Proving $\rm \frac i 2 \ln\frac {x+i}{x-i}=\arctan x$ .

Question

Proving $$\rm \frac i 2 \ln\frac {x+i}{x-i} =\arctan x$$

I'd like to prove this identity without taking the derivatives and integrating, what are some cool ways to prove this?

Answer 1

HINT:

Let $\arctan x=y\implies x=\tan y$

$\dfrac{x+i}{x-i}=\dfrac{\sin y+i\cos y}{\sin y-i\cos y}=\dfrac{i(\cos y-i\sin y)}{-i(\cos y+i\sin y)}=-e^{-2iy}$

Answer 2

Identity is : $$\arctan z=\frac{i}{2} \ln\frac{i+z}{i-z}$$

Proof :

$$\tan z = \frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}$$

$$e^{i\frac i2 \ln\frac{i+z}{i-z}}=(\frac{i+z}{i-z})^{-\frac 12} $$ $$e^{-i\frac i2 \ln\frac{i+z}{i-z}}=(\frac{i+z}{i-z})^{\frac 12}$$

$$\tan(\frac i2\ln\frac{i+z}{i-z}) = \frac{1-\frac{i+z}{i-z}}{i(1+\frac{i+z}{i-z})} = z$$

Answer 3

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. $$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i} \quad \cos(x) = \frac{e^{ix}+e^{-ix}}{2}$$ which comes from using $e^{ix} = \cos(x)+i\sin(x)$. So $$\tan(x) = \frac{1}{i}\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}} = \frac{1}{i}\frac{e^{2ix}-1}{e^{2ix}+1}$$ Now set $$y = \frac{1}{i}\frac{e^{2ix}-1}{e^{2ix}+1}$$ and solve for $x$ to get a formula for arctangent...