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

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

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

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

Ok, here's an outline

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

by de Moivre, Þ 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)

Þ e^{i y}=f(z)

Þ i y=ln(f(z))

Þ 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^iarg(f(z)))=ln(|f(z)|)+iarg(f(z))

which is then in the form a+i b.

Hope this helps - it's a bit rushed!

Mike