i^iDate: 04/03/97 at 21:13:32 From: Anonymous Subject: i^i What is i to the power of i -- i^i? Thanks, Guillemo Date: 04/04/97 at 05:42:23 From: Doctor Mitteldorf Subject: Re: i^i Dear Guillemo, One of the exciting things you discover when you explore complex numbers is that expressions like this have meaning! Expanding our number system from the reals to the complex numbers wouldn't have made much sense if we LOST some of the ability to do arithmetic operations. After all, we did it in order to GAIN the ability to take the square root of -1. A general theorem assures you that arithmetic operations on imaginary and complex numbers are all meaningful and well-defined - with exceptions like ln(0) and 1/0. The Rosetta stone is the Euler equation e^ix = cos(x) + i sin(x) You can verify that this equation makes a sensible definition by expanding the two sides as Taylor series in x. You can also differentiate both sides and see that the answer is self-consistent. Thirdly, you can use the formula for cos(2x) and sin(2x) to show that the right side has the property you expect from an exponential, so that e^i(2x) = (e^ix)^2. Once you believe the Euler equation, you've got a pretty easy time of it. If you let x=pi/2, then the real part is zero e^i (pi/2) = cos(pi/2) + i sin(pi/2) = i So i^i = (e^i*pi/2)^i = e^(i*i*pi/2) = e^(-pi/2). You may find it odd that not only is i^i well-defined, it's real! -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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