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
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