Calculators and complex numbers (Part 9)

In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.

To begin, we recall that the trigonometric form of a complex number z = a+bi is

z = r(\cos \theta + i \sin \theta)

where r = |z| = \sqrt{a^2 + b^2} and \tan \theta = b/a, with \theta in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.

There’s a shorthand notation for the right-hand side (r e^{i \theta}) that I’ll justify later in this series.

In the previous three posts, we discussed De Moivre’s Theorem:

Theorem. If n is an integer, then \left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta).

Yesterday, we used factoring to show that there are three solutions to z^3 = -27, namely, z = -3 and z = \displaystyle \frac{3}{2} \pm \frac{3\sqrt{3}}{2} i. Let’s now use De Moivre’s Theorem to take on the same task. As we’ll see, De Moivre’s Theorem provides a geometrical interpretation of this result that isn’t readily apparent using solely algebra.

Let z = r(\cos \theta + i \sin \theta), so that

z^3 = -27

becomes

r^3 (\cos 3\theta + i \sin 3 \theta) = 27 (\cos \pi + i \sin \pi)

We now match the corresponding parts. The distances from the original have to match, so that r^3 = 27, or r = 3. (Notice that there is only one answer because r must be a positive real number.) Also, the angles 3\theta and \pi must be coterminal. They do not necessarily have to be equal, but the angles must point in the same direction. Therefore,

3 \theta = \pi + 2\pi k, or \theta = \displaystyle ]frac{\pi}{3} + \frac{2\pi k}{3}

for any integer k. At first blush, it appears that there are infinitely many solutions z since there are infinitely many possible angles \theta. However, it turns out that there are only three answers, as expected.

Let’s now plug in numbers for k. Let’s use the easiest three numbers, k = 0, 1, 2.

If k = 0, then \theta = \displaystyle \frac{\pi}{3}, so that

z = 3 \displaystyle \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)

= 3 \displaystyle \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right)

= \displaystyle \frac{3}{2} + \frac{3\sqrt{3}}{2} i.

If k = 1, then \theta = \displaystyle \frac{\pi}{3} + \frac{2\pi}{3} = \pi, so that

z = 3 \displaystyle \left( \cos \pi + i \sin \pi \right) = 3(-1+0i) = -3.

If k = 2, then \theta = \displaystyle \frac{\pi}{3} + \frac{4\pi}{3} = \displaystyle \frac{5\pi}{3}, so that

z = 3 \displaystyle \left( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right)

= 3 \displaystyle \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right)

= \displaystyle \frac{3}{2} - \frac{3\sqrt{3}}{2} i.

Not surprisingly, we obtain the same three answers that we did using algebra.

What if we keep increasing the value of k? Let’s find out with k = 3:

If k = 3, then \theta = \displaystyle \frac{\pi}{3} + \frac{6\pi}{3} = \displaystyle \frac{7\pi}{3}, so that

z = 3 \displaystyle \left( \cos \frac{7\pi}{3} + i \sin \frac{7 \pi}{3} \right)

= 3 \displaystyle \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right)

= \displaystyle \frac{3}{2} + \frac{3\sqrt{3}}{2} i.

In other words, since \displaystyle \frac{\pi}{3} and \displaystyle \frac{7\pi}{3} are coterminal, we end up with the same answer. This resolves the apparent paradox of having infinitely many possible angles \theta but only three solutions z.

Using De Moivre’s Theorem certainly appears to be much more difficult than just factoring! However, this solution provides a geometric interpretation of the three roots that isn’t otherwise apparent. Let’s using the trigonometric form of these three solutions to plot them in the complex plane:

complex roots

All three points lie a distance of 3 from the origin, and so they lie on the same circle. Also, the angle from the origin increases by \displaystyle \frac{2\pi}{3} as we shift from point to point. This divides the circle (with a total angle of 2\pi into three equal parts, and so the points are evenly spaced around the circle. Also, the points form the vertices of an equilateral triangle inscribed within this circle. Again, none of this would have been apparent by strictly factoring the polynomial z^3 + 27.

All of the above can be repeated for finding the roots of any equation z^n = w. This equation has n roots which lie a distance of |w|^{1/n} from the origin on the points of a regular n-gon inscribed in a circle of this radius. The only tricky part is determining the placement of the initial point (i.e., finding the initial value of \theta from which all other points can be determined.

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For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

 

 

Calculators and complex numbers (Part 8)

In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.

To begin, we recall that the trigonometric form of a complex number z = a+bi is

z = r(\cos \theta + i \sin \theta)

where r = |z| = \sqrt{a^2 + b^2} and \tan \theta = b/a, with \theta in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.

There’s a shorthand notation for the right-hand side (r e^{i \theta}) that I’ll justify later in this series.

In the previous three posts, we discussed De Moivre’s Theorem:

Theorem. If n is an integer, then \left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta).

Let’s now use this theorem to solve an algebraic equation.

Find all z so that z^3 = -27.

When I present this to my students, their kneejerk reaction is always to answer “-3.” To which I politely point out, “This is a cubic equation. So how many roots does it have?” Of course, they answer “Three.” To which I respond, “OK, so how do we find the other two roots?”

With enough patience, a student will usually volunteer that z^3 + 27 = 0 has a known root of z = -3, and so the remaining roots can be found using synthetic division by dividing z+3 into z^3 + 27, yielding

z^3 + 27 = (z+3)(z^2 - 3z + 9)

So the other two roots can be found using the quadratic formula:

z = \displaystyle \frac{3 \pm \sqrt{9 - 36}}{2} = \displaystyle \frac{3}{2} \pm \frac{3\sqrt{3}}{2} i

So there are indeed three roots. Though I usually won’t take class time to do it, I encourage my students to cube these two answers to confirm that they indeed get -27.

As an aside, before moving on to the use of De Moivre’s Theorem, I usually point out to my students that there is a formula for factoring the sum of two cubes:

a^3 + b^3 = (a+b)(a^2 - ab + b^2)

And there’s a formula for the difference of two cubes:

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

My experience is that even math majors are not familiar with these two formulas. They know the difference of two squares formula, but they’re either not proficient with these two formulas or else they’ve never seen them before. These can be generalized for any odd positive exponent and any positive exponent, respectively.

In tomorrow’s post, I’ll describe how these three roots can be found by using De Moivre’s Theorem.

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For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

 

 

Calculators and complex numbers (Part 7)

In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.

To begin, we recall that the trigonometric form of a complex number z = a+bi is

z = r(\cos \theta + i \sin \theta)

where r = |z| = \sqrt{a^2 + b^2} and \tan \theta = b/a, with \theta in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.

There’s a shorthand notation for the right-hand side (r e^{i \theta}) that I’ll justify later in this series.

This post builds off the previous two posts by completing the proof of De Moivre’s Theorem.

Theorem. If n is an integer, then \left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta).

The proof has two parts:

  1. For n \ge 0: proof by induction.
  2. For n < 0: let n = -m, and then use part 1.

In the previous post, I presented how I describe the proof of step 1 to students. Today, I’ll discuss how I present step 2.

My personal opinion is that the proof of step 2 goes easiest when a numerical example is done first. Let n = -5. Students can usually volunteer the successive steps of this special case:

\left[ r (\cos \theta + i \sin \theta) \right]^{-5} = \displaystyle \frac{1}{ \left[ r (\cos \theta + i \sin \theta) \right]^{5} }

= \displaystyle \frac{1}{r^5 (\cos 5 \theta + i \sin 5 \theta)}

At this point, students usually want to multiply by the conjugate of the denominator. There’s nothing wrong with doing that, of course, but it’s more elegant to write the numerator in trigonometric form:

= \displaystyle \frac{1(\cos 0 + i \sin 0)}{r^5 (\cos 5 \theta + i \sin 5 \theta)}

= r^{-5} (\cos [0-5\theta] + i \sin[0-5\theta] )

= r^{-5} (\cos [-5\theta] + i \sin[ - 5\theta]),

which matches what we would have expected. Students are now prepared for the proof, which I try to place alongside the above computation for n = -5.

Proof for n < 0. Let n = -m. I tell students to imagine that n is -5, so that m is equal to (students volunteer) 5. In other words, n is negative and m is positive. Then

\left[ r (\cos \theta + i \sin \theta) \right]^{n} = \displaystyle \frac{1}{ \left[ r (\cos \theta + i \sin \theta) \right]^{-n} }

I again remind everyone that -n = m is positive, and so (like a good MIT freshman) the previous work applies:

= \displaystyle \frac{1}{ \left[ r (\cos \theta + i \sin \theta) \right]^m }

The remaining steps mirror the calculation above:

= \displaystyle \frac{1}{r^m (\cos m \theta + i \sin m \theta)}

= \displaystyle \frac{1(\cos 0 + i \sin 0)}{r^m (\cos m \theta + i \sin m \theta)}

= r^{-m} (\cos [0-m\theta] + i \sin[0-m\theta] )

= r^{-m} (\cos [-m\theta] + i \sin[ - m \theta])

= r^{n} (\cos [n\theta] + i \sin[ n \theta]),

thus ending the proof. green line

For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

 

 

Calculators and complex numbers (Part 6)

In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.

To begin, we recall that the trigonometric form of a complex number z = a+bi is

z = r(\cos \theta + i \sin \theta)

where r = |z| = \sqrt{a^2 + b^2} and \tan \theta = b/a, with \theta in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.

There’s a shorthand notation for the right-hand side (r e^{i \theta}) that I’ll justify later in this series.

In the previous post, I used a numerical example to justify De Moivre’s Theorem:

Theorem. If n is an integer, then \left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta).

The proof has two parts:

  1. For n \ge 0: proof by induction.
  2. For n < 0: let n = -m, and then use part 1.

In this post, I describe how I present part 1 to students in class. The next post will cover part 2. As noted before, I typically present this theorem and its proof after a numerical example so that students can guess the statement of the theorem on their own.

Proof for n \ge 0.

Base Case: n = 0. This is trivial, as the left-hand side is

\left[ r (\cos \theta + i \sin \theta) \right]^0 = 1,

while the right-hand side is

r^0 (\cos 0 \theta + i \sin 0 \theta) = 1(\cos 0 + i \sin 0) = 1(1 + 0i) = 1.

Assumption. We now assume, for a given integer $latex $n$, that

\left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta).

Inductive Step. We now use the above assumption to prove the statement for n+1. On the board, I write the left-hand side on the top and the right-hand side on the bottom, leaving plenty of space in between:

\left[ r (\cos \theta + i \sin \theta) \right]^{n+1}

\quad

\quad

\quad

\quad

\quad

= r^{n+1} (\cos [n+1] \theta + i \sin [n+1] \theta).

All we have to do is fill in the space to transform the left-hand side into the target on the right-hand side. (I like to call the right-hand side “the target,” as it suggests the direction in which the proof should aim.) I also tell the class that we’re more than two-thirds done with the proof, since we’ve finished the first two steps and have made some headway on the third. This usually produces knowing laughter since the hardest part of the proof is creatively converting the left-hand side into the target.

The first couple steps of the proof are usually clear to students:

\left[ r (\cos \theta + i \sin \theta) \right]^{n+1} = \left[ r (\cos \theta + i \sin \theta) \right]^n \cdot \left[ r (\cos \theta + i \sin \theta) \right]

= \left[ r^n (\cos n \theta + i \sin n \theta) \right] \cdot \left[ r (\cos \theta + i \sin \theta) \right]

by induction hypothesis. (I’ll also remind students that, as a general rule, when doing a proof by induction, it’s important to actually use the inductive assumption someplace.) At this point, most students want to distribute to get the right answer. This will eventually produce the correct answer using trig identities. However, I again try to encourage them to think like MIT freshmen and use previous work. After all, the distances multiply and the angles add, so the next step can be

=r^n \cdot r (\cos [n \theta + \theta] + i \sin [n \theta + \theta])

= r^{n+1} (\cos [n+1]\theta + i \sin [n+1]\theta).

In tomorrow’s post, I’ll talk about how I present the second part of the proof to my students.

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For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.

 

 

Calculators and complex numbers (Part 5)

In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.

To begin, we recall that the trigonometric form of a complex number z = a+bi is

z = r(\cos \theta + i \sin \theta)

where r = |z| = \sqrt{a^2 + b^2} and \tan \theta = b/a, with \theta in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.

There’s a shorthand notation for the right-hand side (r e^{i \theta}) that I’ll justify later in this series.

The trigonometric form of a complex number permits a geometric interpretation of multiplication, given in the following theorem.

Theorem. \left[ r_1 (\cos \theta_1 + i \sin \theta_1) \right] \cdot \left[ r_2 (\cos \theta_2 + i \sin \theta_2) \right] = r_1 r_2 (\cos [\theta_1+\theta_2] + i \sin [\theta_1+\theta_2]).

While this theorem doesn’t seem all that helpful — just multiplying complex numbers seems easier — this theorem will be a great help for the following problem:

Compute (\sqrt{3} + i)^{2014}. (When teaching this in class, I usually choose the exponent to be the current year.)

Let’s discuss the options for evaluating this expression.

Method #1: Multiply it out. (Students reflexively wince in pain — or knowing laughter — when I make this suggestion.)

Method #2: Use the 2014th row of Pascal’s triangle. (More pain and/or laughter.)

Method #3: Use the above theorem. It’s straightforward to write \sqrt{3} + i as 2 \displaystyle \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right)… for reasons that will become apparent later, I tell my students that I’ll use radians and not degrees for this one. Most students can recognize — and this is important, before I formally prove De Moivre’s Theorem — that they need to multiply 2 by itself 2014 times and add \displaystyle \frac{\pi}{6} to itself 2014 times. Therefore,

(\sqrt{3} + i)^{2014} = \displaystyle 2^{2014} \left( \cos \frac{2014\pi}{6} + i \sin \frac{2014\pi}{6} \right) = \displaystyle 2^{2014} \left( \cos \frac{1007\pi}{3} + i \sin \frac{1007\pi}{3} \right)

I then try to coax my students to compute \displaystyle \cos \frac{1007\pi}{3} without a calculator. With some prodding, they’ll recognize that \displaystyle \frac{1007}{3} = \displaystyle {335}\frac{2}{3}, and so they can subtract 334\pi (not 335\pi) without changing the values of sine and cosine. Therefore,

(\sqrt{3} + i)^{2014} = \displaystyle 2^{2014} \left( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right)

= 2^{2014} \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right)

= 2^{2013} (1-i\sqrt{3})

By this point, students absolutely believe that the trigonometric form of a complex number serves a useful purpose. Also, this numerical example has prepared students for the formal proof of DeMoivre’s Theorem, which will be the subject of the next two posts.

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For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.