Finding the inverse tangent of a complex number

[bsaucer]

TL;DR

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

 

[pasmith]

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

 

[bsaucer]

It's been a while, but I'm back. After some algebraic and trig manipulation, here is what I came up with:

tan^-1(z) = tan^-1(x+iy) = ½ tan^-1(2x/(1-x²-y²)) + i ½ tanh^-1(2y/(1+x²+y²))

Please verify if I'm correct...