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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/