Complex Numbers, Part 5

Blast from the past: remember the quadratic formula

\[
x = \frac{-b \pm \sqrt{{b}^{2}-4ac}}{2a}
\]

and you would look at the discriminant (the b² – 4ac part) to determine if there were 1, 2, or no solutions? You were told that if b² – 4ac < 0, then there were no solutions. Well that was a lie (please don’t ask me about Santa Claus). It turns out that there are always solutions to a quadratic equation

ax² + bx + c = 0, where a, b, and c are real and a ≠ 0.

Let’s look at z² + 4z + 29 = 0 and solve for z. Using the quadratic formula, we get

\[z = \frac{-4 \pm \sqrt{{4}^{2} – 4(1)(29)}}{2(1)} = \frac{-4 \pm \sqrt{-100 }}{2}\]

The square root of -100 used to be a problem but no longer:

\[\frac{-4 \pm \sqrt{-100 }}{2} = \frac{-4 \pm \sqrt{100 }\sqrt{-1}}{2} = = \frac{-4 \pm 10i}{2}= -2 \pm 5i \]

Notice that the answer is a complex conjugate pair. This always happens for any polynomial equation with real coefficients. If a complex number is a root, so is its conjugate. The answer above also means that

z² + 4z + 29 = (z + 2 – 5i)(z + 2 + 5i)

So now if b² – 4ac < 0, the quadratic equation has a conjugate pair as the solution.

If solutions to polynomial equations are complex conjugates, that means that a cubic equation has to have at least one real root because a cubic equation can have at most, three solutions. In fact, any odd powered polynomial equation has to have at least one real root. This makes sense graphically because the highest power of an odd-powered polynomial will eventually dominate all the other terms as x goes very negative or very positive. The highest powered term will have opposite signs for negative and positive values of x, so the polynomial will eventually cross the x-axis.

Let’s do one more problem. If a polynomial has single roots 2 ± i and 3, what is that polynomial? To solve this, just form the factors with these roots:

(z -2 –i)(z -2 +i)(z -3) = (z² – 4z + 5)(z -3) = z³ – 7z² + 17z – 15

Complex Numbers, Part 4

Now let’s do some maths with complex numbers. First, notice what happens with successive powers of i:

i0 = 1 (anything to the 0 power is 1)
i1 = i
i2 = -1 (by definition)
i3 = i2 × i = –i
i4 = i2 × i2 = 1

If you keep increasing the power, you will keep getting the pattern:
1, i, -1, –i. For higher powers, you can find where you are in the pattern by dividing the power by 4 and looking at the remainder. If the remainder is 0, then the answer is 1. If the remainder is 1, the answer is i. A remainder of 2 gives -1, and a remainder of 3 gives –i. For example, i69 , 69÷4 = 17 with a remainder of 1. So i69 = i.

Now lets’ convert between rectangular and polar forms.

Convert the following to polar form:

\[\begin{array}{l}
{z}_{1} = 1 \,- \sqrt{3}i\\
{z}_{2} = -1+ \sqrt{3}i\end{array}\]

First let’s find the modulus r. Both of these complex numbers have the same modulus because of the squaring of each term:

\[
{r}_{1} = \sqrt{{1}^{2}+{(-\sqrt{3})}^{2}} = {r}_{2} = \sqrt{{(-1)}^{2} + {(\sqrt{3})}^{2}} = \sqrt{4} = 2\]

Now let’s compute Arg(z) for each. Using my calculator, I get the same answer for both:

\[\begin{array}{l}
{\theta }_{1} = \text{tan}^{-1}{\left(\frac{-\sqrt{3}}{1}\right)} = -\frac{\pi }{3}\\
{\theta }_{2} = \text{tan}^{-1}{\left(\frac{\sqrt{3}}{-1}\right)} = -\frac{\pi }{3}\end{array}\]

The second one is wrong. In the first one, the real part is positive and the imaginary part is negative. This puts the complex number in the fourth quadrant in an Argand plot, so -𝜋/3 is the correct angle. But in the second one, the real part is negative and the imaginary part is positive. This puts the complex number in the second quadrant, so clearly -𝜋/3 is wrong:

Your calculator does not know where the minus sign is so it’s just programmed to give an answer in the range -𝜋/2 to 𝜋/2, (-90° to 90°). You however, being smarter than your calculator, know that the minus sign in z2, puts it in the second quadrant. You need to adjust the given angle to put it in the second quadrant as illustrated.

So the polar form of z1 is 2 cis (-𝜋/3). The polar form of z2 is 2 cis 2𝜋/3.

Now let’s go from polar to rectangular. Convert

\[4\sqrt{2}\text{cis} \left (-\frac{3𝜋}{4}
\right )\]

into its rectangular form.

In my last post I gave the equations

\[x = r \text{ cos} 𝜃, y = r \text{ sin} 𝜃\]

So z =4√2cos(-3𝜋/4) + 4√2 sin(-3𝜋/4)i = -4 – 4i. You do not need to worry about quadrant issues converting from polar to rectangular.

In my next post, I will solve some quadratic equations that previously did not have any solutions when we were stuck in the real domain. Now life is more complex.

Complex Numbers, Part 3

Complex numbers cannot be plotted on the traditional Cartesian coordinate system because of the imaginary part. A cartesian system is used to plot a pair of real numbers. Since a pair of complex numbers consists of four real numbers, a pair of complex numbers cannot be plotted. However, we can plot a single complex number on a modified version of a Cartesian system where the x-axis is relabelled as the real axis and the y-axis is the imaginary axis. This modified coordinate system is called an Argand diagram after its inventor Jean-Robert Argand, a Swiss mathematician.

Here is a plot of a generic complex number z = x + yi:

An Argand plot can be used to see the results of adding and subtracting complex numbers. These are interesting and you can see these in any textbook but the real power of an Argand plot is to see the relationship between the rectangular form of a complex number (the form we have been using so far) and the polar form.

Remember that polar coordinates are a different way to identify the location of a point in a two-dimensional plane. Instead of giving x and y coordinates, you can give an angle from the positive x-axis to define a direction, then a distance along that direction to define where a point is located. The same is true for an Argand plot of a complex number, and this polar form is more convenient to use than the rectangular form in certain scenarios. Below is the same plot as above, but I’ve added the polar coordinates:

From your past experience with triangles and trigonometry, you can see that the following relationships are true:

\[\begin{array}{l}
r = |z| = \sqrt{{x}^{2}+{y}^{2}}\\
x = r \text{ cos} 𝜃, y = r \text{ sin} 𝜃\\
𝜃 =\text{tan}^{-1}\frac{y}{x}\end{array}\]

From the above, you can see that in terms of r and 𝜃,

z = r cos𝜃 + ir sin𝜃

The polar form notation that is used in VCE Australia is

z = r cis𝜃

where “cis” is shorthand for “cos + i sin”. There are other ways to express a polar complex number as well. In electronics, r∠𝜃 is frequently used.

Some definitions: modulus is the length of a complex number, that is, r. Arg(z) is the principal value of 𝜃. This is pronounced as “the argument of z“. As any particular 𝜃 can have multiples of 2𝜋 radians or 360° added to it, arg(z) restricts 𝜃 to -𝜋 < arg(z) ≤ 𝜋. Be careful when computing 𝜃 with the inverse tangent formula. A calculator will only give values between -𝜋/2 and 𝜋/2 (-90° and 90°), that is the first and fourth quadrants. You need to interpret this result to the correct angle by looking at the signs of x and y to know which quadrant your particular complex number falls in. I will do some sample problems in my next post to illustrate this.

Complex Numbers, Part 2

Addition, subtraction, and multiplication of complex numbers extend the rules for real numbers. Division is done a bit differently but still follow rules you already know for real numbers.

Addition: To add two complex numbers, just add the real parts and the imaginary parts separately. Example:

(2 – 5i) + (-3 +2i) = (2 – 3) + (-5 + 2)i = -1 – 3i

Subtraction: same as addition, you just separately subtract the real parts and the imaginary parts. Example:

(2 – 5i) – (-3 +2i) = (2 – (-3)) + (-5 – (+2))i = 5 – 7i

Multiplication: Just as for real algebraic expressions like (x + y)(2x + 3y), you just multiply two complex numbers the same way then add like terms. Just remember that i² = -1 so a negative sign will appear when i is multiplied by i. Example:

(2 – 5i) × (-3 +2i) = 2 × (-3) + 2 × 2i -5i × (-3) – 5i × 2i
= -6 + 4i + 15i -10i² = -6 + 19i -10(-1) = 4 + 19i

The same method applies when multiplying a real number times a complex number:

2 × (2 – 5i) = 4 – 10i

Before I talk about division, I need to introduce another definition. In my last post, I solved a quadratic equation with the solution -2 ± 3i. So there are two solutions, one with the + and the other with the -. These two solutions are called a complex conjugate pair. -2 + 3i is the complex conjugate of -2 – 3i and vice versa. It turns out that if a polynomial equation has a complex solution, the conjugate of that solution is also a solution. That is, complex solutions to polynomial equations always come in complex conjugate pairs. If z is a complex number, z̄ is used to represent its conjugate.

So now let’s look at complex division. To divide by a complex number (which include real numbers) by a complex number with a non-zero imaginary part, multiply the numerator and denominator by the conjugate of the denominator. Example:

\[(2-5i)\div (-3+2i)=\frac{2-5i}{-3+2i}\times \frac{-3-2i}{-3-2i}=\frac{-6-4i+15i-10}{9+6i-6i+4}\] \[=\frac{-16+11i}{13}=-\frac{16}{13}+\frac{11}{13}i\]

This example illustrates a few more things about complex numbers. First, at the end, dividing a complex number by a pure real number (or multiplying for that matter), you just divide (or multiply) each part of the complex number by the real number.

The other thing you may have noticed is in the multiplication in the denominator. Multiplying a complex number by its conjugate results in a real number. This resulting number has graphical significance which you will see later. In general, if z = x + yi, then z̄ = xyi and z z̄ = x² + y².

In my last post, I found that -2 ± 3i are solutions to z² +4z + 13 = 0. Let’s check one of these solutions:

(-2 + 3i)² + 4(-2 + 3i) +13 = 4 -12i -9 – 8 + 12i + 13 = 0 + 0i

I’ll leave it to you to show that the conjugate also solves the equation.

Note that two complex numbers are equal only if their real parts are equal and their imaginary parts are equal.

In my next post, I’ll look at how we can plot complex numbers.

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.