Suppose $f(z)$ and $g(z)$ are two holomorphic functions in the unit disk $D=z\in \mathbb{C}:|z|<1$ such that $(i)f(0)=g(0)$ and $(ii)$ the equality $$e^{f(z)}=e^{g(z)}$$ holds for every $z\in D$. prove that $f(z)=g(z)$ for every $z\in D$.
This is what i have so for:
Let, $F=f(z)-g(z)$
then, $e^{F(z)}=e^{f(z)}-e^{g(z)}=e^{f(z)-g(z)}=e^{0}=1$
I know i am supposed to use the Cauchy Reimen equaitions, but i can't figure out the next steps.
On
Notice that $f-g$ is a holomorphic function with value in $2i\pi\mathbb{Z}$. If $f-g$ is non-constant, then its image is an open set of $\mathbb{C}$, which is a contradiction. Therefore $f-g\equiv f(0)-g(0)=0$. Whence the result.
It is a bit overkill to use the open mapping theorem as your result hold for continuous functions; indeed, one can notice that the image of $f-g$ must be connected from the intermediate value theorem and since $2i\pi\mathbb{Z}$ is disconnected, then $f-g$ is constant.
$F(z) = f(z) - g(z)$ satisfies $$ e^{F(z)} = e^{f(z)-g(z)} = \frac{e^{f(z)}}{e^{g(z)}} = 1 $$ for all $z \in D$, therefore the derivative vanishes: $$ 0 = \left( e^{F(z)} \right)' = F'(z) e^{F(z)} \, . $$ The exponential function has no zeros, so $F'(z)=0$ in $D$, which means that $F = f-g$ is constant: $$ f(z) - g(z) = f(0) - g(0) = 0 \, . $$