Find the residue of $f(z)$

Question

$$f(z) = \frac{z^{(1/4)}}{z+1}$$

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 $\frac{1+i}{\sqrt2}$

So $(-1)^{1/4} = (e^{i\pi})^{1/4} = e^{i\pi/4} = cos(\frac{\pi}{4}) + i sin(\frac{\pi}{4})$? @kaynex — NMBL, Dec 02 '16 at 04:39
@Bellatrix1106 Yes. What are $\cos(\pi/4)$ and $\sin(\pi/4)$? — Matthew Cassell, Dec 02 '16 at 04:40
Yeah. However, it's also true that $-1 = e^{3πi}$, which gives a different value. Can anyone explain why you wouldn't use this? — Kaynex, Dec 02 '16 at 04:41
$ = \frac{\sqrt2}{2} + \frac{i\sqrt2}{2} = \frac{\sqrt2 (1+i)}{2} = \frac{1+i}{\sqrt2}$ @Mattos Thank you — NMBL, Dec 02 '16 at 04:44
The book also states in the answer $|z|>0,\ 0<arg\ z<2\pi$ so I guess it has something to do with the argument? @Kaynex — NMBL, Dec 02 '16 at 04:45
I'm guessing it has something to do with the branch cut of the principal logarithm. — Kaynex, Dec 02 '16 at 04:49

score 1 · Answer 1 · answered Dec 06 '16 at 17:50

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.

Find the residue of $f(z)$

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