Is $\dfrac{\sin(z)}{z}$ surjective function on $\mathbb{C}\backslash \{0\}$?
I have tried using the power series representation for $\sin(z)$ and also using the formula for $\sin(z)$. But it didn't work. Any hint will be appreciated. Thanks in advance.
One solution using some entire function theory goes like this:
Let $f(w)=\frac{\sin \sqrt w}{\sqrt w}=1-\frac{w}{3!}+\frac{w^2}{5!}-...$.
It is well known (and easy to prove using that $\sin z$ is an entire function of order $1$) that $f$ is an entire function of order $\frac{1}{2}$ and obviously so is $f-a$ for any complex $a$ and all such have infinitely many roots, the number of which grows like $\sqrt R$ in the disc of radius $R$ centered at the origin - as entire functions of finite order with finitely many (possibly none of course) zeroes must be of integral order.
So fixing $a$ and picking some $w \ne 0$ for which $f(w)=a$ and picking a $z$ for which $z^2=w$, we get $\frac{\sin z}{z}=1-\frac{z^2}{3!}+\frac{z^4}{5!}-...=1-\frac{w}{3!}+\frac{w^2}{5!}-...=f(w)=a$ so we are done and the function $\dfrac{\sin(z)}{z}$ is indeed surjective on $\mathbb{C}\backslash \{0\}$
Edit later - let's sketch the proof that a finite order entire function with finitely many zeroes must have integer order (and finite non-zero type) - $f$ is entire of order $q$ with finitely many zeroes, so there is a polynomial $P$ with same zeroes (same multiplicities etc, taking $P=1$, if $f$ has no zeroes) So $g=\frac{f}{P}$ is entire of same order $q$ as polynomials have order zero, and $g$ has no zeroes, so $h=\log g$ is entire. But $g$ order $q$ means $h$ grows at most like $R^{q+\epsilon}$ at infinity for any $\epsilon >0$, which means $h$ is a polynomial $Q$ of degree $d \le q$. It then follows that the order of $g$, hence of $f$ is $d$ integer and also by looking at the leading coefficient of $Q$ which by definition is finite and non-zero, we get that $g$ and $f$ have finite non-zero type.