It is a result that is rather taken for granted

$$ \vert \sin (x)\vert, \vert\cos(x)\vert \leq 1 $$

for all reals x. However, today I wanted to bring up an interesting remark that these are in fact not true in the complex plane. Recall that an analytic continuation of the above two functions are made readily by Euler’s formula

$$ \sin(z):=\frac{e^{iz}-e^{-iz}}{2i}, \space \cos(z) := \frac{e^{iz}+e^{-iz}}{2}, \space z\in \mathbb{C} $$

It is easy to see that both are unbounded with the extended domain as C. Note that

$$ \vert\sin(z)\vert=\vert\frac{-e^{iz}-e^{-iz}}{2}\vert\geq\frac{\vert\vert -e^{iz}\vert-\vert -e^{-iz}\vert\vert}{2}=\frac{\vert e^y-\ e^{-y}\vert}{2} $$

, where we have taken z= x+iy. Now, note that along the rays

$$ z\rightarrow \infty \space \space on\space \arg(z) =\pm \pi/2 $$

The expression on the LHS grows exponentially large as exp (y) → infinity. So, we see that sin (z) is unbounded!

Similiary, can you find a reason why cos (z) is unbounded?