requesting a proof (rationale) check as i'm unsure whether or not my rationale is legitamate or not. Thanks in advance
Consider the function $$f(z) = \frac{\omega}{(z^2 + \omega^2)e^{\alpha\sqrt{z/d}}} $$ where $\alpha,d~\&~\omega$ are held constant. then:
show f has branch points at z = 0 and infinity
locate and identify the order of the poles
we'll start with two first since $e^\sqrt{z}$ is non-zero everywhere on its branch cut with exception at $z=0$ ie the branch point where its undefined.
then considering $g(z) = \frac{\omega}{z^2+\omega^2}$ we have ovbious poles at $z = \pm i\omega$ both of which are simple (we can do this more explicitly by using partial decomposition).
Now the part i'm a little less certain on. Identifying Branch points: we're looking for points of step-discontinuity when passing from one branch to another, this can be found using a variety of techniques but essentially im looking for a point where our function is undefined.
breaking f into a product of two functions we have $$f(z) = g(z)h(z)=\left(\frac{\omega}{z^2+\omega^2}\right)\left(\frac{1}{e^{\alpha \sqrt{z/d}}}\right)$$ we see that it's obvious g is single valued, and although has two poles theres no jump discontinuity. so we consider h(z) to be the source of the branch points. $$h(z) = e^{-\sqrt{z/d}} = \left[e^{\sqrt{z}}\right]^{-\frac{\alpha}{\sqrt{d}}}$$ which we can see has a branch point at $z=0$ with the trick $$e^{\sqrt{z}} = \exp\left(\exp\left(\frac{1}{2}\log(z)\right)\right)$$ which is undefined at $z = 0$ as required.
incase thats not enough, further exploration is that for $z \in \mathbb{C}$ we have $$h(z) = \left[e^{\sqrt{z}}\right]^{-\frac{\alpha}{\sqrt{d}}} = \left[e^{r^{1/2}[\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}]}\right]^{-\frac{\alpha}{\sqrt{d}}}$$ then by considering the path towards the branch point at two different vectors we have
$$\lim_{\epsilon \longrightarrow 0^{+}} e^{r^{1/2}\cos(\epsilon/2)} = e^{r^{1/2}} \neq \lim_{\epsilon \longrightarrow 0^{-}} e^{r^{1/2}\cos(\epsilon/2)} = e^{-r^{1/2}}$$ and so we see a value change depending on how we approach the branch point. as we would expect. thus $z=0$ is a branch point of h(z) and so is also a branch point of f(z).
At this point, is this rationale legitmate? any errors glaring at you?
Next we want to show that f has a branch point at infinity, and so we look for a branch point at $z=0$ for $f(1/z)$
$$f(1/z) = g(1/z)h(1/z) = \left(\frac{\omega}{(1/z)^2+\omega^2}\right)\left(\frac{1}{e^{\alpha/\sqrt{zd}}}\right)$$
$$=\left(\frac{z^2\omega}{1+\omega^2 z^2}\right) \left( \frac{1}{e^{\alpha/\sqrt{zd}}}\right)$$
again, considering $z=0$ we see that $g(1/z)$ has no branch points and is 0 at this point.
my next line of thinking (Rather than repeating myself as above) is to consider the power series expansion of $$e^{1/\sqrt{z}} = \sum_{k=0}^{\infty} \frac{(-1/\sqrt{z})^{k}}{k!}$$ which is obviously undefined at z=0 and so 0 is a branch point of $h(1/z)$ making it a branch point of $f(1/z)$ and so infinity is a branch point of $f(z)$
Where have i gone wrong?
Thanks for taking the time to read this, i appreciate it