Calculators and complex numbers (Part 4)

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.

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.

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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 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,

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

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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 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.

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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.

Calculators and complex numbers (Part 1)

What is $\ln(-5)$ ? Or $(-8)^{1/3}$ ? Easy, right? Well, let’s plug into a calculator and find out. (Click anywhere in the image below to start the movie. The important stuff is the screen at the top; you can see the keystrokes that I used if you following the mouse arrow toward the bottom.)

In this series of posts, I’ll try to explain why the calculator provides these unexpected answers. This series of posts will have 24 posts (!) and will contain some fairly sophisticated mathematics to explain why the calculator does what it does as well as some pedagogical discussion when I present these topics to my class of future secondary teachers. Each post can be thought of as a 5-10 minute portion of one of my lectures.

Fun with combinatorics

I found the following videos through UpWorthy: http://www.upworthy.com/see-this-teachers-amazing-response-to-the-question-but-when-are-we-gonna-have-to-use-this. Hats off to this wonderful middle school math teacher for engaging his students in some surprisingly rich problems.

Part 1 (be sure to read the comments in the original YouTube video to see why the answer isn’t $2^{10} \cdot 10!$ ):

Part 2:

Forgot algebra

Source: http://www.xkcd.com/1050/

Trig identity

Password strength and combinatorics

Source: http://www.xkcd.com/936/

Engaging students: Inverse Functions

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Allison Myers. Her topic, from Algebra II: inverse functions.

CURRICULUM

How can this topic be used in your students’ future courses in mathematics or science?

Functions are a composition of one or more actions that maps one object onto another (each input maps to one output). Inverse functions are a composition of reverse actions that “undo” the actions of the original function.

Inverse functions have real-world applications, but also students will use this concept in future math classes such as Pre-Calculus, where students will find inverse trigonometric functions. Inverse trigonometric functions have a whole new set of real-world applications, such as finding the angle of elevation of the sun, or anything which models harmonic motion.

Students will also see this concept again in Calculus, where they will differentiate inverse trigonometric functions to solve real-world applications involving rate of angular rotation or the rate of change of angular size.

TECHNOLOGY

How can technology be used to effectively engage students with this topic?

In the past, I taught a lesson where the Explore portion of the lesson utilized dry erase markers and transparency sheets to allow students to discover what happens graphically when computing an inverse (trigonometric) function. My goal was for my students to understand why we compute inverses the way we do. To my horror, my theoretical 15-minute, super insightful Explore became messy, full of problems, and confusing to my students.

While reflecting after the lesson, I began to consider how using technology would have better served my students (in their understanding) and myself (in my goals for the lesson). I found Glencoe’s directions for using the TI-Nspire to compute inverse functions (see image below). Using the TI-Nspire, I would start the lesson with a real-world example and data and have my students complete Step 1. Next, I would explain our need to “undo/reverse” the data, and allow the students to come up with different ways to do so. After that, I would ask the students to make conjectures about possible formulas. Using the TI-Nspire would be less messy and time-consuming (as compared to my experience with markers and transparencies), and would also allow the teacher to be within the context of a real-world problem. I believe if we used this (or similar) technology, combined with the constructivist-style teaching, students would come away with not only a better understanding for computing inverse functions but also their real-world applications.

Source: http://glencoe.com/sites/common_assets/mathematics/alg2_2010/other_cal_keystrokes/TI-Nspire/Nspire_423_424_C07L2B_888482.pdf

CULTURE

How has this topic appeared in pop culture (movies, TV, current music, video games, etc.)?

Inverse functions are used every day in real life. For example, when a computer reads a number you type in, it converts the number to binary for internal storage, then it prints the number out again onto the screen that you see – it’s utilizing an inverse function. A basic example involves converting temperature from Fahrenheit to Celsius.

Another example, if one considers music notes on paper to be a function of the sound produced, then the software Sibelius can be considered the inverse function, as it takes a musician’s music and converts it back to music notes.

Proofs Without Words: Alternating Sums of Consecutive Squares

For a visual proof that

$(2n)^2 - (2n+1)^2 + (2n+2)^2 \dots + (4n)^2 = -(4n+1)^2 + (4n+2)^2 \dots + (6n)^2$ ,

see https://www.facebook.com/photo.php?fbid=435352806598223&set=a.416585131808324.1073741827.416199381846899&type=1&theater.