## Complex Numbers, Part 1

Complex numbers extend the real numbers () to include a second element, yi:

z = x + yi

where x and y are real numbers and i is the imaginary unit that is defined in terms of its square, i² = -1. Yes, this does mean that i is the square root of -1. All this time you’ve been told that taking the square root of a negative number is illegal. It is if your calculations are limited to being real numbers, but with this special definition of i, it is now legal. Welcome to the world of complex numbers! The symbol for the set of all complex numbers is ℂ.

With this new definition and this new domain of numbers, there are new definitions and rules for operations. Operations on complex numbers are mostly extending the rules you already know for real numbers. First, let’s look at some definitions.

Just as x is the default variable used for an unknown real number, z is usually used for complex numbers. If z = x + yi, x is called the real part of z and y is called the imaginary part. There is shorthand notation for this:

Re(z) = x, Im(z) = y

where x and y are real numbers: x ∈ ℝ and y ∈ ℝ.

Now if Im(z) = 0, then you just have a real number. So real numbers are a subset of complex numbers. That is ℝ ⊂ ℂ

So were complex numbers created just to give maths students something to do? Absolutely not! (Though it would be nice to have some quiet time by giving my students complex problems). Since there is an extra number in a complex number, there is additional information there that can describe a physical quantity. Most often in science and engineering, this extra information relates to a phase angle or a frequency. I will show examples of this in future posts. Though the imaginary part of a complex number is called “imaginary”, a complex number can describe a real physical quantity.

Expanding our working maths domain to complex numbers, allows for solutions to equations that previously had “no solution”. Consider

x² +4x + 13 = 0

This equation would previously be said to have no solutions. But now, we can find complex solutions. So perhaps we should change the variable to

z² +4z + 13 = 0

Using the quadratic formula to solve this gives

$z = \frac{-b \pm \sqrt{{b}^{2}-4ac}}{2a} =\frac{-4 \pm \sqrt{{4}^{2}-4(1)(13)}}{2(1)} = \frac{-4 \pm \sqrt{-36}}{2}$

Now -36 = (-1)(36) and the rule that the square root of a product is the product of the square roots applies:

$z =\frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm \sqrt{36}\sqrt{-1}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i$

Before we can confirm that -2 ± 3i is a solution to the equation, we need to know how to do basic operations on complex numbers. That will be the subject of my next post.