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?


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

  • $\begingroup$ Unfortunately, proving that some values are attained is not sufficient, but of course it can give ideas about how to prove it for all or for finding the ones that are missed. $\endgroup$ – egreg Feb 28 '16 at 22:44

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 $w\ne-i$.


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.

  • $\begingroup$ So the sine at most omits one value; but does it? $\endgroup$ – egreg Feb 28 '16 at 17:14
  • $\begingroup$ nice question! that is discussed in @egreg's post by the way. $\endgroup$ – AccidentalFourierTransform Feb 28 '16 at 17:19
  • $\begingroup$ Cool theorem but we don't have it our disposal yet. Thanks though $\endgroup$ – RhythmInk Feb 28 '16 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.