Let D be a simply connected domain. If f is analytic and $f(z)\ne0$ on D, then prove that there exists analytic functions g( depending on n) $\forall n\ge1$ such that $g^n=f$
I tried solving it the way I had solved for $ e^g=f $
For $e^g=f$, since $f(z)\ne0$ on D $f'/f$ has a anti-derivative say $h$ which is analytic since$f(z)\ne0\space\space\forall z\space\epsilon$ C
Let $c(z)=fe^-h$ Then $c'=e^-h(f'-fh')=0$ Hence $c(z)=constant=a\space\epsilon$ C
Then $a=fe^-h\Rightarrow f=ae^h$. Let $a=e^b$ and $g(z)=b+h(z)$
Then $e^g=f$