Proving surjectivity of $\cos(z)$ and $\sin(z)$ and find all $z : \cos(z) \in \mathbb R$ and all $z: \sin(z) \in \mathbb R$

I am trying to solve the following two problems:

1) Prove that the functions $\cos(z)$, $\sin(z)$ are surjective over the complex numbers.

2) Find all $z \in \mathbb C$: $cos(z) \in \mathbb R$ and find all $z \in \mathbb C$: $\sin(z) \in \mathbb R$.

For 1),I've tried to prove it for the function $\cos(z)$ (I suppose the other one is analogue) so I've used the fact that $\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}$.

Let $w \in \mathbb C$, I want to show there exists $z \in \mathbb C : f(z)=w$, i.e., $\dfrac{e^{iz}+e^{-iz}}{2}=w$, multiplying by $2$ and then by $e^{iz}$ yields $e^{2iz}+1=2we^{iz}$ iff $e^{2iz}-2we^{iz}+1=0$.

I don't know if this approach is the correct one but here I've replaced $e^{iz}$ by $x$, so the solutions of the equation would be the roots of the polynomial $p(x)=x^2-2wx+1$, by the quadratic formula, I get that $x \in \{\dfrac{2w+w_0}{2},\dfrac{2w-w_0}{2}\}$, where $w_0^2=4w^2-4$.

Then, $e^{iz} \in \{\dfrac{2w+w_0}{2},\dfrac{2w-w_0}{2}\}$. At this point I got lost, I would like to explicitly show that $z$ exists and I can't see existence directly from the fact that $e^{iz}=\dfrac{2w+w_0}{2}$ or $e^{iz}=\dfrac{2w-w_0}{2}$.

First I thought of taking logarithm of both sides of the equation ir order to solve for $z$, but this is not a legitimate operation unless $e^{iz} \in \mathbb R$.

I couldn't go any farther.

For point 2) I have no idea what to do, should I use the identity $\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}$?

I would appreciate some help with the two points (specially with point 2), at least in 1) I could do something).

Btw, Happy new year!

Solution 1:

1) Note that one of $\frac{2w+w_0}{2}, \frac{2w-w_0}{2}$ must be nonzero, and that $e^{iz}$ achieves all values except $0$, since for any $re^{i\theta}\in \Bbb C$ we have $re^{i\theta}=e^{i(\theta-i\ln r)}$ thus we have some $z$ such that $e^{iz}=\frac{2w+w_0}{2}$ or $e^{iz}=\frac{2w-w_0}{2}$.

2) If we write $z=x+iy$ then $$\cos(z)=\frac{e^{ix-y}+e^{-ix+y}}{2}=\frac{e^{-y}e^{ix}+e^y\overline{e^{ix}}}{2}$$ which has conjugate $$\frac{e^{-y}\overline{e^{ix}}+e^ye^{ix}}{2}$$ so $\cos(z)$ is real iff $$e^{-y}e^{ix}+e^y\overline{e^{ix}}=e^{-y}\overline{e^{ix}}+e^ye^{ix}.$$ If $e^{ix}$ is real then $x=n\pi$ for some $n\in\Bbb N$. Otherwise $e^{ix}$ and $\overline{e^{ix}}$ are linearly independent over $\mathbb R$, so $e^{-y}=e^y$ thus $y=0$.

The case of $\sin$ is similar for both problems.

Solution 2:

The complex logarithm is indeed what you can do when you have $e^z$ and you want to get $z$. As you did, the quadratic equation always has at least one solution (repeated if $w \in \{1,-1\}$). Choosing any of them, we only have to "take logs" of $e^{iz} = a$ for some $a \ne 0$ which you have to check. Now $e^{x+yi} = e^x \cdot e^{iy}$ (length times angle) and so we can represent any non-zero complex number as $e^{x+yi}$ for some $(x,y)$ (possible values of $y$ repeating every $2\pi$). This also shows how to get the exact solution set.

Finally, $\sin$ can as you guessed be solved analagously but you can also use the identity involving $\sin$ and $\cos$ being "translations" of one another.