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? 
 A: 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.
A: Feels like the function $f$ you are using is not specified sufficiently. In other words you would have have the same problem evaluating $f(2)$, for example.
So for example, if the idea behind $f$ is to use the smallest positive angle, then you would apply the same principle to computing the residue, choosing $n=0$.
A: In Section#35 The Power Function of the 9th edition, the power function $z^c$ is defined by means of the equation $$z^c = e^{c\log z}\:\:\:\: (z \not= 0)$$
Since they specify $|z|>0$ and $0<\arg z < 2\pi$, this defines a specific branch of the logarithmic function
$$\log z = \ln|z|+i\cdot\arg z \;\;\;\; (|z|>0,0<\arg z < 2\pi)$$
thus $z^{1/4} = e^{\frac{1}{4}\log z}=e^{\frac{1}{4}[\ln|z| +i\cdot\arg z]}$
and since $-1 = 1\cdot e^{i\pi}$, $|z|=|-1|=1$ and $\arg z = \pi$, in this case the residue is $\phi (-1) = e^{\frac{1}{4}[0+i\pi]}= e^{i\pi/4} = \frac{1}{\sqrt{2}}(1+i)$, as required, where $\phi(z) = e^{\frac{1}{4}[\ln|z| +i\cdot\arg z]}$, a specific definition of $z^{1/4}$.
kudos to @Maxim
