# Finding the inverse tangent of a complex number

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• bsaucer
In summary, the formula for finding the arctangent of z is w=arctan(z), where z=x+iy and w=u+iv. The desired results of u and v should be in terms of trigonometric and hyperbolic functions (and their inverses) and not in terms of logarithms. The formula \tan^{-1}z=\frac{i}{2}\log\frac{i+z}{i-z} is suggested but not preferred due to the use of logarithms. An alternative method is to use the identities \cos 2iy = \cosh 2y and \sin 2iy = i\sinh 2y and solve the equations \tan 2x \tanh
bsaucer
TL;DR Summary
Inverse Tangent of complex number in rectangular form.
Let z=x+iy, and w=u+iv. I am looking for a formula to find the arctangent of z, or w=arctan(z). I want the results of u and v to be in terms of trigonometric and hyperbolic functions (and their inverses) and not in terms of logarithms. The values u and v should be functions of x and y.

The formula
$$\tan^{-1}z=\frac{i}{2}\log\frac{i+z}{i-z}$$
seems useful to me but you do not like logarithm.

Last edited:
If you can get as far as $$e^{2iz} = \frac{w + 1}{w - 1}$$ then $$\begin{split} \cos 2z &= \frac{w^2 + 1}{w^2 - 1} \\ \sin 2z &= \frac{2w}{w^2 - 1}\end{split}$$ so the problem is reduced to solving $$\begin{split}\cos (2x + 2iy) &= A \\ \sin (2x + 2iy) &= B\end{split}$$ for $x$ and $y$. The left hand sides can be expanded using the angle sum formulae and the identities $$\cos 2iy = \cosh 2y, \qquad \sin 2iy = i\sinh 2y.$$ By taking ratios of real and imaginary parts we end up with $$\begin{split} \tan 2x \tanh 2y &= - \frac{ \operatorname{Im} A}{\operatorname{Re} A} \\ \tan 2x \coth 2y &= \frac{ \operatorname{Re} B}{\operatorname{Im} B}\end{split}$$ whence $$\begin{split} \tan^2 2x = -\frac{ \operatorname{Im} A \operatorname{Re} B}{ \operatorname{Re} A \operatorname{Im} B} \\ \tanh^2 2y = -\frac{ \operatorname{Im} A \operatorname{Im} B}{ \operatorname{Re} A \operatorname{Re} B}.\end{split}$$

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