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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.

Thanks in advance!

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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)).

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