Finding the inverse tangent of a complex number

[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.

 

[anuttarasammyak]

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

 

[pasmith]

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