Welcome to NRICH.

 
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:-

eiz = 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.
Þz=tan(y)=sin(y)/cos(y)

by de Moivre,
Þz=-i(eiy-e-iy)/(eiy+e -iy)

Let t=eiy 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)

Þeiy = f(z)
Þiy = ln(f(z))
Þy = -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)|ei arg(f(z)))
= ln(|f(z)|) + i arg(f(z))

which is then in the form a + ib.

Hope this helps - it's a bit rushed!

Mike