How to prove the following complex inequalities:
$$|\sin(zx)|\leq e^{|Imz|x}$$ $$|e^{izy}\sin(z(x-y))|\leq e^{|Imz|x}, y\leq x$$ $$\left|\dfrac{\sin(zx)}{z}\right|\leq axe^{|Imz|x}$$
where $a$ is a constant, $z$ is complex number and $x$ is non-negative real number.
I think this can be proved by the Taylor series expansion but I could not figure out why this constant $a$ is appearing and how to separate the imaginary part of higher powers of $z$ in Taylor series.
Any alternative way will be highly appreciated.
For the first one, let $z=u+iv$ where $u=\text{Re}(z)$ and $v=\text{Im}(z)$. Then, we have
$$\begin{align} |\sin(xz)|&=\left|\sin(x(u+iv))\right|\\\\ &=\frac12|e^{-xv}e^{ixu}+e^{xv}e^{-ixu}|\\\\ &\le \frac12\left(e^{xv}+e^{-xv}\right)\\\\ &\le e^{|xv|}\\\\ &=e^{x|v|}\\\\ &=e^{x|\text{Im}(z)|} \end{align}$$
You can handle the rest analogously by just using straightforward analysis.