properties of analytic function $\sin z$ Lets have $f: z \mapsto \sin z$


*

*In what points on a plane is $f$ conformal


I know analytic functions are conformal where $f'(z)\neq 0$. So
$$f'(z)=\frac{ie^{iz}-(-i)e^{-iz}}{2i}=\frac{e^{iz}+e^{-iz}}{2}=\cos z$$
Now $\cos z=0$ when $z=\frac{\pi}{2}+n\pi, \;n\in \mathbb{Z}$ so is $\sin z$ is conformal when $z \neq \frac{\pi}{2}+n\pi, \;n\in \mathbb{Z}$?


*Show that it holds $|\sin z|=\sqrt{\sin^2 x+\sinh^2 y}$.


\begin{align}
|\sin z|&=\Big|\frac{e^{iz}-e^{-iz}}{2i}\Big| \\
&=\Big|\frac{e^{-y}e^{ix}-e^y e^{-ix}}{2i}\Big| \\
&=\Big|\frac{e^{-y}(\cos x+ i\sin x)-e^y (\cos (-x)+i\sin (-x))}{2i}\Big| \\
&=\Big|\frac{e^{-y}(\cos x+ i\sin x)-e^y (\cos x-i\sin x)}{2i}\Big| \\
&=\Big|\frac{\sin x(e^y+ e^{-y})}{2}+ \frac{i\cos x (e^y - e^{-y})}{2}\Big| \\
&=| \sin x \cosh y + i \cos x \sinh y | \\
&=\sqrt{(\sin x)^2 (\cosh y)^2 + (\cos x)^2 (\sinh y)^2} \\
&=\sqrt{(\sin x)^2 ((\cosh y)^2 - (\sinh  y)^2) + (\sinh y)^2} \\
&=\sqrt{(\sin x)^2 + (\sinh y)^2}
\end{align}
But I get stuck at this point. I know I need to get it to $|\operatorname{Re}(z)+i\operatorname{Im}(z)| =\sqrt{\operatorname{Re}(z)^2+\operatorname{Im}(z)^2}$


*Is $f$ bounded function in complex plane. 


Well I know that if $G \subset \mathbb{C}$ is a bounded set then there is point $z\in \mathbb{C}$ and radius $\epsilon > 0$, so that holds $A\subset B(z,\epsilon)$.
But $\sin z$ is entire (analytic in all points $z \in \mathbb{C}$) right? So it can't be bounded by any real sized disk or neighborhood right? I don't know a more sophisticated way to say this.
So any hints for 2. and critique for 1. and 2.?
 A: (1) is fine. For (2), use $e^{ix}=\cos x + i \sin x$ to separate real and imaginary parts. (3) follows from the expression in (2), since $\sinh y$ is unbounded.
A: As in Hans Lundmark's answer, your reasoning for point 1 is correct. For point 2. you're on the right track: now gather up the real and imaginary parts to get:
$$\begin{array}{lcl}|\sin z| &=& \frac{1}{2} |-2\,\sinh y\,\cos x + 2\,i\,\cosh y\,\sin x|\\&=&\sqrt{(\sinh y)^2\,(\cos x)^2 + (\cosh y)^2\,(\sin x)^2}\\&=&\sqrt{(\sinh y)^2+((\cosh y)^2-(\sinh y)^2)\,(\sin x)^2}\\&=&\sqrt{(\sinh y)^2+(\sin x)^2}\end{array}$$
using standard identities $\cosh^2-\sinh^2=\cos^2+\sin^2=1$.
For point 3. reason as in Hans's answer but here is another neat path. The Taylor series for $\sin z = z-z^2/3!+z^5/5!-\cdots$ with its factorials in each term's denominator, converges for all $z\in\mathbb{C}$. $\sin z$ is thus, as you know, an entire function, defined and finite for all $z\in\mathbb{C}$. Liouville's theorem, that the only entire function that is bounded as $z\to\infty$ (bounded and analytic on the extended complex plane $\mathbb{C}\bigcup\{\infty\}$) is a constant, then gives you the rest. $\sin z$ is patently not constant ($\sin 0 = 0$ whereas $\sin (\pi/2) = 1$). So therefore it cannot be bounded as $z\to\infty$. In fact, like $e^z$ it has an essential singularity at infinity. Going well beyond what you need, but interesting, the "Big Picard" theorem is that every function with an essential singularity assumes every value in $\mathbb{C}$, with possibly one exception, infinitely many times in any neighbourhood of the essential singularity. This too tells you that $\sin z$ must be unbounded as $z\to\mathbb{C}$.
