Determine all entire functions $f$ with $\Re f(z) > 0$ for all $z \in \mathbb{C}$ such that, $f(0) = 1$.

Asked Apr 18 '20 at 10:22

Active Apr 18 '20 at 10:22

Viewed 49 times

I am trying to determine all entire functions $f$ with $\Re f(z) > 0$ for all $z \in \mathbb{C}$ such that, $f(0) = 1$.

So far, my attempt is the following. Write $f = u+iv$, and define the function $$ g(z) = \frac{f(z)}{\Re f(z)} = \frac{u(z)+iv(z)}{u(z)} = 1 + i \frac{v(z)}{u(z)}. $$ Since $\Re f(z) > 0$ for all $z \in \mathbb{C}$, this is an entire function. We have $$ 1 = g(0) = 1 + i \frac{v(0)}{u(0)} $$ So $v(0) = 0$. Also, $$ |g(z)|^2 = 1 + \frac{v(z)^2}{u(z)^2}. $$ I want to apply Liousville Theorem somehow, so I'm looking for bounds, i.e. $$ |g(z)| \leq M|z|^m $$ for some $M$, $m \geq 0$, and for large enough $|z|$. I have no progress on this so far, I'm not sure if my attempt is in the right direction.

asked Apr 18 '20 at 10:22

Sigurd

1,649
10
19

1

Hint: little Picard's theorem. If you want to use Liouville's theorem, apply it to $e^{-f(z)}$ – achille hui Apr 18 '20 at 10:27
2

In addition, I think a priori its not clear that your $g$ is analytic – Calvin Khor Apr 18 '20 at 10:31
Thanks, that's helpful! Indeed, I defined the wrong function. It seems that considering $g(z) = e^{-f(z)}$ is a good approach in this setting. – Sigurd Apr 18 '20 at 10:34

Determine all entire functions $f$ with $\Re f(z) > 0$ for all $z \in \mathbb{C}$ such that, $f(0) = 1$.

0 Answers0