Fnding the ln of a Negative Number

Date: 07/22/97 at 12:32:32
From: Cat Fyle
Subject: Modern Algebra

I have a test tomorrow in Modern Algebra, but the test is going to 
consist of complex numbers. I have not had complex numbers, so
I have no idea where to begin. One of the questions is going to have
something to do with finding the ln of a negative number.  Can you 
give me a general formula for this problem?  Like ln(-3)?  

Thank you very much, CAT

Date: 07/22/97 at 12:58:26
From: Doctor Rob
Subject: Re: Modern Algebra

ln(x) = y means x = e^y.  For what y is -3 = e^y?  

You need to know the formula e^(a+i*b) = e^a*[cos(b) + i*sin(b)] 
for real a and b. Then e^a = 3 and cos(b) + i*sin(b) = -1 leads to 
a = ln(3) and b = Pi*(2*k+1) for any integer k.  Thus

   y = ln(-3) = ln(3) + i*Pi*(2*k+1)  for any integer k.

There is not one answer, but many (as with arctangent and other
"functions" you have seen before).

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