# 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.

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.