If $|f(z)| \leq |z|^2+\frac{1}{\sqrt{|z|}}$, show f is quadratic polynomial. Suppose the function f is analytic in the punctured plane $z!=0$ (it means we excluded the zero) and satisfies the above condition, $|f(z)| \leq |z|^2+\frac{1}{\sqrt{|z|}}$, then show f is quadratic polynomial.
I think that if we multiply $z$ on the both side then we get $g(z)=zf(z)$ goes to 0 when z is going to 0. therefore, g is analytic function. Then what we should do to prove this?
 A: First, since $\lim_{z \to 0} z f(z) = 0$, we see that $f$ has a removable singularity at $z=0$. Hence we may as well assume that $f$ is analytic everywhere, and so has an entire  power series expansion at $z=0$.
Second, if $|z| \ge 1$, we have $\frac{1}{\sqrt{|z|}} \le |z|^2$, hence $|f(z)| \le 2 |z|^2$, for $|z|\ge 1$.
Third, if we choose $R \ge 1$, and suppose $|z| = R$, then Cauchy's estimate gives $|f^{(k)}(0)| \le \frac{2 R^2}{R^k}$. Letting $R \to \infty$, it follows that if $k>2$, then $f^{(k)}(0) = 0$. Consequently $f(z) = f(0) + f'(0) z + \frac{1}{2} f''(0) z^2$.
A: As you already noted $zf(z)$ is analytic and zero at $z=0$, hence $f$ is (after removing the singularity) entire and we have $|f(z)|<C|z|^2$ for $|z|>R$ with suitable $C$ and $R$.
Then $g(z)=\frac{f(z)-f(0)}{z}$ is entire and has the property $|g(z)|<C'|z|$ for $|z|>R$ with suitable $C'$.
Finally, $h(z)=\frac{g(z)-g(0)}{z}$ is entire and $|h(z)|<C''$ for $|z|>R$ with suitable $C''$. But then $h$ is bounded on all of $\mathbb C$, hence constant, $h(z)=h(0)$
We conclude 
$$ f(z)=(h(0)z+g(0))z+f(0).$$
Remark: If we only had  the weaker condition $|f(z)|\le |z|^2+\frac1{|z|}$, we could only conclude that $zf(z)$ is a cubic polynomial, so $f$ would be a rational funciton but might have a pole at $0$.
