Lesson: The Arithmetic of Complex Numbers

Question 2

1 point

Transcript

The real numbers, being the numbers that we find in the real world, do a great many things for us, but there are some things they can’t do. The complex numbers were developed to be a number system in which polynomials can be completely factored. That is, if a polynomial is of degree n, then it will have n roots, counting multiplicity, in the complex numbers. In the real numbers, it might not have had any. In the end, it wasn’t so complicated to create the complex numbers, as it turns out that we only needed to add the solution to one polynomial to get them all—a solution to the polynomial x-squared = -1.

A complex number is a number of the form z = x + yi, where x and y are real numbers and i is an element such that i-squared = -1. The set of all complex numbers is denoted by C.

One of the first things we want to remember about the complex numbers is that they contain the real numbers. That is to say, any real number x can be thought of as a complex number, simply with the y in the +yi set equal to 0. As we extend definitions from the real numbers to the complex numbers, we will make note of the fact that in the case when we end up dealing with real numbers, the definitions are the same.

If z = x + yi, we say that the real part of z is x, and we write Re(z) = x, and we say that the imaginary part of z is y, and we write Im(z) = y. If Im(z) = 0, then z is a real number, and we sometimes say that z is purely real. If Re(z) = 0, then we say that z is purely imaginary.

Now, the part of a complex number attached to i is called imaginary since it doesn’t exist in the real world, but do not discount these numbers as mathematical nonsense, since they do lead to many useful results in our real world. It would take several courses to explore the full power of the complex numbers, so our goal for this course will simply be to familiarize you with them, and make some natural extensions of our study of real vector spaces to the new complex numbers. Of course, the first thing we need to do is define our basic arithmetic steps.

Addition of complex numbers z1 = x1 + y1i and z2 = x2 + y2i is defined by z1 + z2 will equal (x1 + x2) + (y1 + y2)i. Now, as promised, we see that if y1 = 0 and y2 = 0, then we have that (x1 + 0i) + (x2 + 0i) will equal (x1 + x2) + 0i, so addition on real numbers is unchanged. We also see that addition is basically done componentwise, as if C is simply a 2-dimensional vector space. We will explore that possibility later.

For now, we also need to define multiplication, and here we’ll see more parallels with the vector space P1 than with the vector space R2. Multiplication of complex numbers z1 = x1 + y1i and z2 = x2 + y2i is defined by z1z2 will equal (x1 + y1i)(x2 + y2i). Again, multiplying as though these were polynomials, we distribute, and we get that this equals x1x2 + x1y2i + x2y1i + y1y2(i-squared). Now we can collect our i terms in the middle, so we’ll have an x1x2 + (x1y1 + x2y1)i, but our last term is actually not an i term because i-squared = -1. And so now if we collect up our real parts, we get (x1x2 – y1y2), and then we add our imaginary part (x1y2 + x2y1)i.

Again, in the case when z1 and z2 are actually real numbers, and we have that y1 and y2 = 0, we would get that (x1 + 0i)(x2 + 0i) would equal (x1x2 – 0) + (0 + 0)i, which is x1x2 + 0i, so multiplication of real numbers is still the same. Also worth noting is the ease with which we can multiply a complex number z2 = x2 + y2i by a real number x1, since our formula would give us that x1(x2 + y2i) = (x1x2 – 0) + (x1y2 + 0)i, which is simply x1x2 + x1y2i. So multiplying by x1 is the same as distributing by x1.

Now let’s look at some examples. Using our definitions, we can compute the following: (4 + 5i) + (3 – 2i) will equal (4 + 3) + (5 – 2)i, which is 7 + 3i. (-3 + 2i) + 4(2 – i) = (-3 + 2i) + (8 – 4i), again using the fact that it is easy to distribute a real number, which equals (-3 + 8) + (2 – 4)i, which is 5 – 2i.

To look at a complex product, (5 + 2i)(-3 – 6i) will equal -15 – 30i – 6i – 12(i-squared), which equals (-15 + 12) + (-30 – 6)i, which equals -3 – 36i. In general, I don’t recommend simply memorizing the formula for complex multiplication—I always distribute it as though it were a polynomial.

Now let’s look at a full-on combination. (1 + i)(2 + 5i) + (2 + i)(7 – 2i). Well, first we’ll do our two products, so the first one equals 2 + 5i + 2i + 5(i-squared), and then our, add our second product, which is 14 – 4i + 7i – 2(i-squared). Doing the calculations inside, this will equal (-3 + 7i) + (16 + 3i), and this equals (-3 + 16) + (7 + 3)i, which is 13 + 10i.

Now, note that, as usual, we will write (x1 + y1i) – (x2 + y2i) to mean (x1 + y1i) + (-1)(x2 + y2i), which, of course, is the same as (x1 + y1i) + (-x2 – y2i).

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