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 ?