If $e^{f(z)}$ is analytic then $f(z)$ is analytic I want to know how to prove that if $e^{f(z)}$ is analytic in some open disk then $f(z)$ is analytic in that open disk. Can someone guide me through this? thanks!
Let $g(z)=e^{f(z)}$ where $g$ is analytic. Therefore, $$g'(z)=f'(z)e^{f(z)}\implies f'(z)=\frac{g'(z)}{e^{f(z)}}$$ is analytic (because quotient of two analytic function and $e^{f(z)}\neq 0$ for all $z$). Therefore $f'$ is analytic and thus $f$ is analytic.
Let $g(z)=e^{f(z)}$ where $g$ is analytic. Therefore, $$g'(z)=f'(z)e^{f(z)}\implies f'(z)=\frac{g'(z)}{e^{f(z)}}$$ is analytic (because quotient of two analytic function and $e^{f(z)}\neq 0$ for all $z$). Therefore $f'$ is analytic and thus $f$ is analytic.