Complex trig functions and images

Question

Does $\sin(z)$ omit any values in $\mathbb{C}$?

My thoughts

Obviously it takes on all real value between $-1$ and $1$. Furthermore, I can show that it takes on another real value such as $4$.I believe $\sin(z)$ will not omit any values since I will always end up with a quadratic which is always solvable in $\mathbb{C}$. Is this correct?

Does $\tan(z)$ omit any values in $\mathbb{C}$?

My guess is similar to above.

Am I on the right track?

Edit

Here is some of previous work as my justification

Find all values of $z$ such that $\tan(z)=3i$

$$\begin{align*} \frac{i(e^{iz}+e^{-iz})}{e^{iz}-e^{-iz}}&=3i\\ e^{iz}+e^{-iz}&=3(e^{iz}-e^{-iz})\\ -2e^{iz}&=-4e^{-iz}\\ e^{iz}&=2e^{-iz}\\ e^{-iz}(e^{2iz}-2)&=0\\ e^{2iz}&=2 \end{align*}$$

So we have $e^{-2y}=2$ so $y=-\frac{\ln(2)}{2}$ and $x=0+2\pi k$ since $\arg(2)=0$ and $\arg (e^{2iz})=2x$

Find all values of $\sin(z)=2$

We consider the exponential formulation of $\sin(z)$ and we have

$$\begin{align*} \frac{e^{iz}-e^{-iz}}{2i}&=3\\ e^{iz}-e^{-iz}&=6i\\ e^{-iz}(e^{2iz}-1)&=6i\\ e^{2iz}-1&=6ie^{iz} \tag{let $w=e^{iz}$}\\ w^2-6iw-1&=0 \end{align*}$$

From the quadratic equation we have

$$\frac{6i \pm \sqrt{-36+4}}{2}=3i \pm 2\sqrt{2}i$$

so $e^{iz}=(3 \pm 2\sqrt{2})i$

Answer 1

You want to solve the equation $\sin z=w$. Write $$ \sin z=\frac{e^{iz}-e^{-iz}}{2i} $$ so you want to see whether the equation $$ e^{2iz}-2iwe^{iz}-1=0 $$ has a solution for every $w\in\mathbb{C}$. Since $0$ is not a root of $t^2-2iwt-1=0$, you are done, because $\exp$ only omits the value $0$.

For the tangent, you want to solve $$ \frac{e^{2iz}-1}{e^{2iz}+1}=iw $$ that is $$ (1-iw)e^{2iz}=1+iw $$ which is solvable provided $iw\ne-1$ and $iw\ne1$, that is $w\ne i$ and $w\ne-i$.

Answer 2

Picard theorem:

Little Picard Theorem: If a function $f : \mathbb C \to \mathbb C$ is entire and non-constant, then the set of values that $f(z)$ assumes is either the whole complex plane or the plane minus a single point.