A First Look at the Complex Modulus

Transcription

1 A First Look at the Complex Modulus After expanding the field of arithmetic by introducing complex numbers, and also defining their properties, Gauss introduces the complex modulus. We have treated the concept of the modulus earlier. The character of each number was discovered to be not self-evident, but, rather, depended on its relation with other numbers, i.e. different moduli. Now on the plane of complex numbers, how will we find the character of complex numbers, in other words, how do we determine whether a number is a quadratic, cubic, biquadratic, or non-residues? As a beginning in this investigation, let us take as an example a complex prime number 4 i. Given 4 i as a modulus, how will we determine the congruency of other number s relative to a complex modulus? How can we find least residues? We will begin by hearing from Gauss, in his 2nd treatise on biquadratic residues: Gauss writes: the schema of complex numbers is a system of equidistant points which are laid on equidistant straight lines such that the infinite plane is decomposed into infinitely many squares. Every number which is divisible by a complex number a + bi = m will likewise form infinitely many squares, whose sides are equal to a 2 +b 2 or whose area is equal to a 2 +b 2 ; the latter squares will have an inclined position to the former if neither of the two numbers a, b are equal to zero. 1 Let s see how this works for 4 i. How do the associates making up this grid relate to one another? Say 4 i is the modulus. Is 4 + i congruent to zero? That this is true is easily seen. Take a different pair of associates, say 1 + 4i and 4 i. 1 The geometric representation given here is one way to represent the arithmetic of complex 1

2 Geometrically it would appear that 1 + 4i is congruent to zero, modulus 4 i, as they both reach a point of zero, and yet these numbers combined do not equal zero like the last two. How about the simpler case of real numbers that are not equally distant from zero, as in the case of 2 and 37 mod. 5? There it is a question of divisibility, that the distance between the two numbers is a multiple of 5. What about in this case? Is 1 + 4i divisible by 4 i? Try this out geometrically on the grid. The modulus defines a given pathway through all other numbers, and a particular viewpoint of them all. Here, we can think of the spaces on the new grid we created as units. Proceeding from the origin, any of the associates could be considered our new unit. Say 1 + 4i is our unit. How many of repetitions of the unit is 4 i from 1 + 4i? What is the proportion between the units? What did you find? In this case 4 i is considered our downward vertical direction, akin to i on the former grid of unit one squares, thus it is one unit vertically down we need to go. The complex modulus presents, even in the simple case of two associates, a problem of complex multiplication. If (4 i)(i) = 1 + 4i, then 1 + 4i, our number, is divisible by the modulus, right? Therefore, 1 + 4i is congruent to zero modulus 4 i. If we want to demonstrate this arithmetically, we need to make the denominator a whole (1 + 4i)(4 + i) number. In this way it is easy to see that (4 i)(4 + i) = i. numbers. There are other physical forms which complex numbers can assume. 2

3 But further still, are all the points on the red grid divisible by any of the associates? Let s pick a new number, say 3+5i. Is it divisible by the associates? Try it geometrically first. If 4 i was our modulus, how many spaces in the red grid would we need to go to get to 3 + 5i? After you find this, show that it is true arithmetically as well. The following animation shows modulus 4 i being multiplied by 1 + i. [ANIMATION:multiplication.swf] Try this for some other numbers on the grid. After we have done this, we have shown that every number that corresponds to a point of a square is congruent to zero. This geometrically validates what Gauss said before, that every number which is divisible by a complex number a + bi = m will likewise form infinitely many squares. Therefore, every number, such as 4 i, and its associates will all be congruent to zero if any one of them is taken as modulus. 3

4 What about other numbers that don t lie on the points of the squares? Gauss writes: Every number which is not divisible with respect to modulus m will correspond to a point, which lies either within such a square, or on the boundary line of two squares; the latter case, however, can only occur if a, b have a common divisor... What are these numbers congruent to, relative to the modulus? Example : 6 + 2i? mod 4 i Think about this geometrically. Just now we found all the numbers which represented points of the squares of the red grid were congruent to zero. Here, relative to 6 + 2i again, is there another line that can be drawn to this point from a different zero? We are not watching a recent presidential campaign debate, but there are a lot of zeros to choose from. Lets pick a zero. How about 9 + 2i, i.e. (4 i)(2 + i). 2 The line drawn from our number divisible by the modulus, 9 + 2i, is 3. 2 Arising as a corollary from this, we here have a way to determine whether uneven complex numbers like 2 + 9i are divisible by any number and thus not prime. It is not hard to demonstrate that any complex number which connects two points on this grid will not be prime. 4

5 Since we are looking at the same point, the two numbers should be congruent relative to this number. Well, is 6 + 2i 3 mod 9 + 2i? Our arithmetic seems to be following along nicely. Try this same number 6 + 2i with a different modulus, say 4 i. The point drawn from 4 i would be 2 + 3i. 5

6 What about if our modulus is 1 + 4i? Again we have a third side of the triangle, 5 2i. Looking at all the associates, we have: 6 + 2i 5 2i mod 1 + 4i 6 + 2i 5 2i mod 1 4i 6 + 2i 2 + 3i mod 4 i 6 + 2i 2 + 3i mod 4 + i Of these two numbers that are congruent relative to the modulus, one of them is smaller. This brings us to a new question. What is the least possible length for any one of the points lying within a square? Given what we have discovered about these triangles, now let s look at one particular point in the square from the standpoint of many lines. 6

7 All the numbers, 6 4i, 2 3i, 3 6i, 2 2i, 5+3i, 1+2i, 6i, 4+5i, 3+i, 7 are congruent to one of the others relative to one of the associates of 1+4i. Which is one the least residue of 1+4i or one of its associates? Here the least residue is 1 + 2i. All other numbers are congruent to 1 + 2i relative to a modulus which is either one of the 4 associates of 1 + 4i, or some multiple of it. Thus, every number in the plane will be congruent to some least residue which is made up of one of the points within the square. We also have figured out something about a whole class of triangles. Notice that one side of each the triangle is the same. Therefore, in each triangle, the two sides are congruent to each other relative to the modulus. 7

8 How are all of these similar? Now, before going on, let s apply what we ve seen with the complex modulus and triangles on the plane, in a different way. What about a triangle which does not have one of its sides as one of the lines on our red grid. Look at these triangles. With all of these triangles, none of their sides are congruent with our modulus 4-i and its associates. This brings us to a more general case of the plane. Which side of these triangles would be the modulus? Couldn t any one of them be the modulus? Therefore, what does this say about triangles in the complex plane? In the complex domain, two sides of any triangle are always congruent to the third. Moving along, we have shown that every number in the plane will be congruent to one least residue. Extend this fact to the powers of complex numbers. Likewise any complex number taken to any power will be congruent to one of these least residues inside the square, zero, or the sides of the square in the case mentioned by Gauss where a and b have a common divisor. An example of this would be (1 + 2i) 2 i mod 1 + 4i [INSERT ANIMATION: FullPartOne8.swf] To continue with our investigation of least residues: how many are there? We started by finding what was congruent to 0, so we have one to start with. It is easy to count the rest inside the square: 16. We have a total of 17 least residues for modulus 1+4i and its associates. Before continuing forward, let 8

9 us remark what we have achieved in the little time playing around with the geometrical representation mentioned by Gauss: for any complex modulus, we have succeeded in reducing the investigation of every possible complex and real number to one of the points inside, or for the special case mentioned by Gauss, or on the boundary line of a square. Lastly, note that 17 is also the product, called the norm,of 1 + 4i and 1 4i, its conjugate. We will refer to the norm as p. Its general form is thus a 2 + b 2, the sum of two squares. Is every norm, the sum of two squares? Also, what would the area of the square be? If our two sides are 1 + 4i and 4 i, what is their product? Wait a second, how can an area be i? Hmmmm. That doesn t seem to work. Rather, if we want to describe the length of the sides of these squares, they would be the a 2 + b 2, and the area would thus be a 2 + b 2. What does this tell us about the nature of these complex numbers, that they cannot represent areas? Let us now proceed onward to investigate the least residues of a given complex modulus. 9