Show that $$\tan^{-1}(z) = \frac{i}{2}\ln\left(\frac{i + z}{i - z}\right)$$
I tried this approach: $$\tan(w) = z$$ $$\tan(w) = \frac{\sin(w)}{\cos(w)}$$ $$\tan(w) = \frac{\frac{e^{iw}-e^{-iw}}{2i}}{\frac{e^{iw}+e^{-iw}}{2}}$$ let $$ u = e^{iw}$$ $$\tan(w) = \frac{u - u^{-1}}{i(u + u^{-1})}$$
But I don't see a way from there
A useful trick: $\frac{a}{b}=\frac{c}{d} \iff \frac{a+b}{a-b}=\frac{c+d}{c-d}$
\begin{align*} i\tan w &= \frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}} \\ \frac{iz}{1} &= \frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}} \\ \frac{1+iz}{1-iz} &= \frac{e^{iw}}{e^{-iw}} \\ e^{2iw} &= \frac{1+iz}{1-iz} \\ 2iw &= \ln \frac{1+iz}{1-iz} \\ w &= \frac{1}{2i} \ln \frac{1+iz}{1-iz} \\ &= \frac{i}{2} \ln \frac{1-iz}{1+iz} \\ \tan^{-1} z &= \frac{i}{2} \ln \frac{i+z}{i-z} \end{align*}
For any $y\in\mathbb{R}$, we have $$ \log(1+iy)=\log\left(\sqrt{1+y^2}\right)+i\tan^{-1}(y)\tag{1} $$ Therefore, $$ \begin{align} \tan^{-1}(y) &=\frac1i\log\left(\frac{1+iy}{\sqrt{1+y^2}}\right)\\ &=\frac1{2i}\log\left(\frac{(1+iy)^2}{1+y^2}\right)\\ &=\frac1{2i}\log\left(\frac{1+iy}{1-iy}\right)\\ &=\frac i2\log\left(\frac{i+y}{i-y}\right)\tag{2} \end{align} $$ Since both sides of $(2)$ are analytic functions of $y$, $(2)$ can be extended at least to $y\in\mathbb{C}$ where $\left|y\right|\lt1$.
Using the identity $$ \tan^{-1}\left(\frac1y\right)=\frac\pi2-\tan^{-1}(y)\tag{3} $$ $(2)$ can be extended to $y\in\mathbb{C}$ where $\left|y\right|\gt1$.
Except for $y=\pm i$, $(2)$ can be extended to $\left|y\right|=1$ by continuity.
Thank you Ng Chung Tak for an excellent answer. I am trying to answer the Stack Exchange Mathematics Question, "Help with the indefinite integral $\int \frac{dx}{2x^4 + 3x^2 + 5}$", and I needed to put a few more steps into the answer to make it clear how you go from line to line. Otherwise, the answer does not makes sense to me, and I think also that a broader audience needs to be able to reference the results step-by-step since these answers are addressing new ground of how to compute the natural logarithm of a complex number in its simplest form.
In the process of answering that question, the difficult situation of "Taking the Inverse Tangent of a Complex Number" addressed here comes in to play, since the integrals involve $\int{\frac{1}{x^2+a^2}}$ where $a^2$ is a complex number, with two solutions for $x$, namely $x^2=(x^2)_+$ and $x^2=(x^2)_-$.
There are some conventions where $\int{\frac{1}{x}} = ln (x)$ is only defined for $x>0$. If $x$ is a complex number, this creates a problem, since the equality greater than zero is hard to define for a complex number. Other conventions have $\int{\frac{1}{x}} = ln (\left\lvert x \right\rvert)$. This answer provides a route for finding $\tan^{-1} z$ when $z<0$, as the final result does not depend on an absolute value of $z$. Indeed, it is well-know that $\tan(\pi)=-1$ so $\tan^{-1} (-1) = \pi = \frac{i}{2}ln\frac{1+z}{1-z}$ so $-{2 \pi}i=ln\frac{1+z}{1-z}$ and $1=\frac{1+z}{1-z}$ and $z=0$.
Since I am just providing some editing so I can understand your text, I am attempting to block-quote your contribution as well as the small edits going from step-to-step. Thank you for your help and insight!
Unfortunately, block quoting of the equations is not formatting correctly. So I will convey the quote start and finish as thus:
Beginning of Quote of Ng Chung Tak's Answer with Minor Step-by-Step Edits
A useful trick: $\frac{a}{b}=\frac{c}{d} \iff \frac{a+b}{a-b}=\frac{c+d}{c-d}$
\begin{align*} \tan w &= \frac{\sin \theta}{\cos \theta} \\ \sin \theta &= \frac{1}{2}(e^{(i \theta)} -e^{(i \theta)}) \\ \cos \theta &= \frac{1}{2}(e^{(i \theta)} +e^{(i \theta)}) \\ i\tan w = i \frac{\sin \theta}{\cos \theta} &= \frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}} \\ z=\tan \theta \underset{implies}{\implies} i z = i \tan \theta = \frac{iz}{1} &= \frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}} \\ 1+iz=\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}}+1 &= \frac{2*e^{iw}}{e^{iw}+e^{-iw}} \\ 1-iz=-\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}}+1 =\frac{-e^{iw}+e^{-iw}}{e^{iw}+e^{-iw}}+1 &= \frac{2*e^{-iw}}{e^{iw}+e^{-iw}} \\ \frac{1+iz}{1-iz} &= \frac{e^{iw}}{e^{-iw}} \\ e^{2iw} &= \frac{1+iz}{1-iz} \\ 2iw &= \ln \frac{1+iz}{1-iz} \\ w &= \frac{1}{2i} \ln \frac{1+iz}{1-iz} \\ &= \frac{i}{2} \ln \frac{1-iz}{1+iz} \\ \tan^{-1} z &= \frac{i}{2} \ln \frac{i+z}{i-z} \end{align*}
Finish of Quote of Ng Chung Tak's Answer with Minor Step-by-Step Edits
