Show that $f(z) = \frac{\sinh (\sqrt z)}{\sqrt z}$ is holomorphic on $\mathbb C$

Question

I have to show that $f(z) = \frac{\sinh (\sqrt z)}{\sqrt z}$ is holomorphic on $\mathbb C$. My attempt is the following:

1. Let $f(z) = \frac{g(z)}{h(z)}$, where $g(z) = \sinh (\sqrt z)$ and $h(z) = \sqrt z$.
2. Show that $h(z) = \sqrt z$ is holomorphic by showing that the Cauchy-Riemann equations hold.
3. Show that $g(z) = \sinh \sqrt z$ is holomorphic by showing that the C-R equations hold.

For the 2. step, altough a bit tedious, I managed to show that $\sqrt z$ satisfies the C-R equations. My question is, if I write $g(z) = \sinh \sqrt z = \frac{e^{\sqrt z} - e^{- \sqrt z}}{2}$ and make the substitution $\sqrt z = w = a + ib$, can I show that the C-R equations hold for $\sinh \sqrt z$?

Or, formulated differently, if a function has an another function which is holomorphic as it's argument, can I show the holomorphicity of this function by just making a substitution?

Answer 1

Let $f(z)$ be holomorphic and even (i.e., $f(-z)=f(z)$). Then $f(\sqrt z)$ is holomorphic.

To see this, we can cut the plane from $0$ to the point at infinity. On opposite sides of the branch cut, $\sqrt{z}$ changes sign. For example, suppose we cut the plane along the positive real axis such that on the upper part of the branch cut $\sqrt {1+i0^+}=1$. Then on the lower part, we see that we have$\sqrt{1+i0^-}=-1=-\sqrt {1+i0^+}$.

For an even function, $f(\sqrt{z})=f(-\sqrt z)$ and $f$ is continuous across the branch cut. Moreover, inasmuch as $f(z)$ is even and holomorphic function, it has a Taylor series representation

$$f(z)=\sum_{n=1}^\infty a_n z^{2n}$$

Hence, $f(\sqrt{z})$ also has a Taylor series representation

$$f(\sqrt z)=\sum_{n=1}^\infty a_n z^{n}$$

and we see that $f(\sqrt z)$ is holomorphic. Hence, inasmuch as $\frac{\sinh(z)}{z}$ is even and holomorphic, so is $\frac{\sinh(\sqrt z)}{\sqrt z}$