Proving $g(z)=z^{2z}\sin{z}$ is not an entire function

Question

This is one part of a three-part question. So far, I have shown $f(z)=2^{z^2}$ is entire (since it is essentially the composition of an exponential and polynomial, both of which are entire), and that $h(z)=\sum_{i=0}^{\infty}z^i$ is not entire (since it only converges on the open unit disc). In both of these cases, I have not had to resort to getting the equations in the form $u(x,y)+iv(x,y)$ in order to apply the Cauchy-Riemann equations. For $g(z)=z^{2z}\sin{z}$ I wouldn't even know how to get it in such a form, the methods above don't work, and don't know of any alternative methods. Could someone explain why this is non-entire, or perhaps give some hints? Thanks

Answer 1

The definition of exponentiation for complex numbers $a,b$ is $a^b = \exp(b \log(a))$. The trouble is that $\log$ is a multivalued function: if $L$ is one logarithm of $a$ then $L + 2\pi i n$ is another for any integer $n$. Unless $b$ happens to be an integer, the result will be that $a^b$ is multivalued. When $a$ and $b$ are functions of $z$, you typically get logarithmic branch points when $a = 0$.

EDIT: Consider $a(z)^{b(z)} = \exp(b(z) \log a(z))$, where $a$ and $b$ are analytic and non-constant near $z_0$, and $a(z_0) = 0$.

If the zero of $a(z)$ at $z_0$ has order $m$, we have $a(z) = (z-z_0)^m g(z)$ where $g(z)$ is analytic and nonzero near $z_0$. Suppose we start at $z_1$ near $z_0$ and go in a circle counterclockwise around $z=z_0$, attempting to maintain continuity. Note that $$ \dfrac{d}{dz} \log a(z) = \dfrac{a'(z)}{a(z)} = \dfrac{m}{z-z_0} + \dfrac{g'(z)}{g(z)}$$ The $g'(z)/g(z)$ term is harmless, because an analytic logarithm of $g(z)$ exists in the neighbourhood of $z_0$. But the $m/(z-z_0)$ means when we come back to the starting point $z_1$, we will have added $2 \pi m i$ to $\log a(z)$. Correspondingly, we will have multiplied $a(z)^{b(z)}$ by $\exp(2 \pi m i b(z_1))$ (which is not $1$ because $m b(z_1)$ is not an integer for $z_1$ sufficiently close to, but not equal, $z_0$). Thus $a(z)^{b(z)}$ is not analytic at $z_0$.