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

So let $\displaystyle 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 $\displaystyle f(1) < 0 $. For $\displaystyle z = 1 $ we know that $\displaystyle r = 1, \ \theta = 0 $. How do we find the value of $\displaystyle k $?