# On the image of the complex sine and cosine

I have a question on the sine and cosine functions.

We know that in the complex world they are defined by

$$\cos(z) = \frac{e^{iz}+e^{-iz}}{2} \hspace{1cm} \mbox{ and } \hspace{1cm} \sin(z) = \frac{e^{iz}-e^{-iz}}{2i}.$$

as functions from $$\mathbb{C}$$ to $$\mathbb{C}$$.

My question is, are these functions surjective? I think the answer is yes, but i am not able to prove it.

If we want to check whether a function is surjective, we need to find some kind of "bound" between the value and the argument.

What do I mean, let's firstly look at $$cos(z) = \frac{e^{iz} + e^{-iz}}{2}$$

$$cos : C \to C$$, so we need to take any $$\omega \in C$$, and determine whether we can find some $$z_\omega \in C$$ such that: $$cos(z_\omega) = \omega$$

That means$$e^{iz_\omega} + \frac{1}{e^{iz_\omega}} = 2\omega$$

Let $$\psi = e^{iz} \ \$$and multiply everything by $$\psi$$ : $$\psi^2 - 2\omega\psi +1 = 0$$

By that we found 2 solutions for $$\psi$$ :

$$\psi_1 = \frac{2\omega - 2\sqrt{\omega^2 - 1}}{2} = \omega - \sqrt{ \omega^2-1}$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \psi_2 = \omega + \sqrt{ \omega^2 - 1 }$$

What finally leads us to : $$e^{iz_{\omega}} = \omega \pm \sqrt{ \omega^2 - 1 }$$

Which has a solution, because function $$\exp: C \to C$$ can take all values except 0. $$\ \ (*) \ \$$

With sine function u proceed really similar.

$$Let \ \omega \in C$$ $$\\$$ And we're looking for some $$z_\omega$$ such that: $$sin(z_\omega) = \omega$$

That is: $$e^{iz_\omega} - \frac{1}{e^{iz_\omega}} = 2i\omega$$

The same substitution ( just to have little easier part of solving ) and multiplying leads us to:

$$\psi^2 - 2i\omega\psi - 1 = 0$$

$$\psi = \frac{2i\omega \pm \sqrt{4i^2\omega^2 + 4}}{2} = i\omega \pm \sqrt{1-\omega^2}$$

And that $$\psi$$ cannot take value 0 (in $$\pm$$ case simultaneously), so we're done, cause there exist $$z_\omega$$ such that :

$$e^{iz_\omega} = i\omega \pm \sqrt{1-\omega^2}$$ ( Obviously even if both $$\pm$$ cases aren't equal to 0, u choose only one, but that's enough ).

So both cosine and sine are surjective (as a functions from $$C \to C$$ ), because we find a way, how to assign an argument for every point $$\omega$$.

PS: I've used the fact (in both sine and cosine part), that function $$\exp: C \to C$$) can take any value except 0 (which u might try to prove as an excercise (if u'll fail, I can), or if u already know that, that's really fine, and we're done)).