Complex analysis / Using analyticity of f to prove f is constant

[AxiomOfChoice]

Homework Statement

I'm supposed to show that, if $f$ is analytic and $|f|$ is constant on a domain $D \subset \mathbb{C}$, $f$ is constant.

Homework Equations

The hint is to write $f^* = |f|^2 / f$. I might also need to use the fact that if $f^*$ is analytic too, then $f$ is constant.

The Attempt at a Solution

Well, I followed the hint, and I fail to see how it helps at all. Given the hypotheses of the problem, I guess we know $f^* = A / f$ for some $A > 0$, but this doesn't strike me as particularly useful. Writing $f = u(x,y) + i v(x,y)$ only seems to complicate things, but don't I eventually have to do this? I'm guessing I'm supposed to use the Cauchy-Riemann Equations together, in some way, with the fact (proved in my text) that if $h(x,y)$ is a real-valued function that satisfies $\nabla h = 0$ on a domain, then $h$ is constant on that domain. But taking partial derivatives and trying to use $u_x = v_y$ and $u_y = -v_x$ just makes things messy.

 

[Dick]

You've got all the ingredients in front of you. If f=u+iv is analytic then f*=(u-iv)=A/f is also analytic. Write out the Cauchy-Riemann equations for both f and f*. What do they tell you about the partial derivatives of u and v?

 

[AxiomOfChoice]

Right now, I've got a system of equations for $f = u+iv$: (1) $uu_x = vv_x$ and (2) $uu_y = vv_y$. These came from $u^2 + v^2 =$ Constant. But I've got to show that $u_x = u_y = 0$ from this and $u_x = v_y$, $u_y = -v_x$. Any hints? I still can't quite get there...

 

[Dick]

I already gave you a hint. f=u+iv and f*=u-iv are analytic. Use Cauchy-Riemann on both of them.

 

[AxiomOfChoice]

I already gave you a hint. f=u+iv and f*=u-iv are analytic. Use Cauchy-Riemann on both of them.

You did. Sorry. I'll do that :)

 

[AxiomOfChoice]

I already gave you a hint. f=u+iv and f*=u-iv are analytic. Use Cauchy-Riemann on both of them.

I figured it out. Thanks! That was very helpful.

 

[bluenickel]

well i figured it out using the hint too,but i didn't have to use the fact that modulus of f is constant on D.i just used the cauchy-riemann eqns from f=u+iv;
du/dx=dv/dy and from f*=u-iv; du/dx=-dv/dy hence du/dx =0, and found the partial derivatives to be zero and so f has to be constant.so where do we use the modulus of f being constant?is it when we show the analyticity of f*?
maybe it's something like that if f is analytic then 1/f is also analytic so f*=a.1/f is also analytic where a is modulus of f.but i don't know anything about the analyticity of 1/f and can't we just prove it using u and v?
-it became kind of long and sort of complicated so sorry if it's not understandable.

 

[Dick]

well i figured it out using the hint too,but i didn't have to use the fact that modulus of f is constant on D.i just used the cauchy-riemann eqns from f=u+iv;
du/dx=dv/dy and from f*=u-iv; du/dx=-dv/dy hence du/dx =0, and found the partial derivatives to be zero and so f has to be constant.so where do we use the modulus of f being constant?is it when we show the analyticity of f*?
maybe it's something like that if f is analytic then 1/f is also analytic so f*=a.1/f is also analytic where a is modulus of f.but i don't know anything about the analyticity of 1/f and can't we just prove it using u and v?
-it became kind of long and sort of complicated so sorry if it's not understandable.

Yes, you use |f|=constant to prove f* is analytic. The problem is that |f| isn't generally an analytic function. A constant IS analytic.

 

[Office_Shredder]

but i don't know anything about the analyticity of 1/f and can't we just prove it using u and v?p

You should know something about the analyticity of f! Just from the normal quotient rule that you have from real analysis

In general: If |f| is not constant, then it's probably not analytic. Also f* is almost never analytic, you just have a very special case here (in fact you can show that the function that takes z to its complex conjugate is not analytic)

 

[bluenickel]

yes,this is really easy with quotient rule.thanks for the help!