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e^(pi*i) = -1: A Contradiction?

Date: 8/17/96 at 20:28:0
From: Anonymous
Subject: e^(pi*i) = -1: A Contradiction?

I have seen that e^(i*Pi) = -1. But if you square both sides of 
the equation, you get e^(2*i*Pi) = 1. Taking the ln of both sides, 
you get 2*i*Pi = 0. But 2*i*Pi does not equal 0! Please explain.

Date: 8/18/96 at 16:13:10
From: Doctor Anthony
Subject: Re: e^(pi*i) = -1: A Contradiction?

No.  But the real part of 2*i*pi does equal 0.

We have e^(ix) = cos(x) + i*sin(x)

This is the equation that you start with.  Now if x = pi this gives

             e^(i*pi) = cos(pi) + i*sin(pi)

this gives   e^(i*pi) = -1, 

Taking logs,     i*pi = ln(-1)

  So           2*i*pi = 2*ln(-1)         
 Returning now to the apparent contradiction when you said that on 
squaring we get e^(2*i*pi) = 1  so taking logs  2*i*pi = ln(1)  you 
are correct so far, but now in the complex domain, ln(z) is a 
multivalued function having an infinity of 'branches', just as for 
example sin^(-1){x} is multivalued.

In complex numbers we have 

     e^w = z = r*e^(i*theta)

Now since e^(i*theta) = cos(theta) + i*sin(theta)                           

                      = cos(theta+2k*pi) + i*sin(theta+2k*pi)  k = 0, 
+or-1, +or-2, +or-3,..etc., etc.

So we can write  e^w = z = r*e^(i*(theta+2k*pi))   Now taking logs
     w = ln(z) = ln(r) + i*theta + i*2k*pi

So the full expression for ln(1) is found by putting r=1 and theta=0. 
This gives  ln(1) = i*2k*pi   with k = +or-1, +or-2 and so on.  

You can see the real part of ln(1) is 0, but there is the complex part 
which is not zero.                                                 
-Doctor Anthony,  The Math Forum
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Associated Topics:
College Imaginary/Complex Numbers
High School Imaginary/Complex Numbers
High School Transcendental Numbers

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