Defining a single valued w(z) and finding w'(8i) I  have a homework question which wants me to find a single-valued branch $w(z)$ of the function $5z^{1/3}$. Find the region where $w(z)$ is analytic. and find $w'(z)$
here I go as,
$f(z)=5z^{1/3}=5 r^{1/3}e^{\frac{i \theta +2n \pi}{3}}$
$0 \leq \frac{\theta+2n \pi}{3} \leq 2\pi$ which means $-2 \pi\leq\theta\leq4 \pi$
So function is not analytical at $z=\frac{-2\pi}{3}$
Then, $f(z)=5z^{1/3} f'(z)=\frac{5}{3} z^{-2/3}$
at $z=8i$
$ \frac{\theta}{3}=\frac{\pi}{2} $
$\theta=\frac{3 \pi}{2}$
$f'(8i)=\frac{5}{3}(e^{8i \frac{3\pi}{2}})^{\frac{-2}{3}}=\frac{5}{3}e^{-i8\pi}=\frac{5}{3}$
Am I right? Cuz it looked me like I made a mistake somewhere
 A: You aren't going about this the right way, I'm afraid.  When you say $-2\pi\leq\arg z\leq 4\pi$, you allow several representations for a given complex number.  Then your formula for $5z^{1/3}$ gives several possible values for $f(z)$, just the reverse of what is required.
Instead, pick a single range of length $2\pi$ for the argument, say $-\pi<\arg z<\pi$.  Note that you can't have a $\leq$ sign in either place, for that would mean there was no branch cut.  With the argument of $z$ restricted as above, the branch cut is at $z\leq 0$.  Then you can proceed along the lines you followed.
As outlined above, $f(z)$ is not defined if $z$ is a non-positive real number.  If we want $f$ to be analytic on the non-zero reals, we can choose a different cut, say the negative imaginary axis.  The one cut we can't choose is the positive imaginary axis, since the problem later asks for the derivative at $z=8i$.  (It is possible to use a curve other than a ray from the origin as the cut, but this is a needless complication, I think.)
