Prove a function is constant. (complex analysis) Suppose function $f$ is holomorphic on $\mathbb C-\{0\}$ and satisfies $$|f(z)| \le \sqrt {|z|} + \frac{1}{\sqrt {|z|}}.$$ Prove that $f$ is a constant function.
I think it's related to Laurent series representation and ML inequality but I have no idea how to prove it. Can anyone give me some hints? Thanks in advance!
 A: Let
$$
 f(z) = \sum_{n=-\infty}^\infty a_n z^n
$$
be the Laurent series for $f$ in $\Bbb C \setminus \{ 0 \}$. The
coefficients $a_n, n \in \Bbb Z,$ can be computed as an integral
(compare https://en.wikipedia.org/wiki/Laurent_series)
$$
  a_n = \frac{1}{2 \pi i} \int_\gamma \frac{f(z)}{z^{n+1}} \, dz
$$
where $\gamma$ is any closed curve which surrounds $z=0$ exactly once.
Taking $\gamma$ as a circle with radius $r > 0$ gives the estimate
$$
 |a_n| \le \frac{1}{r^n} \max \{ |f(z)| : |z| = r \} 
 \le  \frac{ \sqrt {r} + \frac{1}{\sqrt {r}}}{r^n} \, .
$$
For fixed $n \ge 1$  and $r \to \infty$  the RHS tends to zero,
and the same is true if $n \le -1$ and $r \to 0$.
Therefore $a_n = 0$ for all $n \ne 0$.

Alternatively, you can argue as suggested in the comments.
$g(z) = z f(z)$ has a removable singularity at $z=0$ and is
therefore an entire function. From
$$
  |g(z)| \le r^{1/2} + r^{3/2}
$$
the usual estimates of the derivative with the Cauchy integral formula show that
$g$ is a polynomial of degree at most one.
So we have
$$
 f(z) = \frac{a z + b}{z}
$$
for some constants $a, b$, and from the growth restriction for $r \to 0$ it
follows that $b = 0$ must hold.
