Prove that for any $z \in \Bbb{C} \backslash \Bbb{R}$
$$\limsup_{r \to\infty}|\sin(rz)| = \infty,$$ where $r \in \Bbb{R}^+$.
In the previous part of the question I showed that $$\sin(x+iy)=\sin(x)\cosh(y)+\cos(x)\sinh(y)i.$$ I'm supposed to use this to prove this question, but I'm not really sure how to.
Any help is greatly appreciated.
On
You have $$\sin(z)=\dfrac{e^{iz}-e^{-iz}}{2}$$
Hence $$\sin(iz)=\dfrac{e^{-z}-e^{z}}{2}=-\sinh(z)$$
Yet you know that :
$$\limsup\limits_{r \to \infty} f(rz)=\infty \Leftrightarrow\exists(r_n)\in\mathbb{(R^+)}^{\mathbb{N}},\lim\limits_{r \to \infty} f(r_nz)=\infty $$
You also know that $$\forall z \in \mathbb{C}-\mathbb{R}, z=x+iy \\ \text{and} \ \sin(rz)=\sin(rx)\cosh(ry)+\cos(rx)\sinh(ry)$$
Take $$r_n=2n\pi/x$$ can you end ?
We have that $$|\sin(rz)| = \left|\frac{e^{irz}-e^{-{irz}}}{2}\right| \geq \left|\frac{|e^{irz}|-|e^{-irz}|}{2}\right| = \left| \frac{e^{r\text{Re}(z)}-e^{-r\text{Re}(z)}}{2}\right|.$$
One of those converges to infinity and the other to $0$ when $r\to \infty$, the sign of $\text{Re}(z)$ determines which term converges where.