I am able to show that $\displaystyle \mid (sin z)^2 \mid = sin^2 x + sinh^2 x$ but I am a hard time trying to prove that the complex sine function is not bounded using this.
could someone please help
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?