# Proving \frac{i}{2}\lsn\frac{x+i}{x-i}=\arctan x

Proving $$\displaystyle{\frac{{{i}}}{{{2}}}}{l}{s}{n}{\frac{{{x}+{i}}}{{{x}-{i}}}}={\arctan{{x}}}$$

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usaho4w
HINT:
Let $$\displaystyle{\arctan{{x}}}={y}\Rightarrow{x}={\tan{{y}}}$$
$$\displaystyle{\frac{{{x}+{i}}}{{{x}-{i}}}}={\frac{{{\sin{{y}}}+{i}{\cos{{y}}}}}{{{\sin{{y}}}-{i}{\cos{{y}}}}}}={\frac{{{i}{\left({\cos{{y}}}-{i}{\sin{{y}}}\right)}}}{{-{i}{\left({\cos{{y}}}+{i}{\sin{{y}}}\right)}}}}=-{e}^{{-{2}{i}{y}}}$$
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John Koga
I don't know if I'd call it ''cool'' but you could just do it by using the complex exponential representations of trig functions.
$$\displaystyle{\sin{{\left({x}\right)}}}={\frac{{{e}^{{{i}{x}}}-{e}^{{-{i}{x}}}}}{{{2}{i}}}}$$ $$\displaystyle{\cos{{\left({x}\right)}}}={\frac{{{e}^{{{i}{x}}}+{e}^{{-{i}{x}}}}}{{{2}}}}$$
which comes from using $$\displaystyle{e}^{{{i}{x}}}={\cos{{\left({x}\right)}}}+{i}{\sin{{\left({x}\right)}}}$$. So
$$\displaystyle{\tan{{\left({x}\right)}}}={\frac{{{1}}}{{{i}}}}{\frac{{{e}^{{{i}{x}}}-{e}^{{-{i}{x}}}}}{{{e}^{{{i}{x}}}+{e}^{{-{i}{x}}}}}}={\frac{{{1}}}{{{i}}}}{\frac{{{e}^{{{2}{i}{x}}}-{1}}}{{{e}^{{{2}{i}{x}}}+{1}}}}$$
Now set
$$\displaystyle{y}={\frac{{{1}}}{{{i}}}}{\frac{{{e}^{{{2}{i}{x}}}-{1}}}{{{e}^{{{2}{i}{x}}}+{1}}}}$$
and solve for $$\displaystyle{x}$$ to get a formula for arctangent...
Vasquez

Identity is :
$$\arctan z=\frac{i}{2}\ln\frac{i+z}{i-z}$$
Proof :
$$\\\tan z=\frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})} \\e^{i\frac{i}{2}\ln\frac{i+z}{i-z}}=(\frac{i+z}{i-z})^{-\frac{1}{2}} \\e^{-i\frac{i}{2}\ln\frac{i+z}{i-z}}=(\frac{i+z}{i-z})^{\frac{1}{2}} \\\tan(\frac{i}{2}\ln\frac{i+z}{i-z})=\frac{1-\frac{i+z}{i-z}}{i(1+\frac{i+z}{i-z})}=z$$