The Log of a Negative NumberDate: 03/27/98 at 12:37:55 From: Brett Witty Subject: Complex powers Hi, I ran into a revision problem with logs which wasn't the problem itself, just a gaping hole in my learning. They had given us what turned out to be a log of a quadratic, with a x-3 factor or something. I was checking my answers like the good student I was, but my calculator couldn't do log (-3). I guessed that the answer had to be complex. I consulted my Maths coordinator and he agreed with me and said that (for example), log -10 equalled i. I had a play around at home and devised that x^i = -x (where x is a real number) and a few other rules for complex numbers for powers. Am I right? I tried to look up several maths databases for verification, but they didn't have anything. I congratulate you all for all the helpful info and hard work you guys put into the site. It's great help. Cya, Brett W Date: 04/01/98 at 09:44:56 From: Doctor Barrus Subject: Re: Complex powers Hi, Brett! You're partially right. To start out with, the logarithm of a negative number in base 10 is complex. I ran into a problem in verifying what your coordinator said, though. Here's what I did, using the math software program Maple V (release 5): (In the following, I'll use log[a](b) to mean the logarithm with base a of b. For example, log[2](8) = 3, since 2^3 = 8.) First I used the base-conversion rule of logarithms: log[a](b) = log[c](b) / log[c](a) to write: log[10](-10) = log[e](-10) / log[e](10) = ln(-10) / ln(10) (I did this because Maple can only work with complex logs when the logarithm base is e.) Next I found the value of ln(-10): ln(-10) = ln(10 * -1) = ln(10) + ln(-1) (Using the sum rule of logs) To deal with the ln(-1), I used Euler's Identity: e^(i*x) = cos(x) + i*sin(x) In particular, e^(i*pi) = cos(pi) + i*sin(pi) = -1 + i*0 = -1 (I'm not sure how much math you've learned. The Euler identity is something that you can derive using Taylor series, which you learn about in calculus. Also, if you haven't covered trigonometry yet, you'll have to take my word for it that when you're measuring angles in radians (not degrees), cos(pi) = -1 and sin(pi) = 0). Now, since e^(i*pi) = -1, we can write: -1 = e^(i*pi)) ln(-1) = ln(e^(i*pi)) ln(-1) = i*pi So, back to our equation: ln(-10) = ln(10) + ln(-1) = ln(10) + (i*pi) = 2.303 + 3.1412*i (about) which is the numerical answer Maple gives. Now ln(10) is about 2.303, so (going back to our first formula): log[10](-10) = ln(-10)/ln(10) ln(10) + pi*i = --------------- ln(10) = 1 + [pi/ln(10)]*i = 1 + 1.364*i (about) Now, raising 10 to this power gives -10, but this is not equal to i. So log(-10) is not equal to i. Using Maple, I calculated 10^i and got -0.668 + 0.744*i (approximately), which, as you can see, is not equal to -10. So your rule isn't quite correct, but you're right about the logs of negative numbers being complex. Using the steps I used to find out log(-10), you can find out log[a](-a) for any a. You might give it a shot. Good luck! -Doctor Barrus, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/