 0$. What is $f(-1-i)$?

So let $f(z) = z^{-1/4}(z^2+1) = \frac{z^2+1}{r^{1/4} \ \text{cis} \left(\frac{\theta + 2k \pi}{4} \right)}$. Now $f(1) < 0$. For $z = 1$ we know that $r = 1, \ \theta = 0$. How do we find the value of $k$?