Trig functions with complex variables

By Anonymous on Tuesday, March 2, 1999 - 02:18 pm :

While doing calculus, i came across the following problem.
how does one evaluate trig functions and their inverses, with complex variables.
e.g. arctan(xi) ; i=sqr

By Anonymous on Thursday, March 4, 1999 - 10:34 am :

Most of these problems can be sorted out if you start with DeMoivre's Theorem:-

e^iz = cos(z) + isin(z).

Start from there, and you should find it's a matter of algebraic manipulation. Have a go!

Mike Pearson

By Anonymous on Tuesday, March 9, 1999 - 02:47 pm :

i haven't had much success with the algebraic manipulation, and i wondered if you could show me it. how one does evaluate arctan(xi); i=sqrt(-1)
thanks

By Mike Pearson (Gmp26) on Tuesday, March 9, 1999 - 05:20 pm :

Hi anonymous!

A name would be nice, but no matter...

Ok, here's an outline

Let $y = \atan (z)$ , where both $z$ and $y$ are complex. $\Rightarrow z = \tan (y) = \sin (y) / \cos (y)$

by de Moivre, $\Rightarrow z = - i (e^{i y} - e^{- i y}) / (e^{i y} + e^{- i y})$

Let $t = e^{i y}$ and rewrite the above equation.

Multiply through by $t$ to yield a quadratic in $t$ .

Solve this for $t$ .

You should then have a couple of solutions of the form $t = f (z)$

$\Rightarrow e^{i y} = f (z)$

$\Rightarrow i y = \ln (f (z))$

$\Rightarrow y = - i (\ln (f (z)))$

So it reduces to the problem of taking the log of a complex number.

Use polar coordinates for $f (z)$ , so

$\ln (f (z)) = \ln (| f (z) | e^{i \arg (f (z))}) = \ln (| f (z) |) + i \arg (f (z))$

which is then in the form $a + i b$ .

Hope this helps - it's a bit rushed!

Mike