Monthly Archives: October 2017

Multiplying complex numbers

Today I discussed what it looks like when we multiply complex numbers on the plane. In this entry, I’m going to give some more examples of how that works.

We already know how to multiply complex numbers algebraically. For example, what is the product of  \(z_1 = (4 – 2i)\) and \(z_2 = (3 + i)\)? First we use the distributive law: \(z_3 = (4 – 2i)(3 + i) = 12 + 4i – 6i – 2i^2\). We replace \(i^2\) with \(-1\) then simplify this to \(z_3 = 14 – 2i\).

Let’s look at these three numbers on the complex number plane.

On the standard complex plane, there doesn’t seem to be much relationship between the two multiplicands and their product. The product is farther away, which we would expect, but it’s not clear why it’s located where it is.

First, let’s look at the absolute values of each point.

  • \(|z_1| = |4-2i| = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\)
  • \(|z_2| = |3+i| = \sqrt{9 + 1} = \sqrt{10}\)
  • \(|z_3| = |13-2i| = \sqrt{196+4} = \sqrt{200} = 10\sqrt{2} \)

The product of the absolute values of the multiplicands is the absolute value of the product. This is the first property of the product of complex numbers.

This explains why \(z_3\) has the distance that it does, but not why it’s located where it is. For this, we need to look at how much each point is rotated from the positive real axis (the x-ray).

We can create three right triangles by connecting each point to the x-ray and to the point \(0\). For instance, looking at the triangle formed by \(z_2\), it has legs of \(3\) and \(1\) and a hypotenuse of \(\sqrt{10}\). This means the angle at the lower left corner is \(\tan^{-1}{1/3} \approx 18.43^o\).

Using the inverse tangents, we can calculate the angle of rotation for each point. Here’s an important detail: If we’re rotating counterclockwise, we’ll call it a positive angle; if we’re rotation clockwise (as with \(z_1\) and \(z_3\)), we’ll call it a negative angle.

Here are the respective angles of rotation, approximately:

  • \(z_1: -26.57^o\)
  • \(z_2: 18.43^o\)
  • \(z_3: -8.13^o\)

What do we get if we add the angles of rotation for \(z_1\) and \(z_2\)? It’s pretty close to what we got for \(z_3\); the difference is because we rounded the values.

So here’s the second property of the product of complex numbers: The angle of rotation for the product will be the sum of the angles of rotations of the multiplicands.

Let’s do another example. Take \(z_1 = (-1 + i)\) and \(z_2 = (1 – i)\). Then \(z_3 = -1 + i + i – i^2 = 2i\). Here are the respective absolute values:

  • \(|z_1| = \sqrt{2}\)
  • \(|z_2| = \sqrt{2}\)
  • \(|z_3| = 2\)

and the angles of rotation:

  • \(z_1: 135^o\)
  • \(z_2: -45^o\)
  • \(z_3: 90^o\)

… which follows the pattern we’ve established: The absolute values are multiplied, the angles of rotation are added.

Key consequences

In general, this is a fun observation that might help you understand what multiplication of complex numbers means. But it’s a very powerful observation to explain two key mathematical truths.

The first and more important of these is that the product of two negative real numbers is positive. Because all negative real numbers are on the negative portion of the real number line, they have an angle of rotation of \(180^o\) from the x-ray. When we multiply two negative real numbers, the angle of rotation of the product will be \(360^o\), that is, the product will be on the x-ray.

The second is that the product of a non-zero complex number and its conjugate will always be a positive real number. The conjugate of a complex number is formed by keeping the real part the same and taking the opposite of the imaginary part.

Let’s look at \(z_1 = 1+2i\) and its conjugate \(z_2 = 1-2i\).

If we create our triangles, we see two congruent triangles (two right triangles with congruent legs). Hence the angle of rotation for \(z_2\) is the clockwise equivalent of \(z_1\)’s, and the angle of rotation for their product with be \(0^o\). A non-zero complex number with an angle of rotation of \(0^o\) is a positive real number.

Advanced section! (Here be dragons)

Incidentally, there is a different way of giving complex numbers. Rather than giving a real part and an imaginary part, we could instead state the absolute value and the angle of rotation. These are called polar coordinates. Our class calculator (the TI-84) even has a  setting that lets us work with them.

If you set the calculator to \(a + bi\), you will be working with complex numbers in the way we’ve discussed in class. This is the most common way to work with complex numbers. However, if you set the calculator to \(re^{\theta i}\), then enter \(\sqrt{-1}\), you’ll get \(1e^{90i}\).

The number before \(e\) is the absolute value of the complex number; “r” stands for “radius”, because in this system you’re working with the radius of the circle that the point is on, and the angle of rotation of the number.

You probably haven’t met \(e\) before. This is a constant called Euler’s number that we’ll discuss later in this course.

The number between \(e\) and \(i\) is the angle of rotation. Depending on your settings, it can be given in either degrees or radians.

Please don’t set your calculator to the \(re^{\theta i}\) setting. It will confuse other users greatly.

Now, why does this work? Recall that when we take the product of complex numbers, the absolute value is the product of the multiplicands, while the angles are the sum. Let’s look at \(re^{\theta i}\) for each of our complex number multiplicands.

  • \(z_1 = r_1e^{\theta_1 i}\)
  • \(z_2 = r_2e^{\theta_2 i}\)
  • \(z_3 = z_1\cdot z_2 = r_1e^{\theta_1 i} r_2e^{\theta_2 i}\)

What is the rule for multiplying when we have the same number (\(e\)) to different powers? We add the powers! So \(z_3 = r_1r_2 e^{(\theta_1 + \theta_2) i}\). This is exactly what we want: Multiply the absolute values (r) and add the angles of rotation (\(\theta\)).

Simplifying Square Roots

Because square roots are usually irrational, we generally don’t want to convert them into decimal form until the very last step, if at all. However, it is typical to simplify square roots by taking out any perfect square factors. To do this, we need to identify these.

One method is to create factor trees. The method in this post, however, involves creating a list of prime factors. We test the target number against each prime number in turn. The first few prime numbers are 2, 3, 5, 7, 11, and 13.

For instance, what are the prime factors of 924?

Can 924 be divided by 2? Yes, 924/2 = 462. Can 462 be divided by 2? Yes, 462/2 = 231.

231 can’t be divided by 2, so now we try 3. 231/3 = 77.

77 can’t be divided by 3, so we try 5. That doesn’t work either, so we try 7: 77/7 = 11, which we also know is a prime.

This gives us our list of factors: 924 = 2 * 2 * 3 * 7 * 11. Of these, only 2 * 2 represents a perfect square, so \(\sqrt{924} = 2\sqrt{231}\).

When we’re testing prime numbers, it’s important to know when we can stop. Let’s say we want to know if 113 is prime. Do we need to test all numbers smaller than 101? That’s a lot of work.

It turns out we only need to test all the prime numbers up to \(\sqrt{113}\). Why is this?

To see why, look at 115. The prime factors of 115 are 5 and 23. While there is a prime factor that is greater than \(\sqrt{115}\), there is also a prime factor less than it.

In general, if a number \(n\) is composite, it has at least two factors, \(a\) and \(b\). Since \(n = ab\), then \(a = n/b\). Let \(m = \sqrt{n}\), so \(n = m\cdot m\) and \(m = n/m\). If \(a > m\), then \(n/b > n/m\).

Solve this for \(b\): \(n > nb/m \Rightarrow nm > nb \Rightarrow m > n\).

In other words, if there is a factor that is greater than \(\sqrt{n}\), there is another factor that is less than \(\sqrt{n}\). So we only need to try the prime numbers less than the square root of a number to see if it’s prime.

For 113 specifically: Since 113 is less than 121, we only need to test prime numbers less than 11, that is, 2, 3, 5, and 7. None of these are factors of 113, so we can conclude that 113 is prime (which it is).

Since 17 * 17 = 289, that means that the list of primes provided above (2, 3, 5, 7, 11, 13) are enough to test any number less than 289 for factors. Adding 17 and 19 to the list lets us test less that 529.

Week of Oct 23, 2017

Powerpoints: Oct 27 Oct 26 Oct 25 Oct 24 Oct 23

Monday
Topic: Adding complex numbers
Homework: Khan Academy exercises (2) OR p. 279 3-6, 12-17

Tuesday
Topic: Multiplying complex numbers
Classwork: Adding and multiplying complex numbers

Wednesday:
Topic: Plotting, absolute value of complex numbers
Homework: Absolute value of complex numbers

Thursday:
Topic: Absolute value and conjugate

Friday
Topic: Rationalizing (Division)
Homework: Complex number review worksheet (4-8 G)

Week of Oct 23, 2017

Powerpoints: Oct 27 Oct 26 Oct 25 Oct 24 Oct 23

Monday
Topic: Adding complex numbers
Homework: Khan Academy exercises (2) OR p. 279 3-6, 12-17

Tuesday
Topic: Multiplying complex numbers
Classwork: Adding and multiplying complex numbers

Wednesday
Topic: Plotting, absolute value of complex numbers
Homework: Absolute value of complex numbers

Thursday
Topic: Absolute value and conjugate

Friday
Topic: Rationalizing (Division)
Homework: Complex number review worksheet (4-8 K)