The function $f$ defined by
$$
f(z) = \frac{\sin z}{z}
$$
is nonconstant, entire, of order $1$, and of genus either $0$ or $1$. Since $f(z)$ is real for real $z$ and all zeros of $f$ are real, Laguerre's theorem on separation of zeros implies that all zeros of $f'$ are real (and separated by the zeros of $f$).
I figured I'd give a more self-contained answer, essentially replicating the proof of Laguerre's theorem.
We have
$$
f(z) = \frac{\sin z}{z} = \prod_{n \neq 0} \left(1 - \frac{z}{\pi n}\right) e^{z/(\pi n)}.
$$
From this we get
$$
\frac{f'(z)}{f(z)} = \sum_{n \neq 0} \left(\frac{1}{z - \pi n} + \frac{1}{\pi n}\right),
$$
so that, if $z = x+iy$,
$$
\operatorname{Im}\left(\frac{f'(z)}{f(z)}\right) = -y \sum_{n \neq 0} \frac{1}{(x-\pi n)^2+y^2},
$$
which cannot vanish unless $y = 0$. Thus $f'$ has only real zeros.
(Additionally, by showing that $[f'(x)/f(x)]' < 0$ where $f(x) \neq 0$, one can see that $f'/f$ has exactly one zero between each pair of zeros of $f$.)
The Taylor series for $\frac{\sin(z)}{z}$ has all its terms real and positive when $z=i$, and the corresponding terms are the same in magnitude for any $z$ on the unit circle, so the maximum has to occur for $z=i$.
Best Answer
The image is the whole complex plane minus the interval $[-1,1]$ (on the real axis) and the negative imaginary axis.
Here is a figure illustrating the mapping. (The color-coding is explained at the top of that page.)