How to show the only branch cut needed is the imaginary axis. For the function 
$f(z) = z^{\frac{1}{5}} (z^2 + 4) ^{ \frac{2}{5} } $ 
show how a branch can be chosen such that the only branch cut needed is along the imaginary axis between $\pm2i$
I feel like I conceptually understand branch cuts but I'm simply not sure where to begin on this problem. Any help is appreciated.
 A: For $$f(z)  = z^{1/5} (z^2+4)^{2/5}$$ we  introduce $\mathrm{LogA}$, a
branch of the logarithm with argument between $-3/2\pi$ and $\pi/2$
and put
$$f(z) = \exp(2/5\mathrm{LogA}(z-2i))
\\ \times \exp(1/5\mathrm{LogA}(z))
\\ \times \exp(2/5\mathrm{LogA}(z+2i)).$$
We have  three intervals  that we must  examine for continuity. First,
take $z=it$ with $t\gt 2.$ To the right of the imaginary axis we get
$$\exp(2/5(\log(t-2)+\pi i/2))
\\ \times \exp(1/5(\log(t)+\pi i/2))
\\ \times \exp(2/5(\log(t+2)+\pi i/2))
\\ = t^{1/5} (t^2-4)^{2/5} \exp(\pi i/2)
= i t^{1/5} (t^2-4)^{2/5}.$$
On the left we get
$$\exp(2/5(\log(t-2)-3\pi i/2))
\\ \times \exp(1/5(\log(t)-3\pi i/2))
\\ \times \exp(2/5(\log(t+2)-3\pi i/2))
\\ = t^{1/5} (t^2-4)^{2/5} \exp(-3\pi i/2)
= i t^{1/5} (t^2-4)^{2/5}.$$
We have continuity across the cut and may then use Morera's theorem to
conclude  analyticity   in  $\Im(z)  >   2$  same  as  in   this  MSE
link.   Next  we have
the segment $2\gt t\gt 0.$ We get on the right
$$\exp(2/5(\log(2-t)-\pi i/2))
\\ \times \exp(1/5(\log(t)+\pi i/2))
\\ \times \exp(2/5(\log(t+2)+\pi i/2))
\\ = t^{1/5} (4-t^2)^{2/5} \exp(\pi i/2/5).$$
and on the left
$$\exp(2/5(\log(2-t)-\pi i/2))
\\ \times \exp(1/5(\log(t)-3\pi i/2))
\\ \times \exp(2/5(\log(t+2)-3\pi i/2))
\\ = t^{1/5} (4-t^2)^{2/5} \exp(-11\pi i/2/5).$$
No continuity here. Finally do  the segment $0\gt t\gt -2$, getting on
the right
$$\exp(2/5(\log(2-t)-\pi i/2))
\\ \times \exp(1/5(\log(-t)-\pi i/2))
\\ \times \exp(2/5(\log(2+t)+\pi i/2))
\\ = (-t)^{1/5} (4-t^2)^{2/5} \exp(-\pi i/2/5)$$
and on the left
$$\exp(2/5(\log(2-t)-\pi i/2))
\\ \times \exp(1/5(\log(-t)-\pi i/2))
\\ \times \exp(2/5(\log(2+t)-3\pi i/2))
\\ = (-t)^{1/5} (4-t^2)^{2/5} \exp(-9\pi i/2/5)$$
No continuity  here either.  This  establishes the branch  cut between
$\pm 2i$ and  analyticity in the slit plane. Observe  that for a point
$x$ on  the positive  real axis and  with $-\pi/2\lt \theta\lt  0$ the
argument of $x-2i$ we obtain
$$\exp(2/5(\log\sqrt{x^2+4}+i\theta))
\\ \times \exp(1/5(\log(x))
\\ \times \exp(2/5(\log\sqrt{x^2+4}-i\theta))
\\ = x^{1/5} (x^2+4)^{2/5}$$
so this branch  matches the one produced by the  real logarithm on the
positive real axis.
