Holomorphic functions and limits of a sequence

Solution 1:

You can apply the hypothesis, the quotient rule, and 2. to show that $\left(\frac{f}{g}\right)'$ is the zero function. This means that $\frac{f}{g}$ is constant, and the result follows.

Solution 2:

By the identity theorem, the functions $f'(z)\over{f(z)}$ and $g'(z)\over{g(z)}$ are equal on all of $D$. These functions are holomorphic (since the denominators have no roots in $D$) so they locally have an antiderivative at any point: $log$ $f(z)$ and $log$ $g(z)$ respectively. When two functions are equal their if antiderivatives differ by a constant so we have $log$ $f(z)$=$log$ $g(z)$ + $k$. Once again by the identity theorem, if $log$ $f(z)$=$log$ $g(z)$ + $k$ on an open set, then this also holds on all of $D$. It now follows that $f$ and $g$ differ by a multiplied constant.