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How do i find domain of $\tan(z)$ that is differentiable??

I haven seen forums where people showed until this part $$ \tan(z) = \frac{\tan(x)+i\tanh(y)}{1−i\tan(x)\tanh(y)} , $$ where $z$ is complex.

what should I do next or is that any other method??

EditPiAf
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kevin
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  • I'm not sure what exactly you mean. Do you want the maximal connected open set containing a given point such that tan remains holomorphic there? Or do you actually want everywhere in $\mathbb{C}$ where tan is holomorphic? The latter is where cos is nonzero, which is everywhere except the obvious points on the real axis. – Ian Sep 26 '17 at 16:14
  • @Ian I just want to make it into a form such that I can apply the Cauchy-Riemann equation or if possible, other methods to prove tan(z) to be differentiable – kevin Sep 26 '17 at 16:25

2 Answers2

1

For complex differentiability, you need to check the Cauchy-Riemann equations.

You need to find the real part $U$ and the imaginary part $V$:

$$\mathrm f(x+\mathrm iy) = U(x,y) + \mathrm i V(x,y)$$

Then you need to check that both $U_x = V_y$ and $U_y = -V_x$.

In a post last year, the real and imaginary parts were shown to be \begin{eqnarray*} U(x,y) &=& \frac{\sin 2x}{\cos 2x + \cosh 2y} \\ \\ V(x,y) &=& \frac{\sinh 2y}{\cos 2x + \cosh 2y} \end{eqnarray*}

You now want to check where both $\frac{\partial U}{\partial x} = \frac{\partial V}{\partial y}$ and $\frac{\partial U}{\partial y} = -\frac{\partial V}{\partial z}$.

You also need $\tan z$ to be well-defined for it to be differentiable, i.e. $\cos 2x + \cosh 2y \neq 0$.

Fly by Night
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It is not necessary to invoke the CR equations. For real $x$ the $\tan$ is formally defined by $$\tan x:={\sin x\over\cos x}\ ,\tag{1}$$ and this does give indeed a real value, unless $\cos x=0$. The latter is the case when $x$ is an odd multiple of ${\pi\over 2}$. The natural extension of $\tan$ to ${\mathbb C}$ therefore is $$\tan z:={\sin z\over\cos z}=-i\>{e^{iz}-e^{-iz}\over e^{iz}+e^{-iz}}\ ,$$ and this coincides with $(1)$ when $z\in{\mathbb R}$. The differentiation rules for analytic functions then allow to conclude that $\tan$ is analytic in all points $z\in{\mathbb C}$ where $e^{iz}+e^{-iz}\ne0$. Now $$e^{iz}+e^{-iz}=-e^{-iz}\bigl(e^{i(2z-\pi)}-1\bigr)\ .$$ Here the RHS is $=0$ iff $$i(2z-\pi)=2k\pi i,\quad k\in{\mathbb Z}\ ,$$ or $$z=k\pi+{\pi\over2},\quad k\in{\mathbb Z}\ .$$ It follows that the extended $\tan$ is complex differentiable at all points of ${\mathbb C}$ minus the already known singular points on the real axis.

Christian Blatter
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