Associated Topics || Dr. Math Home || Search Dr. Math

### Why does ln(-x) x>0 equal ln(x)+pi*i?

Date: 6/5/96 at 23:19:55
From: Lucas W Tolbert
Subject: ln(-x) x>0

Could you please explain why the ln(-x) x>0 equals ln(x)+pi*i?

Date: 6/6/96 at 4:31:55
From: Doctor Anthony
Subject: Re: ln(-x) x>0

We have  -x = x(-1)
= x{cos(pi) + i.sin(pi)}
= x.e^(i.pi)

Take logs:  ln(-x) = ln(x) + ln(e^(i.pi)
= ln(x) + i.pi

In fact we could have found the general solution giving all the
possible branches for ln(-x), by putting -1 = cos(k.pi) + i.sin(k.pi)
with k taking all ODD integer values.

If we do this we get  ln(-x) = ln(x) + i.k.pi      k = 1, 3, 5, ....

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

Associated Topics:
High School Exponents
High School Imaginary/Complex Numbers

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search