Here comes the question:

(a) Solve the equation $\sin z = 2$.

(b) Express $\arcsin = \sin^{-1}$, $\arccos = \cos^{-1}$, $\arctan = \tan^{-1}$ in terms of $\ln$.

(c) Describe the ranges (images) of $\sin$, $\cos$ and $\tan$.

I have solved parts (a) and (b) by letting $z=\sin w = {e^{iw}-e^{-iw}\over2i}$, $z=\cos w = {e^{iw}+e^{-iw}\over2i}$ and $ z=\tan w={e^{iw}-e^{-iw}\over e^{iw}+e^{-iw}}$ and solving for $w$. And I obtained

\begin{align}\arcsin z &= -i\ln(iz\mp \sqrt{1-z^2})\\ \arccos z &= -i\ln(iz\mp \sqrt{-1-z^2})\\ \arctan z &= -{i\over2}\ln\left({z + 1 \over 1-z}\right)\end{align}

But I am stuck in part c. I think I am supposed to use part (b) in some way but I couldn't figure it out. Any help is appreciated.


In the case of $\cos$ and $\sin$, you already have an answer here.

It turns that $\tan\mathbb{C}=\mathbb{C}\setminus\{\pm i\}$.Take $w\in\mathbb{C}\setminus\{\pm i\}$. Let $w'$ be a square root of $w^2+1\neq0$. Since $\cos$ is surjective and $w'\neq0$, there is a $z_0\in\mathbb C$ such that $\dfrac1{\cos z_0}=w'$. Therefore, $\dfrac1{\cos^2z_0}=w^2+1$. But$$\frac1{\cos^2z_0}=w^2+1\iff 1+\tan^2z_0=w^2+1\iff\tan^2z_0=w^2.$$Therefore, $\tan z_0=w$ or $\tan z_0=-w$ and, in the second case, $\tan(-z_0)=w$, since $\tan$ is an odd function.

Note that we can't have $\tan z=\pm i$, because it would follow that $1+\tan^2z=0$, which is impossible, since $1+\tan^2z=\dfrac1{\cos^2z}$.

  • $\begingroup$ So is the image of tan $\mathbb{C}\setminus\{\pm i\}? I didn’t quite understand what you explained here can you please be more expository? $\endgroup$ Nov 28 '17 at 6:59
  • $\begingroup$ @CemSarıer My second sentence was “It turns out that $\tan\mathbb{C}=\mathbb{C}\setminus\{\pm i\}$.” Is there something ambiguous or unclear about it? And, no, I cannot be more expository than this. But if you have a specific doubt, I will try to explain what I wrote. $\endgroup$ Nov 28 '17 at 7:11


The range of the cosine is the domain of the arc cosine.

From your formula, the arc cosine is defined everywhere, except when the argument of the logarithm is zero.

But $$iz\pm \sqrt{1-z^2}=0$$ is not possible (multiplying by the conjugate you get $-z^2-(1-z^2)=0$.)

This answers for the cosine.

  • $\begingroup$ I read something about Branch Cuts which seem relevant to this but is it neccessary? Or is the way that you showed enough? $\endgroup$ Nov 28 '17 at 6:51
  • $\begingroup$ @CemSarıer: this is a local analysis, so branches have no impact. $\endgroup$
    – user65203
    Nov 28 '17 at 8:26

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