8.1. Review of Complex NumbersFor a more detailed treatment of complex numbers, see ch. 3 of
James Nearing's free book at Figure A. Visualizing complex numbers as points in a plane. Figure C. A complex number and its conjugate. We assume there is a number, i, such that i^{2} = 1. The square roots of 1 are then i and i. (In electrical engineering work, where i stands for current, j is sometimes used instead.) This gives rise to a number system, called the complex numbers, containing the real numbers as a subset. Any complex number z can be written in the form z = a + bi, where a and b are real, and a and b are then referred to as the real and imaginary parts of z. A number with a zero real part is called an imaginary number. The complex numbers can be visualized as a plane, Figure A, with the real number line placed horizontally like the x axis of the familiar xy plane, and the imaginary numbers running along the y axis. The complex numbers are complete in a way that the real numbers aren't: every nonzero complex number has two square roots. For example, 1 is a real number, so it is also a member of the complex numbers, and its square roots are 1 and 1. Likewise, 1 has square roots i and i, and the number i has square roots \(1/\sqrt{2}+i/\sqrt{2}\) and \(1/\sqrt{2}i/\sqrt{2}\). Complex numbers can be added and subtracted by adding or subtracting their real and imaginary parts, Figure B. Geometrically, this is the same as vector addition. The complex numbers a+bi and abi, lying at equal distances above and below the real axis, are called complex conjugates. The results of the quadratic formula are either both real, or complex conjugates of each other. The complex conjugate of a number z is notated as \(\bar{z}\) or \(z^*\). The complex numbers obey all the same rules of arithmetic as the reals, except that they can't be ordered along a single line. That is, it's not possible to say whether one complex number is greater than another. We can compare them in terms of their magnitudes (their distances from the origin), but two distinct complex numbers may have the same magnitude, so, for example, we can't say whether 1 is greater than i or i is greater than 1. Example 88◊ Prove that \(1/\sqrt{2}+i/\sqrt{2}\) is a square root of i. ◊ Our proof can use any ordinary rules of arithmetic, except for ordering. \begin{align} (\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})^2 &= \frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\cdot\frac{i}{\sqrt{2}} \\ &+ \frac{i}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\cdot\frac{i}{\sqrt{2}} \\ &= \frac{1}{2}(1+i+i1) \\ &= i \end{align} Example 88 showed one method of multiplying complex numbers. However, there is another nice interpretation of complex multiplication. We define the argument of a complex number, Figure D, as its angle in the complex plane, measured counterclockwise from the positive real axis. Multiplying two complex numbers then corresponds to multiplying their magnitudes, and adding their arguments, Figure E. Figure D. A complex number can be described in terms of its magnitude and argument. Figure E. The argument of uv is the sum of the arguments of u and v. SelfCheck:Using this interpretation of multiplication, how could you find the
square roots of a complex number? Example 89The magnitude z of a complex number z obeys the identity \(z^2=z\bar{z}\). To prove this, we first note that \(\bar{z}\) has the same magnitude as z, since flipping it to the other side of the real axis doesn't change its distance from the origin. Multiplying z by \(\bar{z}\) gives a result whose magnitude is found by multiplying their magnitudes, so the magnitude of \(z\bar{z}\) must therefore equal z^{2}. Now we just have to prove that \(z\bar{z}\) is a positive real number. But if, for example, z lies counterclockwise from the real axis, then \(\bar{z}\) lies clockwise from it. If z has a positive argument, then \(\bar{z}\) has a negative one, or viceversa. The sum of their arguments is therefore zero, so the result has an argument of zero, and is on the positive real axis [1]. [1] I cheated a little. If z's argument is 30 degrees, then we could say \(\bar{z}\)'s was 30, but we could also call it 330. That's OK, because 330+30 gives 360, and an argument of 360 is the same as an argument of zero. This whole system was built up in order to make every number have square roots. What about cube roots, fourth roots, and so on? Does it get even more weird when you want to do those as well? No. The complex number system we've already discussed is sufficient to handle all of them. The nicest way of thinking about it is in terms of roots of polynomials. In the real number system, the polynomial x^{2}1 has two roots, i.e., two values of x (plus and minus one) that we can plug in to the polynomial and get zero. Because it has these two real roots, we can rewrite the polynomial as (x1)(x+1). However, the polynomial x^{2}+1 has no real roots. It's ugly that in the real number system, some secondorder polynomials have two roots, and can be factored, while others can't. In the complex number system, they all can. For instance, x^{2}+1 has roots i and i, and can be factored as (xi)(x+i). In general, the fundamental theorem of algebra states that in the complex number system, any nthorder polynomial can be factored completely into n linear factors, and we can also say that it has n complex roots, with the understanding that some of the roots may be the same. For instance, the fourthorder polynomial x^{4}+x^{2} can be factored as (xi)(x+i)(x0)(x0), and we say that it has four roots, i, i, 0, and 0, two of which happen to be the same. This is a sensible way to think about it, because in real life, numbers are always approximations anyway, and if we make tiny, random changes to the coefficients of this polynomial, it will have four distinct roots, of which two just happen to be very close to zero. I've given a proof of the fundamental theorem of algebra on page 162.
