# TrigonometryInverse tangent function

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Do we have identities for the following

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

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

#### MarkFL

Staff member
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

##### Well-known member
MHB Math Helper
I am looking for the real and imaginary part of

$$\arctan(x+iy)$$

#### Prove It

##### Well-known member
MHB Math Helper
-

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

##### Well-known member
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

$\chi$ $\sigma$

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