Let $f,g$ be nonvanishing entire functions such that $f + g = 1$ for every $z\in \mathbb{C}$. Do $f$ and $g$ have to be constants themselves.
My attempt: $$f = 1-g \implies \frac{1}{f} = \frac{1}{1-g}$$ Since $f$ is non zero, $\frac{1}{f}$ is holomorphic and therefore defined everywhere. Thus $\frac{1}{1-g}$ is defined everywhere. That means that $1-g \neq 0$ and $g \neq 1$ for any $z\in \mathbb{C}$. But $g$ is an entire function that is non vanishing and $g \neq 1$. By Picard's little theorem if an entire function misses 2 points, it must be constant so $g$ and therefore $f$ must be constants as well.
I am not sure if my solution is correct and complete, so I would appreciate any feedback.
