Asin/acos/atan for Complex NumbersDate: 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/ |
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