Entire function satisfying $f(z)=f(zi)$ for all $Z\in \mathbb{C}$

Question

let $f$ be an entire function such that $|f(z)|\leq \alpha |z|^3 $ for $|z|\geq 1$ and some constant $\alpha$ and $ \forall z\in \mathbb{C}:f(z)=f(zi) $ then

1. f is constant
2. $f(z)$ is a quadratic function
3. No such $f$ exist
4. $ \forall z \in \mathbb{C}:f(z)=\alpha z^3$

zero function satisfying this condition. How to prove cubic and quadratic function doesnot satisfy this condition?

Answer 1

Let $g(z):=f(z)-f(0).$

Then we have $g(0)=0.$ From $f(z)=f(zi)$ we get $g'(0)=f'(0)=if(0)$ and $g''(0)=f''(0)=-f''(0).$ Hence $g'(0)=g''(0)=0.$

Then there is an entire funktion $h$ such that

$$g(z)=z^3h(z).$$

This shows that $\frac{g(z)}{z^3}$ has a removable singularity at $0$.Hence $\frac{g(z)}{z^3}$ is bounded for $|z|<1.$

From $|f(z)|\leq \alpha |z|^3$ for $|z| \ge 1$, we see that $g(z)/z^3$ is bounded on $ \mathbb C.$ It follows by Liouville that $g(z)/z^3$ is constant.

Hence, there is $c$ such that

$$f(z)=cz^3+f(0)$$

for all $z$.

We then get

$$cz^3+f(0)=f(z)=f(iz)=-icz^3+f(0)$$

for all $z$. It follows that $c=0.$

Conclusion: $f$ is constant.