Find the residue of $f(z)$ Find the residue of $f(z)$
$$f(z) = \frac{z^{(1/4)}}{z+1}$$
So this is a pole of order 1 with a singularity at $z=-1$
$$z^{1/4}\Big|_{-1}\ = (-1)^{1/4}$$
And I'm not sure what to do with that. The book says the answer is $\frac{1+i}{\sqrt2}$

The book is not being fair because it didn't give you enough information to solve the problem.  You have a multivalued function.  What is the branch cut?  If the branch cut is the negative real axis, then $\arg{z} \in (-\pi,\pi)$ and you need to know what side of the branch cut you are on.  On the one hand, if you are just above the branch cut, then $-1=e^{+i \pi}$ and the book's answer is correct.  On the other hand, if you are just below the branch cut, the $-1=e^{-i \pi}$  and the answer is instead $\frac{1-i}{\sqrt{2}}$.  
If the branch cut is taken anywhere else, then $\arg{z}=+\pi$ and the book's answer is correct.
