# Homework Help: Prove f is constant - liouville's theorem (?)

1. May 19, 2012

### jinsing

1. The problem statement, all variables and given/known data
Let f=u+iv, where u(z)>v(z) for all z in the complex plane. Show that f is constant on C.

2. Relevant equations

none

3. The attempt at a solution

Here's my attempt (just a sketch):

Since f is entire, then its components u(z), v(z) are also entire <- is this necessarily true?

Since f is entire, the CR equations hold.

v(z) is bounded for all z, so since v is entire and bounded, by liouville's theorem v is constant. <- kinda shaky on this one, too.

Since CR equations hold, we know v_y = u_x = 0 and -v_x = u_y = 0.

So, f' = u_x - iu_y = 0, so f is constant.

Does this seem okay.. or am I way off base?

2. May 20, 2012

### jinsing

I hate to be this guy.. but does this look okay? I still need help..

3. May 20, 2012

### gopher_p

Don't know if you have Little Picard at your disposal, but that would be the route that I would take.

It is not true that $u$ and $v$ are entire. They are real analytic, though. Liouville doesn't apply.

Even if $v$ were entire, it isn't bounded.

4. May 20, 2012

### Citan Uzuki

No, this doesn't look okay at all. In general, neither the real nor imaginary parts of a holomorphic function are themselves holomorphic. In fact, it's not difficult to show that they cannot be holomorphic unless f is constant (basically, note that both u and v are real-valued, and apply the open mapping theorem).

As for this problem, the way I would go about it is by looking at the function g(z) = 1/(f(z) - i).

5. May 20, 2012

### jinsing

gopher p: no I can't use Little Picard, unfortunately. I looked it up, though, and it did seem pretty helpful! Thanks for the suggestion!

Citan Uzuki: yeah, that's what I thought.. my method definitely seemed a tad suspicious to me. I don't quite understand how 1/(f(z)-i) would help.. could you please elaborate a little more?

Thank you so much, so far!

6. May 20, 2012

### jinsing

Or, rather, how'd you get 1/(f(z)-i)?

7. May 20, 2012

### Dick

If u(z)>v(z) then f(z)=u+iv only has values in half of the complex plane. Sketch the region. i isn't in that region. So |f(z)-i| has a lower bound. Is that enough of a hint?

8. May 20, 2012

### jinsing

That's perfect. Thank you very much you guys!