1. ## Complex Analysis

Hi! Help me please with this one.

Let's $f(z)$ be an entire function
and $Re f(z)\geq 0$for all $z$
Prove that $f$ is constant.

Thanks.

2. ## Re: Complex Analysis

Originally Posted by sinichko
Hi! Help me please with this one.

Let's $f(z)$ be an entire function
and $Re f(z)\geq 0$ for all $z$
Prove that $f$ is constant.
Hint: What can you say about the function $e^{-f(z)}$ ?

3. ## Re: Complex Analysis

Originally Posted by Opalg
Hint: What can you say about the function $e^{-f(z)}$ ?
I don't understand how it can help?
$e^{-f(z)}=\frac{1}{e^{u(x,y)}(\cos v(x,y)+i\sin v(x,y))}$

4. ## Re: Complex Analysis

Originally Posted by sinichko
I don't understand how it can help?
$e^{-f(z)}=\frac{1}{e^{u(x,y)}(\cos v(x,y)+i\sin v(x,y))}$
$-f(z) = -u(x,y) - iv(x,y)$. So $e^{-f(z)}=e^{-u(x,y)}(\cos v(x,y)-i\sin v(x,y))$. You are told that $u(x,y)\geqslant0$ for all x,y. What does that tell you about $|e^{-f(z)}|$ ?

5. ## Re: Complex Analysis

Originally Posted by Opalg
$-f(z) = -u(x,y) - iv(x,y)$. So $e^{-f(z)}=e^{-u(x,y)}(\cos v(x,y)-i\sin v(x,y))$. You are told that $u(x,y)\geqslant0$ for all x,y. What does that tell you about $|e^{-f(z)}|$ ?

$|e^{-f(z)}|$ is bounded?

6. ## Re: Complex Analysis

Originally Posted by sinichko
$|e^{-f(z)}|$ is bounded?
Yes! So what do you deduce from that?

7. ## Re: Complex Analysis

Originally Posted by Opalg
Yes! So what do you deduce from that?
Thanks.
Maybe I should apply Liouville's theorem
, but I have to prove that function $e^{-f(z)}$- an entire, given that $f(z)$ - entire function...how can i prove that $e^{-f(z)}$ entire function also?

8. ## Re: Complex Analysis

Originally Posted by sinichko
Thanks.
Maybe I should apply Liouville's theorem
, but I have to prove that function $e^{-f(z)}$- an entire, given that $f(z)$ - entire function...how can i prove that $e^{-f(z)}$ entire function also?
Yes, that is exactly what you need to do. In fact, a differentiable function of a differentiable function is always differentiable (that is basically what the chain rule says).
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