Complex Analysis

**sinichko** · November 6th 2011, 12:27 AM

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.

**Opalg** · November 6th 2011, 01:23 AM

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 $e^{-f(z)}$ ?

**sinichko** · November 8th 2011, 07:42 AM

Originally Posted by Opalg

Hint: What can you say about the function $e^{-f(z)}$ ?

I don't understand how it can help?
$e^{-f(z)}=\frac{1}{e^{u(x,y)}(\cos v(x,y)+i\sin v(x,y))}$

**Opalg** · November 8th 2011, 08:40 AM

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

**sinichko** · November 8th 2011, 09:53 AM

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?

**Opalg** · November 8th 2011, 10:36 AM

Originally Posted by sinichko

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

Yes! So what do you deduce from that?

**sinichko** · November 8th 2011, 11:07 AM

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?

**Opalg** · November 8th 2011, 11:46 AM

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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