$\sin z$ and $\cos z$ are defined in terms of $e^{iz}$ if $z$ is a complex number . We know that $\sin$ and $\cos$ have range $[-1,1]$ in real system. How can one find the range of complex functions?
EDIT: Your comment suggests that both $\sin z$ and $\cos z$ have full complex plane as range. What is wrong in the following reasoning?
$\cos z = {(e^{iz} + e^{-iz})/2}$ , so $|\cos z| = |(e^{iz} + e^{-iz})|/2 $
$\leqslant(1 + 1)/2 = 1$ as $|e^{iz}|=1$
Best Answer
The range of $\cos$ is $\mathbb C$. In order to prove that, take a $w\in\mathbb C$ and solve the equation $\cos z=w$. Then\begin{align}\cos z=w&\iff e^{iz}+e^{-iz}=2w\\&\iff e^{2iz}-2we^{iz}+1=0\\&\iff \bigl(e^{iz}\bigr)^2-2we^{iz}+1=0.\end{align}So, solve the equation $Z^2-2wZ+1=0$ with respect to $Z$. If $Z$ is a solution, then $Z\neq0$ (because $0$ is not a solution) and now you take $z\in\mathbb C$ such that $e^{iz}=Z$.
The problem with your argument agains the assertion that the range of $\cos$ is $\mathbb C$ is that $|e^{iz}|=1$ is false in general, although it is true when $z\in\mathbb R$.