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 Check out our web site! http://mathforum.org/dr.math/ |
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