Why is this function entire? $f(z) = z^{-1} \sin{z} \exp(i tz)$ In problem 10.44 of Real & Complex Analysis, the author says $f(z) = z^{-1} \sin{z} \exp(i tz)$ is entire without explaining why. My guess is that $z = 0$ is a removable singularity, $f(z) = 1$ and $f'(z) = 0$, but I cannot seem to prove it from the definitions of limit and derivative. The definition of derivative gives:
$$
\left|\dfrac{\sin z \exp(itz)}{z^2}\right|
$$
Is my intuition correct? How can I prove that the above goes to $0$ as $z \to 0$?

Note that $u:z\mapsto\sin(z)/z$ is indeed entire since $u(z)=\sum\limits_{n=0}^{+\infty}(-1)^nz^{2n}/(2n+1)!$ has an infinite radius of convergence. Multiplying $u$ by the exponential, also entire, does not change anything.
