Complex Numbers

Date: 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/