Proving f(z)=e^(g(z)) on a Convex Set Omega

michael.wes
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Homework Statement



Suppose that f is analytic on a convex set omega and that f never vanishes on omega. Prove that f(z)=e^(g(z)) for some analytic function g defined on omega.
Hint: does f'/f have a primitive on omega?

Homework Equations



f(z)=\sum_{k=0}^\infty a_k(z-p)^k

The Attempt at a Solution



I was able to prove that f'/f has a primitive on omega by the Cauchy-Goursat theorem, but I'm not sure where to go from here. Any help is appreciated!
 
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Let F be a primitive of f'/f. Now, consider the function G(z)=e^{F(z)}/f(z). What is it's derivative? What can you conclude from that?
 


I got that e^{g(z)}=cf(z), for some complex constant c and some analytic function g. It's usually easy in these problems to show that the constant is 1, but this is not a concrete function, so I'm not sure how to do that.
 


Well, the constant is not necessairly 1, so you'll have to find something else. You'll have to modify your function g in some way such that the equation is right...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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