How can I show the complex sine function is unbounded? My question is really simple. How can I show intuitively to my complex analysis students that the sine function is unbounded? What kind of behavior makes the complex sine function different from the real one in this sense?
 A: The function $$\sin : z \mapsto \sin(z)$$ is holomorphic on $\mathbb{C}.$ Hence, if it is bounded, it is constant by Liouville's theorem. It is of course a contradiction. 
A: Here is a picture displaying $|\sin(x+iy)|$ as a function of $x$ and $y$ that may be appealing to students: the $x$ variable conveys the circular trigonometry part ($-2 \pi < x < 2 \pi$)  and the $y$ variable ($-1 < y < 1$)  accounts for the hyperbolic trigonometry part:
]1
Explanation: using Euler formulas: $$\sin(x+iy)=\sin(x)\cosh(y)+i \cos(x)\sinh(y)$$
The square of the modulus of this expression is $$(\cosh(y)\sin(x))^2+(\cos(x)\sinh(y))^2$$
which is arbitrary large: for example, if $x=\pi/2$, it is equal to $$\cosh(y)^2$$
A: This is one approach you can provide as far as boundedness of sine function is concerned. We know that $$\sin^{2}z + \cos^{2}z = 1\tag{1}$$ for all $z\in\mathbb{C}$ and if $z \in \mathbb{R}$ then $\cos z, \sin z$ are also real and then from the above equation it follows that $\sin z, \cos z$ are bounded.
If $z$ is not real then $\cos z, \sin z$ are not real and hence the equation $(1)$ does not really lead us to the conclusion that $\cos z, \sin z$ are bounded. In fact more generally if $a \in \mathbb{C}$ then $b = \sqrt{1 - a^{2}} \in \mathbb{C}$ no matter how large $|a|$ is. So one thing is clear: if $z \in \mathbb{C} - \mathbb{R}$ then one should not expect that $\sin z, \cos z$ are bounded.
Next one should be able to show that $\sin z, \cos z$ are actually unbounded and this is easily done by noting that $|\sin ix| = \sinh x, |\cos ix| = \cosh x$ for all real $x > 0$ and these tends to $\infty$ as $x \to \infty$.  
