Determine $\arctan{e^i}$

Question

In this answer, the quantity $\arctan(e^i)$ must be determined.

It's obviously $e^i = \cos(1) + i\sin(1)$, but there is no formula for $\arctan(x + y)$ like $\sin(x + y)$, for example, and I'm stuck.

In the comments, the author suggests:

$$\arctan(e^i) = \frac{1}{2} \arctan \left( \frac{\cos (1)}{1 - \sin (1)} \right) + \frac{1}{2} \arctan \left( \frac{\cos (1)}{1 + \sin (1)} \right) + i \left[ \frac{1}{4} \log \left( 2 - 2 \sin(1)\right) - \frac{1}{4} \log \left( 2 + 2 \sin(1)\right) \right]$$

But, again, I am not able to obtain this from $\arctan(e^i) = \arctan \left[ \cos(1) + i\sin(1) \right]$.

1. What relation has been used?

And then, if I consider the real part of $\arctan(e^i)$ above, and use the arctan angle-addition formula, I obtain:

$$\arctan \left( \frac{\cos (1)}{1 - \sin (1)} \right) + \arctan \left( \frac{\cos (1)}{1 + \sin (1)} \right) = \arctan(w)$$

$$w = \frac{ \frac{\cos (1)}{1 - \sin (1)} + \frac{\cos (1)}{1 + \sin (1)} }{1 - \frac{\cos (1)}{1 - \sin (1)} \frac{\cos (1)}{1 + \sin (1)} } = \frac{ \frac{\cos (1) \left[ 1 + \sin (1) \right] + \cos(1) \left[ 1 - \sin (1) \right]}{1 - \sin^2(1) } }{\frac{1 - \sin^2(1) - \cos^2 (1)}{1 - \sin^2(1)}} = \frac{\cos (1) \left[ 1 + \sin (1) \right] + \cos(1) \left[ 1 - \sin (1) \right]}{1 - \sin^2(1) - \cos^2 (1)}$$

But this denominator is $0$.

2. How to proceed?

3. Any other method (not necessarily using the above steps) to obtain this result is ok.

Answer 1

You want to solve $\tan z=a$ where $a=e^{i}$. Note first that one expects the solution to be unique only modulo $\pi$.

$$\tan z=\frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}=\frac{w-1}{i(w+1)}$$ where $w=e^{2iz}$. $$\tan z=a\implies\frac{w-1}{w+1}=ia\implies \frac{2}{1+w}=1-ia \implies w+1=\frac{2}{1-ia}\implies w=\frac{1+ia}{1-ia}.$$ In your case, $$e^{2iz}=\frac{1+ie^i}{1-ie^i}=\frac{1+e^{i(1+\pi/2)}}{1-e^{i(1+\pi/2)}} =\frac{e^{-i(1/2+\pi/4)}+e^{i(1/2+\pi/4)}}{e^{i(1/2+\pi/4)}-e^{i(1/2+\pi/4)}} =i\frac{\cos(1/2+\pi/4)}{\sin(1/2+\pi/4)}=i\cot(1/2+\pi/4).$$ So $$2iz=2n\pi i+\frac{\pi i}2+\log\cot(1/2+\pi/4)$$ etc.

Answer 2

Let $$e^i=\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{1}{i}(\frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}})$$ which gives us $$ie^i=\frac{e^{2i\theta}-1}{e^{2i\theta}+1}$$and now solving for $e^{2i\theta}$ gives us $$e^{2i\theta}=\frac{i-e^i}{i+e^i}\to 1$$ and taking complex logarithm on both sides of equation 1 gives us $$\theta=\arctan{e^i}=\frac{1}{2i}\log(\frac{i-e^i}{i+e^i})$$which on simplification with the help of simple trigonometry gives $$\frac{i-e^i}{i+e^i}=\frac{i-\cos1-i\sin1}{i+\cos1+i\sin1}=\frac{i(1-\sin1)-\cos1}{i(1+\sin1)+\cos1}={\frac{\sin(\frac{\pi}{4}-\frac{1}{2})}{\cos(\frac{\pi}{4}-\frac{1}{2})}(\frac{i\sin(\frac{\pi}{4}-\frac{1}{2})-\cos(\frac{\pi}{4}-\frac{1}{2})}{i\cos(\frac{\pi}{4}-\frac{1}{2})+\sin(\frac{\pi}{4}-\frac{1}{2})}})=i\tan(\frac{\pi}{4}-\frac{1}{2})$$and the final answer converts to $$\theta=\arctan{e^i}=\frac{1}{2i}\log(i\tan(\frac{\pi}{4}-\frac{1}{2})$$ Hope that helps!