|


Log of a Negative NumberDate: 11/26/2002 at 18:41:03 From: Corey French Subject: Log of a negative number I have heard my Alg II teacher and numerous other people say that the Log of any negative number is not a real solution. Therefore, logically, it must be a complex number. When I put ln(-1) into my calculator it gives me 3.14159i (pi i). This intrigued me and I began to wonder how my calculator arrived at this answer. Can you explain how to find the log of a negative number (using complex numbers)? Thanks! Corey French
Date: 11/26/2002 at 21:27:53
From: Doctor Wigton
Subject: Re: Log of a negative number
Hi Corey,
The natural logarithm of a complex number, z, is defined as:
ln (z) = ln |z| + i * arg(z)
Where arg(z) (also known as the phase) is given by:
z = x + iy
arg(z) = arg(x + iy) = arctan(y / x)
In other words, arg(z) is the angle made when x and y are plotted on
the unit circle, with the x axis as the reals and the y axis
representing imaginary values. It looks like this:
imag(z)
|
|
|
---------------- real(z)
|
|
|
Now, z = -1 in the case of ln(-1). The real part of this is -1 because
it is all real. Similarly, the imaginary part of -1 is 0, because
there is nothing imaginary about it. So x = -1 and y = 0. We now plot
that on our set of axes for complex numbers.
imag(z)
|
|
(-1, 0) |
O--------------- real(z)
|
|
|
Now look at the angle our point z makes with the x axis. It is 180
degrees, or pi radians.
Thus arg(-1) is pi. Your calculator may also be able to give you this
if you have an "angle" function.
Back to our original definition of natural log of a complex number.
ln(z) = ln |z| + i * arg(z)
ln(-1) = ln |-1| + i * arg(-1)
ln(-1) = ln (1) + i * pi
ln(-1) = 0 + i * pi
ln(-1) = i * pi
This should be able to get you started. Feel free to write again
with more questions.
- Doctor Wigton, The Math Forum
http://mathforum.org/dr.math/
|
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]


Ask Dr. MathTM
© 1994-2008 The Math Forum
http://mathforum.org/dr.math/