Question

Suppose that ƒ is a nowhere vanishing holomorphic function in a simply connected region $\Omega$ .Prove that there exists a holomorphic function g on $\Omega$ such that $f(z)=e^{g(z)}$ .

Answer 1

Remember we deal with branching through the exponential.

f(z)−−−−√n=exp(lnf(z)n)=exp(ln|f(z)|+iargf(z)n)=exp(ln|f(z)|n)exp(i(Arg f(z)+2πk)n)

f(z)n=exp⁡(ln⁡f(z)n)=exp⁡(ln⁡|f(z)|+iarg⁡f(z)n)=exp⁡(ln⁡|f(z)|n)exp⁡(i(Arg f(z)+2πk)n)

where ln|f(z)|ln⁡|f(z)| is well defined since |f(z)|>0|f(z)|>0 on ΩΩ and k∈[0,n−1]k∈[0,n−1], thus we've found nn possible g(z)g(z)'s with the property