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Let f,g and h be entire functions such that following equation holds: $e^f+e^g=h$. Then show that either h has no zeroes or has infinitely many zeroes.

There is a hint to the question: Apply Little Picard theorem on the function f-g.

Here is my attempt: from the equation we get $e^g(e^{f-g}+1)$=h. Now suppose there is a zero of h at suppose a $\in C$ . Now $e^g$ is never zero and so $e^{f-g}$=-1 at a $\in$ C. But since $e^{f-g}$ is not a polynomial entire function , a corollary of great Picard theorem will say that $e^{f-g}$=-1 for infinitely many complex numbers which will imply that h has infinitely many zeroes.

Is my approach fine?

And also I don't get the hints and I don't get how this problem is an application of Little Picard Theorem. Thanks in advance for any help!!!

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$h(z) = 0$ if and only if $f(z) - g(z) = (2n+1) \pi i$ for some $n \in \mathbb N$.

Can you apply Little Picard now?

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    Yeah I think I got this. If h has finitely many zeroes then f-g will omit infinitely many values of the form (2n+1)$\pi \iota$. By Liitle Picard f-g is constant. If f-g is the zero constant then h=2$e^g$ and hence never zero, if it is such a constant that $e^{f-g}$=-1 then h is identically 0 or else again h is nonzero. Am I right? – 2017-02-26
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    Yes, pretty much. Although, if $f-g$ is any constant that is not of the form $(2n+1)\pi i$, then $h$ is never zero - it doesn't have to be the zero constant. – 2017-02-26
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    yeah sure!! Thank you very much!! – 2017-02-26