Complex NumbersDate: 11/23/97 at 19:53:47 From: Stuart Zykorie Subject: Complex numbers Do all complex numbers have a multiplicative inverse? In other words, if you have a region R defined by R=(a+b(square root of -5)) where a and b are defined by all real integers, does every element of R have a multiplicative inverse. In my findings, it would seem that this is true. However, I've been told that some do not and those that do are called the units of the complex numbers. Stuart Z. Date: 11/24/97 at 16:20:23 From: Doctor Bruce Subject: Re: Complex numbers Hello Stuart, You are combining two very different questions here. You asked if all complex numbers have a multiplicative inverse. Except for 0, the answer is yes. Let x + iy be a complex number, so x and y are real numbers and at least one of x,y is not zero. Then the number x/(x^2 + y^2) - iy/(x^2 + y^2) is the multiplicative inverse of x + iy. But you also mentioned the set R = {a + b*sqrt(-5) | a,b are real integers}. Every nonzero element of R has a multiplicative inverse as a complex number, by what we just said. But in general that inverse is not itself an element of R. The only elements of R with this property are +1 and -1; there are no others. It is pretty easy to prove this: just show that if a/(a^2 + b^2) - ib/(a^2 + b^2) is an element of R, then we must have b = 0 and a = +1 or -1. In the study of Algebraic Number Theory, we introduce the concept of the Norm of a complex number. The Norm of a + bi, for example, is a^2 + b^2. We show that the Norm operates on elements of sets like your R and gives ordinary integers for values. Furthermore, the Norm is multiplicative. This means Norm(a + b*sqrt(-5)) * Norm(c + d*sqrt(-5)) = Norm[(a + b*sqrt(-5)) * (c + d*sqrt(-5))]. Elements of a set of complex numbers which have Norm equal to 1 are called "units" of the set. So we see that elements of a set which have a multiplicative inverse in the set are just the units of the set. -Doctor Bruce, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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