Is f constant if limf(z) exists and is nonzero as z tends to z0?

[AcC]

Homework Statement

Let f:C\rightarrowC be differentiable, with f(z)\neq0 for all z in C. Suppose limf(z) is exist and nonzero as z tends to z0. Prove that f is constant.

 

[snipez90]

Are you stating the problem correctly? The complex exponential e^z satisfies all of those conditions but is obviously not constant.

 

[AcC]

Are you stating the problem correctly? The complex exponential e^z satisfies all of those conditions but is obviously not constant.

Yes you are true!?? but I found this question in complex analysis book, I tried to solve but I hadnt thought that it could be false..!

 

[AcC]

Take limit e^z as z→-infinity it equal to zero, so it not satisfy the statement. I proved this statement with using Liouville Theorem.

 

[Uncle_John]

In my books it's written that any integral over a closed loop \gamma equals zero:

\oint f(z)dz = 0

But at the same time it says

\oint \frac{dz}{z-a} = i 2 \Pi I

where I is an index number saying how many times loop \gamma goes around point a.

Aren't they contradicting each other?

PS: what can I do to stop Latex going to new line every time I use it?

 

[grey_earl]

Uncle John, welcome to Physicsforums! Please open a new thread for your question.

AcC, in that case you should state that C includes the point at infinity, which most authors denote as C* or something similar, i.e. the extended complex plane.