Simple Complex Analysis Question

[tylerc1991]

Homework Statement

Suppose f(z) is entire and the harmonic function u(x,y) = Re[f(z)] has an upper bound u_0. (i.e. u(x,y) <= u_0 for all real numbers x and y). Show that u(x,y) must be constant throughout the plane.

The Attempt at a Solution

Since f(z) = u(x,y) + iv(x,y) is entire, then the component functions u(x,y) = Re[f(z)] and v(x,y) = Im[f(z)] are entire also. Since u(x,y) has an upper bound and is entire, by Liouville's theorem, u(x,y) is constant.

Is this correct? Thank you for your time!

 

[Dick]

How do you figure u(x,y) is 'entire'? Entire is connected with being analytic. Analytic functions are almost always complex. They need to satisfy Cauchy-Riemann. You need to think a little harder about this.

 

[tylerc1991]

How do you figure u(x,y) is 'entire'? Entire is connected with being analytic. Analytic functions are almost always complex. They need to satisfy Cauchy-Riemann. You need to think a little harder about this.

The suggestion in the book was to apply Liouville's theorem to the function g(z) = exp[f(z)]

Trying this, I found that since f(z) is entire then g(z) is entire (composition of entire functions is entire). and that the modulus of g(z) = |g(z)| = |exp[f(z)]| = |exp[u(x,y)]*exp[iv(x,y)]| = exp[u(x,y)] <= exp[u_0]. So I can say that g(z) is constant by Liouville's theorem. If g(z) is constant then f(z) is constant and hence u(x,y) and v(x,y) are constant. I believe that this works, but I thought that my original way was quicker. Since analyticity is connected to differentiability in some domain, and since u(x,y) is differentiable in the entire plane, I thought I could say that u(x,y) was entire in the plane. But I see the point about complex valued functions versus real valued. In any case, was my first (and somewhat abandoned) attempt correct now? Thank you very much!

 

[Dick]

The suggestion in the book was to apply Liouville's theorem to the function g(z) = exp[f(z)]

Trying this, I found that since f(z) is entire then g(z) is entire (composition of entire functions is entire). and that the modulus of g(z) = |g(z)| = |exp[f(z)]| = |exp[u(x,y)]*exp[iv(x,y)]| = exp[u(x,y)] <= exp[u_0]. So I can say that g(z) is constant by Liouville's theorem. If g(z) is constant then f(z) is constant and hence u(x,y) and v(x,y) are constant. I believe that this works, but I thought that my original way was quicker. Since analyticity is connected to differentiability in some domain, and since u(x,y) is differentiable in the entire plane, I thought I could say that u(x,y) was entire in the plane. But I see the point about complex valued functions versus real valued. In any case, was my first (and somewhat abandoned) attempt correct now? Thank you very much!

The first attempt is still wrong. The second one is correct. |g(z)| is bounded by exp(u_0). So it's constant. And if g(z) is constant then f(z) is constant. Just to be sure you might want to say why that makes f(z) constant. There are functions where exp(f(z))=constant, where f(z) is not constant, but those functions aren't continuous. Just to make sure that isn't the point of this question.

 

[tylerc1991]

The first attempt is still wrong. The second one is correct. |g(z)| is bounded by exp(u_0). So it's constant. And if g(z) is constant then f(z) is constant. Just to be sure you might want to say why that makes f(z) constant. There are functions where exp(f(z))=constant, where f(z) is not constant, but those functions aren't continuous. Just to make sure that isn't the point of this question.

It should be fine for me to say that since g(z) is constant then f(z) is constant. What type of function would make exp[f(z)] = constant and f(z) still not constant. I know this is just a tangent but I am curious about that. Since I would think that exp[f(z)] = c => f(z) = log(c) which is still just some constant?

 

[Dick]

It should be fine for me to say that since g(z) is constant then f(z) is constant. What type of function would make exp[f(z)] = constant and f(z) still not constant. I know this is just a tangent but I am curious about that. Since I would think that exp[f(z)] = c => f(z) = log(c) which is still just some constant?

Define g(z)=exp(f(z)) with z=x+iy and f(z)=0 if x<0 and 2*pi*i if x>=0. g(z)=1, a constant. f(z) isn't constant. But it's not continuous, much less analytic. Just checking they aren't trying to trip you up on a technicality. I'd agree it's probably ok to just say f(z) is constant if g(z) is.