e^(i*pi) = -1: pi = 0 ?Date: 10/17/97 at 11:29:15 From: John K. Koehler Subject: Weird tricks Dr. Math, I know that from a certain trig identity we get the equation e^(i*pi) = -1. I have been playing around with this equation and have found some disturbing things. I hope that you can help me. First: e^(2*i*pi) = 1 e^(-2*pi) = 1 (raised both sides to i and 1^n is 1) -2*pi = 0 (took the ln of both sides and ln(1) = 0) pi = 0 ! Second: i*pi = ln(-1) {1} ln(-n) = ln(-1) + ln(n) therefore by {1} ln(-n) = i*pi + ln(n) Finally: i^2 = -1 e^(i*pi) = i^2 e^(-pi) = i^(2i) e^(-pi/2) = i^i e^(-pi/2) is a real number and therefore so is i^i Can you please tell me why I get such ridiculus results? I asked my calc2 teacher last year - but all he could tell me was that imaginary numbers don't work like reals, and I still don't see how that would change anything. Thanks in advance, John Date: 10/17/97 at 13:08:36 From: Doctor Bombelli Subject: Re: Weird tricks Well, John, not all the results you have are ridiculous! In fact, you and your teacher are both right; complex numbers don't always work like real numbers, but sometimes they do. Complex numbers can be written in many equivalent ways: z = x+iy z = re^*(it) where t is the angle the point (x,y) makes with the positive x axis. t is called the argument of z--arg(z) z = r[cos(t)+i sin(t)] In fact, the definition of e^(it) is cos(t)+i sin(t). This is why we have e^(-ipi) = cos(pi)+i sin(pi)= -1. We would like w = log(z) and z = e^w to match up, just as for real numbers. Let z = re^(it) and w = u+iv. z = e^w = e^(u+iv) = e^u*e^(iv) and z is also re(^it) Match the real part and the "i part" (the imaginary part). r = e^u and e^(it) = e^(iv) So u = ln(r) and v can be any value with cos(v) = cos(t) [since e^(it) is cos(t)+i sin(t)]. So we say, for complex numbers, ln(z)=ln(r)+i arg(z)+i2kpi (k an integer). In fact, ln(z) has many values. We also define, in this way, that z^a = e^[a ln(z)] when a is complex. (Check that this works for real numbers a, also.) Note that 1^a = e^[ln(1)] = e^[0+i2kpi], so only one of the answers is e^0 = 1. Here is your problem: e^(2*i*pi) = 1 e^(-2*pi) = 1 (raised both sides to i and 1^n is 1) -2*pi = 0 (took the ln of both sides and ln(1) = 0) pi = 0 In steps 2 and 3 you don't have one-to-one functions any more. (The reason e^x = 1 implies x = 0 is because the real logarithm function is one-to-one. It is kind of like why x^3 = 1 gives x = 1 and x^2 = 1 gives x = 1 or -1. The cube root is one-to-one, but the square root isn't.) Now in your second experiment: i*pi = ln(-1) {1} ln(-n) = ln(-1) + ln(n) therefore by {1} ln(-n) = i*pi + ln(n) This is okay. ln(-n) = ln(n) + i pi + i 2kpi from the definition above. In your third experiment: i^2 = -1 e^(i*pi) = i^2 e^(-pi) = i^(2i) e^(-pi/2) = i^i e^(-pi/2) is a real number and therefore so is i^i i^i is e^[i ln(i)], and ln(i) is ln(1) + i pi/2 + i 2kpi So... i^i = e^[0 + i^2 pi/2+ i^2 2kpi] = e^[- pi/2 - 2kpi] which is a real number. I commend you on playing around with this stuff. That is how new mathematics is learned, on all levels. -Doctor Bombelli, 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- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/