Asin/acos/atan for Complex Numbers

Date: 3/27/96 at 2:43:43
From: Jordan
Subject: asin/acos/atan for complex numbers

Dr. Math,

I read a letter from cjjones where he asked, if Sin(a) = 5 then what 
is a?  From that letter I figured out that 
  sin(x+iy) = sin(x)cosh(y) + i*cos(x)sinh(y) 
and that 
  cos(x+iy) = cos(x)cos(y) - i*sin(x)sinh(y).  

But now I'm interested in going the other way.  How do you find 
asin(x+iy), acos(x+iy), and atan(x+iy)?

  Jordan

Date: 7/14/96 at 21:49:33
From: Doctor Jerry
Subject: Re: asin/acos/atan for complex numbers

For convenience, let z=x+iy.  To calculate arcsin(z) we solve the 
equation

  sin(w) = z, where w=u+iv.

Since sin(w) = (e^(iw)-e^(-iw))/(2i), we solve

  (e^(iw)-e^(-iw))/(2i) = z. 

Multiplying both sides of this equation by (2i)*e^(iw) we obtain

  e^(2iw)-1 = 2ize^(iw).

Letting p = e^(iw), this equation becomes the quadratic

  p^2 - (2iz)p - 1 = 0.

Solving for p = e^(iw),

  e^(iw) = iz + (1-z^2)^(1/2) or

  w = -i*ln(iz + (1-z^2)^(1/2)).

This is a formula for w = arcsin(z) = arcsin(x+iy).

I've given a "formal" argument and have skipped some details.  For 
example, (1-z^2)^(1/2) can have more than one solution. 

I hope my answer satisfies some of your curiosity.  To understand the 
argument down to the last detail, it probably would be necessary for
you to read an elementary book on complex variables.

-Doctor Jerry,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/