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 ?