Logs of Complex NumbersDate: 02/11/2004 at 22:27:22 From: Brittany Subject: Log with complex numbers Give an example showing that Log(z1/z2) does not always equal Log(z1) - Log(z2) where z1 and z2 are complex numbers. Each time I try to plug in different complex numbers, such as z1 = e^i*pi/3 and z2 = e^2i*pi/3, I find that the two calculations are equal. I can't find a counterexample. Date: 02/12/2004 at 09:28:33 From: Doctor Vogler Subject: Re: Log with complex numbers Hi Brittany, The complex logarithm is an interesting function. Simpler complex functions are defined everywhere except at certain points called "poles" where they behave like divisions by zero. But the logarithm has a thing called a "branch" where you have to make a jump. Because the function f(z) = e^z has the property f(z + 2*pi*i) = f(z), there isn't only one inverse (or logarithm). Well, we pick one of those inverses for the function Log, but it causes there to be a jump. Think of it this way. If you make a 3-D graph with the complex plane on the x-y coordinates and the value of the imaginary part of the logarithm on the z coordinate, then what we really *should* get is a spiral that is flat when you go straight out from the center x = y = 0 but winds upwards as you go counterclockwise around the center, and increases by 2*pi each time around. But that's not a function because a function can only have one value, so we limit the domain by just taking one loop. That means the spiral starts on the negative x axis at z = -pi, does one counterclockwise loop and ends on the same negative x axis but higher, at z = pi. So there is this break where the complex log jumps by 2*pi along the negative x axis. So in order to make Log (z1/z2) different from Log(z1) - Log(z2), you just have to go over this break. In other words, find a Log(z1) whose imaginary part is negative near -pi and a Log(z2) whose imaginary part is positive near pi (or vice-verse), and then subtracting them will be on some other part of the logarithm spiral that was cut off in order to make Log a function. If you have any questions or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 02/12/2004 at 12:16:43 From: Brittany Subject: Log with complex numbers I tried z1 = e^i5pi/3 and z2 = e^i*pi/3 so that Log(z1) = -i*pi/3 and Log(z2) = i*pi/3, and when you subtract these, you get -i2pi/3, which is what I get when I calculate Log(z1/z2). Am I not looking at the correct z1 and z2? Date: 02/12/2004 at 12:42:29 From: Doctor Vogler Subject: Re: Log with complex numbers Hi Brittany, You need imaginary parts closer to pi and -pi. Try making Log(z1) = -i*2*pi/3 Log(z2) = i*2*pi/3 If you have any questions or need more help, please write back. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/