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.

Area of a circle (Part 3)

Math majors are completely comfortable with the formula $A = \pi r^2$ for the area of a circle. However, they often tell me that they don’t remember a proof or justification for why this formula is true. And they certainly don’t remember a justification that would be appropriate for showing geometry students.

In this series of posts, I’ll discuss several ways that the area of a circle can be found using calculus. I’ll also discuss a straightforward classroom activity by which students can discover for themselves why $A = \pi r^2$ .If $R$ denotes a circular region with radius $a$ centered at the origin, then

$A = \displaystyle \iint_R 1 \, dx \, dy$

This double integral may be computed by converting to polar coordinates. The distance from the origin varies from $r=0$ to $r=a$ , while the angle varies from $\theta = 0$ to $\theta = 2\pi$ . Using the conversion $dx \, dy = r \, dr \, d\theta$ , we see that

$A = \displaystyle \int_0^{2 \pi} \int_0^a r \, dr \, d \theta$

$A = \displaystyle \int_0^{2\pi} \left[ \frac{r^2}{2} \right]_0^a \, d\theta$

$A = \displaystyle \int_0^{2\pi} \frac{a^2}{2} \, d\theta$

$A = \displaystyle 2 \pi \cdot \frac{a^2}{2}$

$A = \displaystyle \pi a^2$

We note that the above proof uses the fact that calculus with trigonometric functions must be done with radians and not degrees. In other words, we had to change the range of integration to $[0,2\pi]$ and not $[0^o, 360^o]$ .