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 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/
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