Residue of $\dfrac{z^{1/4}}{z+1}$ at $ z = -1$. This question is from Churchill and Brown's Complex Variables and Applications 8th edition, page 248: 

Find $Res ({f},-1)  $ for $f = \dfrac{z^{1/4}}{z+1}$ given $|z| > 0, 0 < \arg z< 2\pi$

Attempt: 
Since the denominator has a simple zero, the residue is $(-1)^{1/4}$. Now, this has four values: $e^{i\pi\left(\frac{1+2n}{4}\right)}$ for $n = 0,1,2,3$. How to decide which value I should choose? 

The domain of $f$ is four copies of the complex plane.  They are arranged like a four-level carpark.  Every time you go around the origin you end up on the next copy.  $\theta$ increases by $2\pi$ so $z^{1/4}$ is multiplied by $i$.
So there are four functions $f$, each one is consistent and continuous as long as you stay on one side of the origin.
The choice of the residue comes from whichever of the four $f$s you are usibg in the integral.  The path of the integral can't go around the origin.
