If f(z) holomorphic show that f(z) is constant

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In summary: Then, the first derivatives must be bounded, and then the second derivatives must be bounded, and so on. Then use the fact that a bounded holomorphic function must be constant by Liouville's theorem.In summary, to show that a holomorphic function f(z) = u + iv on a domain D, satisfying |f(z)| = constant, is also a constant function, the Cauchy-Riemann equations and the Laplace equation can be used to prove that the first and second derivatives of f must also be bounded, leading to the conclusion that f is constant. Alternatively, Liouville's theorem can be applied to a bounded holomorphic function to show that it is constant.
  • #1
Shackleford
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Homework Statement



If f(z) = u + iv is a holomorphic function on a domain D satisfying |f(z)| = constant, show that f(z) is constant.

Homework Equations



f(z) = u(z) + y(z)i = u(x,y) + y(x,y)i

z = x + yi

The Attempt at a Solution



|f(z)| = √u2+v2) = √u(x,y)2+v(x,y)2)

I know this is a simple problem, but I'm just not seeing the connection. I'm probably missing some obvious relation/equation to consider.
 
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  • #2
Maybe you can just quote Liouville? Maybe extending into an entire function with
constant nodulus.

If not, use Maximum Modulus http://en.wikipedia.org/wiki/Maximum_modulus_principle

Maybe let ##c=u^2+v^2 ## , then ##0= 2uu_x+2vv_x ## and ##0 =2uu_y+ 2vv_y ## and use Cauchy -Riemann.
 
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  • #3
Shackleford said:

Homework Statement



If f(z) = u + iv is a holomorphic function on a domain D satisfying |f(z)| = constant, show that f(z) is constant.

Homework Equations



f(z) = u(z) + y(z)i = u(x,y) + y(x,y)i

z = x + yi

The Attempt at a Solution



|f(z)| = √u2+v2) = √u(x,y)2+v(x,y)2)

I know this is a simple problem, but I'm just not seeing the connection. I'm probably missing some obvious relation/equation to consider.
It seems to me that you might want to start out by using the definition of holomorphic .

You also have a couple of typos in your post.

I assume you mean: f(z) = u(z) + v(z)i = u(x,y) + v(x,y)i
 
  • #4
SammyS said:
It seems to me that you might want to start out by using the definition of holomorphic .

You also have a couple of typos in your post.

I assume you mean: f(z) = u(z) + v(z)i = u(x,y) + v(x,y)i

Shoot. Yes. Sorry.

If f is complex differentiable at any point a ∈ D, f is called a holomorphic function on D.

Then of course f is holomorphic ⇔ Cauchy-Riemann Equations hold.

If f is a holomorphic function, then f'(z) = ux + ivx
 
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  • #5
WWGD, I had that idea last night. I haven't been able to divine a different "proof," so that's likely what I'll turn in.
 
  • #6
use the cauchy-riemann conditions along with the laplace equation
 
  • #7
Or, like occh suggested, use the fact that a harmonic function cannot be bounded. Then, if |f| is bounded, both the Real, Complex part must be bounded.
 

FAQ: If f(z) holomorphic show that f(z) is constant

1. What does it mean for a function to be holomorphic?

A holomorphic function is a complex-valued function that is differentiable at every point in its domain. This means that the function has a well-defined derivative at every point, making it a smooth and well-behaved function.

2. How does holomorphicity relate to the constancy of a function?

If a function f(z) is holomorphic, it means that it is differentiable at every point in its domain. This implies that the function is locally linear, meaning that it can be approximated by a line at every point. If a function is locally linear, it must be constant, as a line has a constant slope.

3. Can a non-constant function ever be holomorphic?

No, a non-constant function cannot be holomorphic. If a function is not constant, it will have points where it is not differentiable. This means that it cannot satisfy the definition of a holomorphic function.

4. How can I prove that a function is holomorphic?

To prove that a function is holomorphic, you must show that it is differentiable at every point in its domain. This can be done by taking the partial derivatives of the function and checking that they satisfy the Cauchy-Riemann equations. If the equations are satisfied, the function is holomorphic.

5. Can a function be holomorphic in some regions but not others?

Yes, it is possible for a function to be holomorphic in some regions of its domain but not others. For example, a function may be holomorphic everywhere except at certain singularities or discontinuities. However, if a function is holomorphic in a region, it will be holomorphic in all subregions of that region.

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