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### Log of a Negative Number

```Date: 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

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/
```
Associated Topics:
College Imaginary/Complex Numbers
High School Imaginary/Complex Numbers
High School Logs

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