Have look at this plot. Is $\sinh^2(x)$ bounded?
You should know that $|f^2(z)|=|f(z)|^2$
So if $\sin(z)$ were bounded how does that yield a contradiction?
Have look at this plot. Is $\sinh^2(x)$ bounded?
You should know that $|f^2(z)|=|f(z)|^2$
So if $\sin(z)$ were bounded how does that yield a contradiction?