MHB Do we have identities for the inverse tangent function for complex numbers?

AI Thread Summary
The discussion focuses on identities for the inverse tangent function, particularly for complex numbers. It explores the relationship between the arctangent of sums and differences, leading to a derived formula involving the tangent of the difference. The participants also seek to determine the real and imaginary parts of the arctangent of a complex number, presenting a logarithmic representation and series expansions for cases where the modulus is less than one. The derived expressions for the real and imaginary parts are presented as infinite series involving cosine and sine functions. Overall, the thread delves into complex analysis and the properties of the arctangent function in the complex plane.
alyafey22
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Do we have identities for the following

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

$$\arctan(x)-\arctan(y) = $$
 
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For the second one, we could write:

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

Take the tangent of both sides:

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

Apply the angle-difference identity for tangent on the left

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

Hence:

$$\tan^{-1}(x)-\tan^{-1}(y)=\tan^{-1}\left(\frac{x-y}{1+xy} \right)+k\pi$$
 
I am looking for the real and imaginary part of

$$\arctan(x+iy) $$
 
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ZaidAlyafey said:
I am looking for the real and imaginary part of

$$\arctan(x+iy) $$

[math]\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*}[/math]
 
ZaidAlyafey said:
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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