Proving Entire Function f(z) and its Image is All of C | Homework Help

[g1990]

Homework Statement

suppose that f is an entire function and that lim z->inf. f(z)=inf.
show that the image of f is all of C

Homework Equations

I know that f(z) must have isolated values

The Attempt at a Solution

I think the approach should be to pick an arbitrary value w in C and show that there is a z such that f(z)=w... but I don't know where to go from there

 

[Dick]

Suppose f(z) is never equal to w. Then what kinds of things can you say about g(z)=1/(f(z)-w)?

 

[g1990]

lets see, if f(z) never equals w, then g(z) is analytic everywhere on C. Then, as z approaches infinity, g(z) approaches 0 because f(z) approaches infinity. I'm not sure why this is a contradiction.

 

[Dick]

lets see, if f(z) never equals w, then g(z) is analytic everywhere on C. Then, as z approaches infinity, g(z) approaches 0 because f(z) approaches infinity. I'm not sure why this is a contradiction.

Think about whether g(z) might be bounded.

 

[g1990]

Lets see... the modulus of f is definitely bounded from below by 0. If I could get it to be bounded from above that would be a contradiction b/c of the maximum modulus principle. I'm not sure how to do that though. You could say that if z if large, f is above a certain value so that g is below a certain value. Then, f has a minimum value on the remaining compact set, so that g has a maximum on that compact set. Then the max of the two would be the bound for g. Is that right?

 

[Dick]

Lets see... the modulus of f is definitely bounded from below by 0. If I could get it to be bounded from above that would be a contradiction b/c of the maximum modulus principle. I'm not sure how to do that though. You could say that if z if large, f is above a certain value so that g is below a certain value. Then, f has a minimum value on the remaining compact set, so that g has a maximum on that compact set. Then the max of the two would be the bound for g. Is that right?

You are garbling that a bit. But you've got the basic idea right. Yes, it's compactness and |f(z)|->infinity. Try and state it a little more carefully and then I'll believe you.

 

[g1990]

Alright, so pick an N. Then, when |z|>M for some value M, |f(z)|>N. Thus, when |z|>M, |g(z)|<1/|N-w|. When |z|<=N, then |g(z)| attains a maximum (on the boundary) because g is continuous and the disc |z|<=N is compact. Call this maximum P. Then, g(z) is bounded by max(P, 1/|N-w|). But, g is entire, so by the max-modulus principle g is constant. Then, f must also be constant because w is fixed. This is a contradiction. Does that sound better?

 

[Dick]

Does |f(z)|>N really imply that 1/|f(z)-w|<1/|N-w|? Suppose N=1, f(z)=1.01*i and w=i. You are just being really sloppy. Shouldn't your choice of N somehow involve |w|?

 

[g1990]

Lets see, if |f|>N, then |g|=1/|f-w|. By the triangle inequality, |f-w|>||f|-|w||>||N|-|w||
I could let N=S|w|, so that then |f-w|>||N|-|w||=||S||w|-|w||=|S-1|
Then, |g(z)|=1/|f-w|<1/(S-1). I don't get why having |w| in the bound of g(z) is a problem.

 

[g1990]

Oh, maybe I need that N>|w|?

 

[Dick]

Oh, maybe I need that N>|w|?

N>2|w| will do it, won't it? Then |g(z)|<1/|w|.

 

[g1990]

Okay, so if I let N>2|w|, then my argument will work?

 

[Dick]

Okay, so if I let N>2|w|, then my argument will work?

Sure.