Trigonometry Inverse tangent function

[ZaidAlyafey]

Do we have identities for the following

\(\displaystyle \arctan(x+y) = \)

\(\displaystyle \arctan(x)-\arctan(y) = \)

 

[MarkFL]

For the second one, we could write:

\(\displaystyle \tan^{-1}(x)-\tan^{-1}(y)=\theta\)

Take the tangent of both sides:

\(\displaystyle \tan\left(\tan^{-1}(x)-\tan^{-1}(y) \right)=\tan(\theta)=\tan(\theta-k\pi)\) where \(\displaystyle k\in\mathbb{Z}\)

Apply the angle-difference identity for tangent on the left

\(\displaystyle \frac{x-y}{1+xy}=\tan(\theta-k\pi)\)

Hence:

\(\displaystyle \tan^{-1}(x)-\tan^{-1}(y)=\tan^{-1}\left(\frac{x-y}{1+xy} \right)+k\pi\)

 

[ZaidAlyafey]

I am looking for the real and imaginary part of

$$\arctan(x+iy) $$

 

[Prove It]

-

I am looking for the real and imaginary part of

$$\arctan(x+iy) $$

\(\displaystyle \displaystyle \begin{align*} \arctan{(z)} &= \frac{i}{2} \ln {\left( \frac{i + z}{i - z} \right) } \\ &= \frac{i}{2} \left[ \ln{ \left| \frac{i + z}{i - z} \right| } + i \arg{ \left( \frac{i + z}{i - z} \right) } \right] \\ &= -\frac{1}{2}\arg{ \left( \frac{i + z}{i - z} \right) } + i \left( \frac{1}{2} \ln{ \left| \frac{i + z}{i - z} \right| } \right) \end{align*}\)

 

[chisigma]

I am looking for the real and imaginary part of

$$\arctan(x+iy) $$

For $|z|<1$ is...

$\displaystyle \tan^{-1} z = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n + 1}\ z^{2n + 1}\ (1)$

... and setting $\displaystyle z= x+ i y = \rho\ e^{i\ \theta}$, if $\rho <1$ You obtain...

$\displaystyle \mathcal{Re}\ \{\tan^{-1} z \} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n + 1}\ \rho^{2n + 1}\ \cos (2n+1)\ \theta\ (2)$

$\displaystyle \mathcal {Im}\ \{\tan^{-1} z \} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n + 1}\ \rho^{2n + 1}\ \sin (2n+1)\ \theta\ (3)$

Kind regards

$\chi$ $\sigma$