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:

For $n \ge 0$ : proof by induction.
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)

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

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

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:

For $n \ge 0$ : proof by induction.
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:

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

$\quad$

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

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.

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 5)

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

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

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.

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 4)

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

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

In the previous post, I proved the following theorem which provides a geometric interpretation for multiplying complex numbers.

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

Perhaps unsurprisingly, there’s also a theorem for dividing complex numbers. Students can using guess the statement of this theorem.

Theorem. $\displaystyle \frac{ r_1 (\cos \theta_1 + i \sin \theta_1) }{ r_2 (\cos \theta_2 + i \sin \theta_2) } = \displaystyle \frac{r_1}{r_2} (\cos [\theta_1-\theta_2] + i \sin [\theta_1-\theta_2])$ .

Proof. The proof begins by separating the $r_1$ and $r_2$ terms and then multiplying by the conjugate of the denominator:

$\displaystyle \frac{ r_1 (\cos \theta_1 + i \sin \theta_1) }{ r_2 (\cos \theta_2 + i \sin \theta_2) }$

$= \displaystyle \frac{r_1}{r_2} \cdot \frac{ \cos \theta_1 + i \sin \theta_1 }{ \cos \theta_2 + i \sin \theta_2 } \cdot \frac{ \cos \theta_2 - i \sin \theta_2 }{ \cos \theta_2 - i \sin \theta_2 }$

$= \displaystyle \frac{r_1}{r_2} \cdot \frac{ (\cos \theta_1 + i \sin \theta_1)( \cos \theta_2 - i \sin \theta_2) } {\cos^2 \theta_2 - i^2 \sin^2 \theta_2}$

$= \displaystyle \frac{r_1}{r_2} \cdot \frac{ (\cos \theta_1 + i \sin \theta_1)( \cos \theta_2 - i \sin \theta_2) } {\cos^2 \theta_2 + \sin^2 \theta_2}$

$= \displaystyle \frac{r_1}{r_2} (\cos \theta_1 + i \sin \theta_1)( \cos \theta_2 - i \sin \theta_2)$

At this juncture in the proof, there are two legitimate ways to proceed.

Method #1: Multiply out the right-hand side. After all, this is how we proved the theorem yesterday. For this reason, students naturally gravitate toward this proof, and the proof works after recognizing the trig identities for the sine and cosine of the difference of two angles.

However, this isn’t the most elegant proof.

Method #2: I break out my old joke about the entrance exam at MIT and the importance of using previous work. I rewrite the right-hand side as

$= \displaystyle \frac{r_1}{r_2} (\cos \theta_1 + i \sin \theta_1)( \cos [-\theta_2] + i \sin [-\theta_2])$ ;

this also serves as a reminder about the odd/even identities for sine and cosine, respectively. Then students observe that the right-hand side is just a product of two complex numbers in trigonometric form, and so the angle of the product is found by adding the angles.

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 3)

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

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

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

Why is this important? When students first learn to multiply complex numbers like $1+i$ and $2+i$ , they are taught to just distribute (or, using the nomenclature that I don’t like, FOIL it out):

$(1+i)(1+2i) = 1 + 2i + i + 2i^2 = 1 + 3i - 2 = -1 + 3i$ .

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

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

Proof. As above, we distribute (except for the $r_1$ and $r_2$ terms):

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

$= r_1 r_2 (\cos \theta_1 \cos \theta_2 + i \cos \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2 + i^2 \sin \theta_1 \sin \theta_2$

$= r_1 r_2 (\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 + i[ \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2])$

$= r_1 r_2 (\cos [\theta_1+\theta_2] + i \sin [\theta_1+\theta_2])$ .

When actually doing this in class, the big conceptual jump for students is the last step. So I make a big song-and-dance routine out of this:

Cosine of the first times cosine of the second minus sine of the first times sine of the second… where have I seen this before?

The idea is for my students to search deep into their mathematical memories until they recall the appropriate trig identity.

For the original multiplication problem, we see that

$1+i = \sqrt{2} \left( \cos 45^\circ + i \sin 45^\circ \right)$

$1 + 2i = \sqrt{5} \left( \cos[\tan^{-1} 2] + i \sin[\tan^{-1} 2] \right) \approx \sqrt{5} \left( \cos 63.435^\circ + i \sin 63.435^\circ \right)$

Therefore, the product of $1+i$ and $1+2i$ will be a distance of $\sqrt{2} \cdot \sqrt{5} = \sqrt{10}$ from the origin, and the angle from the positive real axis will be $45^\circ + \tan^{-1} 2 \approx 45^\circ + 63.435^\circ = 108.435^\circ$ . Indeed,

$-1 + 3i \approx \sqrt{10} \left( \cos 108.435^\circ + i \sin 108.435^\circ \right)$ .

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 2)

In yesterday’s post, I showed a movie (also provided at the bottom of this post) that calculators can return surprising answers to exponential and logarithmic problems involving complex numbers. In this series of posts, I hope to explain why the calculator returns these results.

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.

For example, the point $z = -\sqrt{3} + i$ is in the second quadrant of the complex plane. The modulus is

$r = \sqrt{ (-\sqrt{3})^2 + (1)^2 } = \sqrt{4} = 2$ .

(Notice that $1$ , and not $i$ , appears in the above expression.) Also,

$\tan \theta = \displaystyle \frac{1}{-\sqrt{3}}$ , so that $\theta = \displaystyle -\frac{\pi}{6} + n \pi$

Since $-\sqrt{3} + i$ is in the second quadrant, we choose $\theta = \displaystyle -\frac{\pi}{6} + \pi = \displaystyle \frac{5\pi}{6}$ . Therefore,

$-\sqrt{3} + i = \displaystyle 2 \left( \cos \frac{5\pi}{6} + \sin \frac{5\pi}{6} \right)$

This can be checked by simply evaluating the right-hand side and distributing:

$\displaystyle 2 \left( \cos \frac{5\pi}{6} + \sin \frac{5\pi}{6} \right) = \displaystyle 2 \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = -\sqrt{3} +i$

When teaching this in class, I’ll run through about 2-4 more examples to make sure that this concept is stuck in my students’ heads.

Notes:

The angle $\theta$ is not uniquely defined… any angle that is coterminal with $\frac{5\pi}{6}$ would also have worked. For example,

$-\sqrt{3} + i = \displaystyle 2 \left( \cos \frac{17\pi}{6} + \sin \frac{17\pi}{6} \right)$

and

$-\sqrt{3} + i = \displaystyle 2 \left( \cos \frac{-7\pi}{6} + \sin \frac{-7\pi}{6} \right)$

It’s really important to remember that $\theta$ need not be equal to $\displaystyle \tan^{-1} \frac{b}{a}$ . After all, the arctangent of an angle must lie between $-\pi/2$ and $\pi/2$ , which won’t work for complex numbers in either the second or third quadrant. That said, it is true that

$-\sqrt{3} + i = \displaystyle -2 \left( \cos \frac{-\pi}{6} + \sin \frac{-\pi}{6} \right)$

The above procedure is also the essence of converting from rectangular coordinates to polar coordinates (or vice versa), which is a function pre-programmed on many scientific calculators.
When teaching this topic, I often use physical humor to get the above points across.

I’ll pick the direction parallel to the chalkboard to be the positive real axis, and the direction perpendicular to the chalkboard (i.e., pointing toward my students) as the positive imaginary axis. I’ll pick some convenient spot in front of the class to be the origin.
Standing at the origin, I’ll face the positive real axis, spin in an angle of $5\pi/6 = 150^\circ$ , and take two steps to arrive at the point $-\sqrt{3} + i$ .
Returning to the origin, I’ll face the positive real axis, spin the other direction in an angle of $-210^\circ$ , and take two steps to arrive at the same point.
Returning to the origin, I’ll face the positive real axis, spin in an angle of $510^\circ$ (getting more than a little dizzy while doing so), and take two steps to arrive at the same point.
Returning to the origin, I’ll face the positive real axis, spin in an angle of only $-30^\circ$ , and take two steps backwards (while doing the moonwalk) to arrive at the same point.

green line

We will need to use this concept of writing a complex number in trigonometric form in order to explain the calculator’s results. For completeness, here’s the movie that I used to begin this series of posts.

Complex knowledge

I took a statistics course at MIT. I would go study and do problems, and have high confidence that I understood the material. Then I’d go to the lecture, and be more confused than I was when I entered the classroom. Thus, I discovered that some teachers were capable of conveying negative knowledge, so that after listening to them, I knew less than I did before.

It was also clear that knowledge varies considerably in quantity among people, and this convinced me that real knowledge varies over a very wide range.

Then I encountered people who either did not know what they were talking about, or were clearly convinced of things that were wrong, and so I learned that there was imaginary knowledge.

Once I understood that there was both real and imaginary knowledge, I concluded that knowledge is truly complex.

– Hillel J. Chiel, Case Western Reserve University

Source: American Mathematical Monthly, Vol. 120, No. 10, p. 923 (December 2013)

Are complex numbers complex?

It’s an unfortunate fact of history that numbers of the form $a+bi$ are called complex numbers. In modern English, of course, the word complex is usually associated with phrases like difficult, inscrutable, time-consuming, hard to solve, and other negative connotations that teachers would prefer to not introduce into a math class.

However, my understanding is that the other meaning of the word complex was in mind when the term complex numbers was coined. After all, in modern English, we still refer to a group of buildings as an apartment complex or maybe an office complex. In this sense, complex means two (or more) things that are joined together to form a single unit, which is precisely what happens as the real part $a$ and the imaginary part $bi$ are joined to form $a + bi$ . Indeed, my understanding is that complex was chosen to be the opposite of simplex, or a single unit (like a real number).

Anyway, hopefully this bit of history can make complex numbers less mystifying for students.

While I’m on the topic, the word imaginary was another unfortunate choice of words by our ancestors, but — like complex — we’re just stuck with it.

Also while I’m on the topic, this is a good chance to review a great piece of showmanship about teaching complex numbers: