Given that $f(z)$ is analytic show that $f(\sqrt{z})$ is also analytic

Question

I am reading Titshmarch book on Riemann zeta function and this is what he says at some point:

where $$\Xi(z)=\xi\left(\frac{1}{2}+iz\right)$$

Now, he says that because $\Xi(z)$ is analytic and even then $\Xi(\sqrt{z})$ is also analytic. Can someone explain to me why that is? Am I missing something obvious?

And secondly the other statements that I don't quite understand is why can we say that $\Xi(\sqrt{z})$ has infinity of zeros based on the fact that it has order of $\frac{1}{2}$. Thank you for any help on either of tghe statements.

Answer 1

It's because $\Xi$ is an even function. This means that $$\Xi(z)=a_0+a_2z^2+a_4z^4+\cdots$$ on the whole complex plane. Therefore $$\Xi(\sqrt z)=a_0+a_2z+a_4z^2+\cdots$$ is also an entire function.

Answer 2

If $f$ is analytic and even on a ball around $0$, then the power series expansion contains only even terms, and hence there is some analytic function $g$ such that $f(z) = g(z^2)$.