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### 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/

```
Associated Topics:
College Imaginary/Complex Numbers
College Trigonometry
High School Imaginary/Complex Numbers
High School Trigonometry

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