# Prove that $\cos(z)$ and $\sin(z)$ are surjective over the complex numbers. [duplicate]

I have an exercise that says:

(a) Prove that $\cos(z)$ and $\sin(z)$ are surjective functions from $\mathbb C \to \mathbb C$.

(b) Find the solutions of the equation $\cos(z)=\dfrac{5}{4}$.

As far as part (a) goes, I have no idea how to show surjectivity. I know by definition that $\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}$ and $\sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}$. I want to show that given $w \in \mathbb C$, there are $z_1, z_2 \in \mathbb C$ such that $\cos(z_1)=w$ and $\sin(z_2)=w$. I've tried to play with the expressions from above but I couldn't get to anything. I need help on this.

For part (b), once I've proved (a), then I know (b) makes sense, i.e., there exist solutions for that equation.

$\cos(z)=\dfrac{5}{4}$ iff $\dfrac{e^{iz}+e^{-iz}}{2}=\dfrac{5}{4}$ Again, I have no idea what can I get out from this equation.

• We have $\sin(x+iy)=\sin x\cosh y+i \cos x\sinh y$, similarly for the cosine.
– Pedro
Jun 19, 2014 at 3:50
• You said $\frac45$ at the top, but later $\frac 54$... Jun 19, 2014 at 4:12
• @ThomasAndrews I've edited, thanks. Jun 19, 2014 at 4:19
• are they injective too? Feb 14, 2022 at 4:46

(a) $$c=\dfrac{e^{iz}+e^{-iz}}{2} \quad\text{iff}\quad 2cw=w^2+1 \quad\text{with}\quad w=e^{iz}$$ Now solve the quadratic to find $w$ from $c$, which is easy, and then find $z$ from $w$, which may be a little more delicate.
For $c=5/4$ we get $w^2-(5/2)w+1=0$ and so $w=1/2$ or $w=2$. Since these are real, the solutions of $w=e^{iz}$ are $z=-i \log(w)+2 k \pi=\mp i \log(2)+2 k \pi$.