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I am studying the following exercise:

Show that the meromorphic functions on the Riemann sphere have the form $p(z)/q(z)$, where $p$, $q$ are coprime polynomials.

Is an exercise in Donaldson Riemann Surface.

I thought the following way to solve:

Let $f$ be a meromorphic function on the Riemann sphere. Let $\lambda_1,\ldots,\lambda_n$ be a points in $f^{-1}(0)$ such that $\lambda_i \neq \infty$. Let $\mu_1,\ldots,\mu_m$ be the points in $f^{-1}(\infty)$ in the same way. Then $$\frac{f(z)(z- \mu_1)\cdots(z-\mu_m)}{(z-\lambda_1)\cdots(z-\lambda_n)} =:g(z)$$ defines a meromorphic function with no zeros (except possibly at $\infty$) and no poles (except possibly $\infty$). Either way $g(z)$ either lacks zeros it lacks poles. So $\deg(g)=0$ and hence $g$ is constant. Then $f=p/q$ for some polynomials $p$ and $q$.

I wonder if you're reasonable. Any adjustment ??

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    I think there are several problems. 1) $g(z)$ may still have zeros or poles. For example, take $f(z)=1/z^2$. But to fix this, you could just add powers to your $z-\mu_i$ and $z-\lambda_i$. 2). I am not sure why $g(z)$ is constant. You could have a holomorphic function on $\mathbb C$ with no poles or zeros. Take $e^z$ for example. – 2017-01-15
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    Do you have any tips to give? – 2017-01-15
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    Well you can manufacture a meromorphic function $g$ on the Riemann sphere that is holomorphic on $\mathbb C$ by killing the poles. Examine the Laurent expansion of $g(1/z)$, and see that it cannot have an infinite tail (because $g$ is meromorphic on the Riemann sphere). So $g(z)$ is a polynomial. – 2017-01-15
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    Your argument is basically correct, except you need to take multiplicities into account. That is, if $\lambda_i$ is a zero of multiplicity $n_i$, you need to divide by $(z-\lambda_i)^{n_i}$, and likewise you need to multiply by $(z-\mu_j)^{n_j}$ for a pole of order $n_j$. – 2017-01-15
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    @Manoel $z\mapsto e^z$ is not meromorphic on all of $\mathbb C\cup\{\infty\}. \qquad$ – 2017-01-15

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To show that a function that is meromorphic in $\mathbb C\cup\{\infty\}$ and has no poles or zeros is constant, you can argue as follows.

First say that it's holomorphic in all of $\mathbb C$ because of the lack of poles. Then, since it's holomorphic at $\infty$, it has neither a pole nor an essential singuarity at $\infty$; hence it has some finite value there. Now you have a function that is continuous on the compact space $\mathbb C\cup\{\infty\}$ and that never takes the value $\infty$, so there is some open neighborhood of $\infty$ in which the function has no values. Thus it is bounded and holomorphic on $\mathbb C.$ Then use Liouville's theorem.