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This is our school's previous Q.E. problem :

Let $p(z)$ be a polynomial. Then $\sin z=p(z)$ has infinitely many solutions in $\mathbb{C}$ iff $p(z)$ is constant.

One direction is easy ($\Leftarrow$), but for the other side I don't have any idea. Can you give some hints? Thanks.

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    Hint: the assertion is wrong. – 2017-01-06
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    @DanielFischer really? – 2017-01-06
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    Yes. If $q(z)$ is any polynomial, then $f(z) = \sin z - q(z)$ is an entire transcendental function. By Picard's theorem, $f$ attains every complex value, with possibly one exception, infinitely often. If $0$ happens to be such an exceptional value for $q$, then $\sin z - p_c(z)$, where $p_c(z) = q(z) + c$, has infinitely many zeros for every $c\in \mathbb{C}\setminus \{0\}$. – 2017-01-06

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Edit

Thanks to a comment I can correct the first answer.

You assertion is wrong, and you can prove that by contraposition.

We want to find a $p$ non constant such that $$\sin(z)=p(z)$$ as infinitely many solutions.

We can simply take $p(z)=z$.

You can look the details of this equation here.

Consider the function

$$f\colon z\in \mathbb C\mapsto \sin(z)-p(z).$$

The function $f$ is entire (holomorphic on $\mathbb C$). You then know that

$$\vert f(z)\vert \xrightarrow[\vert z\vert\to\infty ]{} +\infty.$$

So you can consider $R$ such that for all $z$ such that $\vert z\vert>R$, $f(z)\ne 0$.

Then you know that $f$ as finitely many zeros in $\bar B(0,R)$ because it is a compact and $f$ is a non null holomorphic function.

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    $f$ is transcendental, so we don't have $\lvert f(z)\rvert \to +\infty$ as $z\to \infty$. – 2017-01-06