If $f$ is entire and the composition $f\circ g$ has a simple pole at a finite point, then $f$ is linear 
Let $f(z)$ be entire and $g(z)$ be analytic on $D(0,1)-{0}$. If $f(g(z))$ has a pole of order one at $z=0$, then $f(z)$ is linear and $g(z)$ has a pole of order one at $z=0$. 

This question came up on a qualifying exam I took. I've never really seen anything like it, so I'm anxious to know whether my solution is even relevant. Of particular concern to me is that I didn't really involve the punctured disk domain of $g$. 
SKETCH OF SOLUTION: Since $f(g(z))$ has a pole of order 1 at $z=0$, $zf(g(z))$ can be extended to a function analytic on $D(0,1)$ by the Riemann Removable Singularity Theorem. Thus $\frac{d}{dz} zf(g(z))=zf'(g(z))g'(z)+f(g(z))$ for $z \not = 0$, and some complex number at $z=0$ which is the limit of the former equation as $z$ goes to zero (as derivatives of analytic functions are themselves analytic, hence continuous). 
Therefore (nontechnically, for brevity) the term $\frac{b_{-1}}{z}$, $b_{-1} \not = 0$ that comprises the principal part of $f(g(z))$ must be "added out" by $zf'(g(z))g'(z)$. Hence $f'(g(z))$ has a pole of order 2 and $(g'(z))$ is analytic on $D(0,1)$, or $g'(z)$ has a pole of order two and $f'(g(z))$ is analytic on $D(0,1)$, which could only happen if $f'(g(z))$ and hence $f'(z)$ is constant (since $f(g(z))$ has a pole of order one). If the former case applies, $g'(z)$ analytic on $D(0,1)$ implies $f(g(z)))$ is analytic on $D(0,1)$ contradicting the assumption. Hence $f'(g(z))$ and thus $f'(z)$ is constant, hence $f(z)$ is linear; and $g'(z)$ has a pole of order two at $z=0$ meaning $g(z)$ has a pole of order one at $z=0$. 
Again, this is a rough sketch of my actual answer, but I at least want to know if my solution here made strides to solve the problem (even if there was a more elegant solution down some other lines), or if it is fatally flawed in some way. 
 A: Your solution does make strides, but it does not cover  all possibilities for $ f'(g(z))g'(z)$ to have a pole of order $2$ at $0$. Maybe  both $g'$ and $f'\circ g$ have an essential singularity there? Those can "cancel out" to produce a pole, like $z^{-2}e^{1/z}$ and $e^{-1/z}$. 
There is also the possibility of one of two factors having a zero at $0$. 
To repair the proof, begin by   excluding the possibility of $g$ having essential singularity at $0$. Indeed, if there is a sequence $z_n$ converging to $0$ such that $g(z_n)$ is bounded, then $f(g(z_n))$ is bounded,   contradicting the assumption that it has a pole. Hence, $|g(z)|\to \infty$ as $z\to0$. This means $g$ has a pole at $0$. Let's say its order is $k$.  Then $g'$ has a pole of order $k+1$ at $0$. 
Then follow your approach: $$\frac{d}{dz} zf(g(z))=zf'(g(z))g'(z)+f(g(z))$$
where the function on the left is holomorphic at $0$, hence $f'(g(z))g'(z)$ has a pole of order $2$ there. Since $g'$ has a pole of order $\ge 2$, it follows that $f'(g(z))$ is bounded near $0$. Hence, $f'$ is bounded near $\infty$. Therefore $f'$ is constant, and the conclusion follows.  
