Prove f is constant - liouville's theorem (?)

In summary, to show that f is constant on C, we can use the fact that u(z) is always greater than v(z) for all z in the complex plane. This means that f(z) = u(z) + iv(z) only has values in half of the complex plane. By sketching the region, we can see that i is not included in this region. Therefore, |f(z) - i| has a lower bound. This suggests using the function g(z) = 1/(f(z) - i) to show that f is constant on C.
  • #1
jinsing
30
0

Homework Statement


Let f=u+iv, where u(z)>v(z) for all z in the complex plane. Show that f is constant on C.


Homework Equations



none

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?
 
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  • #2
I hate to be this guy.. but does this look okay? I still need help..
 
  • #3
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
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
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
Or, rather, how'd you get 1/(f(z)-i)?
 
  • #7
jinsing said:
Or, rather, how'd you get 1/(f(z)-i)?

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

Related to Prove f is constant - liouville's theorem (?)

1. What is Liouville's theorem and how does it relate to proving a function is constant?

Liouville's theorem is a fundamental result in complex analysis, which states that a bounded entire function must be constant. This means that if a function is defined and continuous over the entire complex plane and is not allowed to grow infinitely large, then it must be a constant function. Proving that a function is constant using Liouville's theorem involves demonstrating that the function is bounded and entire.

2. How can I determine if a function is bounded?

A function is considered bounded if there exists a number M such that the absolute value of the function is always less than or equal to M, for all inputs. In other words, the function does not exceed a certain value as its inputs change. To prove that a function is bounded, you can use techniques such as finding the maximum or minimum value of the function, or using the definition of a bounded function to show that it satisfies the criteria.

3. What does it mean for a function to be entire?

An entire function is a complex-valued function that is defined and continuous over the entire complex plane. This means that it does not have any singularities or points where it is undefined. In other words, the function is defined and well-behaved for all possible inputs. To prove that a function is entire, you can use techniques such as power series expansions or the Cauchy-Riemann equations.

4. Can Liouville's theorem be used to prove that a function is not constant?

No, Liouville's theorem can only be used to prove that a function is constant. If a function is not bounded or not entire, then Liouville's theorem does not apply and cannot be used to prove that the function is not constant. In these cases, other techniques must be used to determine the behavior of the function.

5. How can Liouville's theorem be applied in real-world situations?

Liouville's theorem has many applications in mathematics and physics, particularly in the study of differential equations. It is also used in analyzing the behavior of physical systems, such as in the study of fluid dynamics or quantum mechanics. In practical terms, Liouville's theorem can be used to prove the stability of certain systems and to determine the long-term behavior of dynamic systems.

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