# My Favorite One-Liners: Part 67

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here are a couple of similar problems that arise in Precalculus:

1. Convert the point $(5,-5)$ from Cartesian coordinates into polar coordinates.
2. Convert the complex number $5 - 5i$ into trigonometric form.

For both problems, a point is identified that is 5 steps to the right of the origin and then 5 steps below the $x-$axis (or real axis). To make this more kinesthetic, I’ll actually walk 5 paces in front of the classroom, turn right face, and then walk 5 more paces to end up at the point.

I then ask my class, “Is there a faster way to get to this point?” Naturally, they answer: Just walk straight to the point. After some work with the trigonometry, we’ll establish that

1. $(5,-5)$ in Cartesian coordinates is equivalent to $(5\sqrt{2}, -\pi/4)$ in polar coordinates, or
2. $5-5i$ can be rewritten as $5\sqrt{2} [ \cos(-\pi/4) + i \sin (-\pi/4)]$ in trigonometric form.

Once this is obtained, I’ll walk it out: I’ll start at the origin, turn clockwise by 45 degrees, and then take $5\sqrt{2} \approx 7$ steps to end up at the same point as before.

Continuing the lesson, I’ll ask if the numbers $5\sqrt{2}$ and $-\pi/4$, or if some other angle and/or distance could have been chosen. Someone will usually suggest a different angle, like $7\pi/4$ or $15\pi/4$. I’ll demonstrate these by turning 315 degrees counterclockwise and walking 7 steps and then turning 675 degrees and walking 7 steps (getting myself somewhat dizzy in the process).

Finally, I’ll suggest turning only 135 degrees clockwise and then taking 7 steps backwards. Naturally, when I do this, I’ll do a poor man’s version of the moonwalk:

# Engaging students: Graphing with polar coordinates

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 comes from my former student Perla Perez. Her topic, from Precalculus: graphing with polar coordinates.

How does this topic extend what your students should have learned in previous courses?

Graphing polar coordinates is usually taught in a Pre-Calculus class. Students have learned about the Cartesian Coordinates and extend their knowledge to polar coordinates. Unlike Cartesian Coordinates, which represent how to get from a specific point to the point of origin (or vice versa), the polar coordinate tells us the direction by the angle, and the distance from that point to the origin. Students will need to know how to take the measure of an angle and how to use the Pythagorean Theorem to solve for the distance which is considered the radius. Most students who are enrolled in a Pre-Calculus class have taken geometry where they have learned about the Pythagorean Theorem and what a radius is. This alongside their algebra 1 and geometry classes means they also know how to graph and plot points.

References:

http://ritter.tea.state.tx.us/rules/tac/chapter111/ch111c.htm

How could you as a teacher create an activity or project that involves your topic?

Polar coordinates use a different type of graph, rather than just an x and y coordinates plane. The polar coordinate plane includes symmetrical circles surrounding the center and is given a radius creating a graph that looks like a dart board. At this point students should know what a polar coordinate is. The next step is actually graphing it.

As an activity to get students excited for the wonderful world of polar coordinates, I have created a dart board game. Using an appropriate dart board, such as a magnetic one, have the students create groups of three to four student each. The point of the game is to have students create polar coordinates. The board must be properly labeled with the angles. There will be four rounds, depending on the number of members in a group. When a member throws a dart at the board it must land on a point. Wherever it lands the students must figure out the radius and the angle of the dart to the origin. This game enables students to practice finding the radius and the angle of the dart with only their previous knowledge, labels, and each other.

What interesting things can you say about the people who contributed to the discovery and/or the development of this topic?

Throughout centuries and in all parts of the world, mathematicians and astronomers have come to shape our understanding of the polar coordinate system. Two Greek astronomers Hipparchus and Archimedes used polar coordinates in much of their work. Though they didn’t commit to the full coordinate plane, Hipparchus first begins by writing a table of chords where he was able to define stellar positions. Archimedes focused on a lot on spirals and developed what now known as the Archimedes spiral, in which the radius depends on the angle. Descartes also used a simpler concept of coordinates, but relating more to the x-axis. In 1671, Sir Isaac Newton was one of the biggest contributors to the elements used in analytic geometry.  The idea of polar coordinates, however, comes from a man named Gregorio Fontana (1735-1803), centuries later. Astronomers now use his polar coordinates to measure the distance of the sky and stars.

References:

History of Mathematics, Vol. II:

https://en.wikipedia.org/wiki/Polar_coordinate_system

# How I Impressed My Wife: Part 5h

Earlier in this series, I gave three different methods of showing that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Using the fact that $Q$ is independent of $a$, I’ll now give a fourth method.
Since $Q$ is independent of $a$, I can substitute any convenient value of $a$ that I want without changing the value of $Q$. As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

The four roots of the denominator satisfy

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| \sqrt{b^2 - 1}$

So far, I’ve handled the cases $|b| = 1$ and $|b| > 1$. In today’s post, I’ll start considering the case $|b| < 1$.

Factoring the denominator is a bit more complicated if $|b| < 1$. Using the quadratic equation, we obtain

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| i \sqrt{1-b^2}$

However, unlike the cases $|b| \ge 1$, the right-hand side is now a complex number. So, To solve for $u$, I’ll use DeMoivre’s Theorem and some surprisingly convenient trig identities. Notice that

$(1-2b^2)^2 + (2|b| \sqrt{1-b^2})^2 = 1 - 4b^2 + b^4 + 4b^2 (1 - b^2) = 1 - 4b^2 + b^4 + 4b^2 - b^4 = 1$.

Therefore, the four complex roots of the denominator satisfy $|u^2| = 1$, or $|u| = 1$. This means that all four roots can be written in trigonometric form so that

$u^2 = \cos 2\phi + i \sin 2\phi$,

where $2\phi$ is some angle. (I chose the angle to be $2\phi$ instead of $\phi$ for reasons that will become clear shortly.)

I’ll begin with solving

$u^2 = \displaystyle 1 - 2b^2 + 2|b| i \sqrt{1-b^2}$.

Matching the real and imaginary parts, we see that

$\cos 2\phi = 1-2b^2$,

$\sin 2\phi = 2|b| \sqrt{1-b^2}$

This completely matches the form of the double-angle trig identities

$\cos 2\phi = 1 - 2\sin^2 \phi$,

$\sin 2 \phi = 2 \sin\phi \cos \phi$,

and so the problem reduces to solving

$u^2 = \cos 2\phi + i \sin 2\phi$,

where $\sin \phi = |b|$ and $\cos \phi = \sqrt{1-b^2}$. By De Moivre’s Theorem, I can conclude that the two solutions of this equation are

$u = \pm(\cos \phi + i \sin \phi)$,

or

$u = \pm( \sqrt{1-b^2} + i |b|)$.

I could re-run this argument to solve $u^2 = \displaystyle 1 - 2b^2 - 2|b| i \sqrt{1-b^2}$ and get the other two complex roots. However, by the Conjugate Root Theorem, I know that the four complex roots of the denominator $u^4 + (4 b^2 - 2) u^2 + 1$ must come in conjugate pairs. Therefore, the four complex roots are

$u = \pm \sqrt{1-b^2} \pm i |b|$.

Therefore, I can factor the denominator as follows:

$u^4 + (4 b^2 - 2) u^2 + 1 = (u - [\sqrt{1-b^2} + i|b|])(u - [\sqrt{1-b^2} - i|b|])$

$\qquad \times (u - [-\sqrt{1-b^2} + i|b|])(u - [-\sqrt{1-b^2} + i|b|])$

$= (u - \sqrt{1-b^2} - i|b|)(u - \sqrt{1-b^2} + i|b])$

$\qquad \times (u +\sqrt{1-b^2} + i|b|)(u +\sqrt{1-b^2} + i|b|)$

$= ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)$

To double-check my work, I can directly multiply this product:

$([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)$

$= (u^2 - 2u \sqrt{1-b^2} + 1 - b^2 + b^2) (u^2 + 2u \sqrt{1-b^2} + 1 - b^2 + b^2)$

$= ([u^2 +1] - 2u\sqrt{1-b^2})([u^2+1] + 2u\sqrt{1-b^2})$

$= [u^2+1]^2 - [2u\sqrt{1-b^2}]^2$

$= u^4 + 2u^2 + 1 - 4u^2 (1-b^2)$

$= u^4 + u^2 (2 - 4[1-b^2]) + 1$

$= u^4 + u^2 (4b^2 - 2) + 1$.

So, at last, I can rewrite the integral $Q$ as

$Q = \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)}$

I’ll continue with this fourth evaluation of the integral, continuing the case $|b| < 1$, in tomorrow’s post.

# Engaging students: Graphing with polar coordinates

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 comes from my former student Laura Lozano. Her topic, from Precalculus: graphing with polar coordinates.

How could you as a teacher create an activity or project that involves your topic?

An activity that I believe will go really well with graphing polar coordinates or any type of graphing lesson will be to convert the classroom floor into a graph. Also, I will have a selection of random objects like, a rubber ducky, boat, toy, etc. The size of the graph will depend on the size of classroom of course. If the classroom is really small then I would have to take this activity outdoors or maybe even the gym or anywhere with enough room for the graph and my students. The graph doesn’t have to be super big but I would use a graph no smaller than 8 feet by 8 feet area. I could create the graph lines with tape on the floor or draw them on big paper and tape the paper on the floor. I would start the activity with first talking about points on a Cartesian graph. An example could be to first have a students plot a couple points like (5, 4), (3, 6), or (-4, 2) on the board. Then transition them from Cartesian to polar coordinates by using the floor graph and have them discover how they relate by using the x and y coordinates to find the radius and the angle. Then later, after they get the hang of it, I would have the class split up into groups of two and let them choose an object, like a rubber ducky, boat, or toy, to set on the graph and have them write and tell me the point of their object.

We see radars in the news almost all the time. One category that it is usually used in is weather. The weather center uses their radars to detect for any water particles, debris, and basically anything that is in the air that could be approaching. The way that they tell if a storm or any other weather change is coming is by the radar’s omitting radio waves. The radar omits waves that then come back to the radar if the waves clash with anything in the air. The radar can detect how far an object is by the time it takes for the wave to come back. It works just like an echo! Also, recently with the search of the Malaysian airplane, we saw it used more. The news will show a clip of aircraft radar or ship radar searching for something in the air or in the ocean. Radars look almost exactly like a polar graph does. On the left is a regular polar graph. On the right is a ship’s radar. Both graphs have angles with circles.

How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic? Note: It’s not enough to say “such-and-such is a great website”; you need to explain in some detail why it’s a great website.

Graphing calculators can be used to discover polar coordinates and polar equations. I would first tell them to take out their calculators and just type in a random number from -10 to 10. I choose this interval because the graphing calculators have this window preset for graphing. I number that I randomly chose was the number 4. So I would go to the “Y=” button and type in 4. Then I would hit “GRAPH” and I should get a straight line horizontal line going through the y-axis at 4. I would then change the calculator mode and change from “FUNC” to “POL”. Then I would tell them to do the exact thing again with whatever number they chose. Once the hit “GRAPH” a circle should then come up. They then see how different polar graphs are from Cartesian graphs. Now, the graphs on a polar coordinate graph will all be circular instead of lines and curved lines like on the Cartesian graph.

Resources:

http://forecast.weather.gov/jetstream/doppler/how.htm

# Inverse Functions: Logarithms and Complex Numbers (Part 30)

Ordinarily, there are no great difficulties with logarithms as we’ve seen with the inverse trigonometric functions. That’s because the graph of $y = a^x$ satisfies the horizontal line test for any $0 < a < 1$ or $a > 1$. For example,

$e^x = 5 \Longrightarrow x = \ln 5$,

and we don’t have to worry about “other” solutions.

However, this goes out the window if we consider logarithms with complex numbers. Recall that the trigonometric form of a complex number $z = a+bi$ is

$z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

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

Over the past few posts, we developed the following theorem for computing $e^z$ in the case that $z$ is a complex number.

Definition. Let $z = r e^{i \theta}$ be a complex number so that $-\pi < \theta \le \theta$. Then we define

$\log z = \ln r + i \theta$.

Of course, this looks like what the definition ought to be if one formally applies the Laws of Logarithms to $r e^{i \theta}$. However, this complex logarithm doesn’t always work the way you’d think it work. For example,

$\log \left(e^{2 \pi i} \right) = \log (\cos 2\pi + i \sin 2\pi) = \log 1 = \ln 1 = 0 \ne 2\pi i$.

This is analogous to another situation when an inverse function is defined using a restricted domain, like

$\sqrt{ (-3)^2 } = \sqrt{9} = 3 \ne -3$

or

$\sin^{-1} (\sin \pi) = \sin^{-1} 0 = 0 \ne \pi$.

The Laws of Logarithms also may not work when nonpositive numbers are used. For example,

$\log \left[ (-1) \cdot (-1) \right] = \log 1 = 0$,

but

$\log(-1) + \log(-1) = \log \left( e^{\pi i} \right) + \log \left( e^{\pi i} \right) = \pi i + \pi i = 2\pi i$.

This material appeared in my previous series concerning calculators and complex numbers: https://meangreenmath.com/2014/07/09/calculators-and-complex-numbers-part-21/

# Calculators and complex numbers (Part 24)

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

$z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

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.

Theorem. If $z = x + i y$, where $x$ and $y$ are real numbers, then

$e^z = e^x (\cos y + i \sin y)$

Definition. Let $z = r e^{i \theta}$ be a complex number so that $-\pi < \theta \le \theta$. Then we define

$\log z = \ln r + i \theta$.

Definition. Let $z$ and $w$ be complex numbers so that $z \ne 0$. Then we define

$z^w = e^{w \log z}$

Technical point: for the latter two definitions, these are the principal values of the functions. In complex analysis, these are usually considered multiply-defined functions. But I’m not going to worry about this technicality here and will only consider the principal values.

This is the last post in this series, where I state some generalizations of the Laws of Exponents for complex numbers.

In yesterday’s post, we saw that $z^{w_1} z^{w_2} = z^{w_1 + w_2}$ as long as $z \ne 0$. This prevents something like $0^4 \cdot 0^{-3} = 0^1$, since $0^{-3}$ is undefined.

Theorem. Let $z \in \mathbb{C} \setminus \{ 0 \}$, $w \in \mathbb{C}$, and $n \in \mathbb{Z}$. Then $(z^w)^n = z^{wn}$.

As we saw in a previous post, the conclusion could be incorrect outside of the above hypothesis, as $\displaystyle \left[ (-1)^3 \right]^{1/2} \ne (-1)^{3/2}$.

Theorem. Let $u \in \mathbb{R}$ and $z \in \mathbb{C}$. Then $(e^u)^z = e^{uz}$.

Theorem. Let $x, y > 0$ be real numbers and $z \in \mathbb{C}$. Then $x^z y^z = (xy)^z$.

Again, the conclusion of the above theorem could be incorrect outside of these hypothesis, as $(-2)^{1/2} (-3)^{1/2} \ne \left[ (-2) \cdot (-3) \right]^{1/2}$.

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 23)

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

$z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

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.

Theorem. If $z = x + i y$, where $x$ and $y$ are real numbers, then

$e^z = e^x (\cos y + i \sin y)$

Definition. Let $z = r e^{i \theta}$ be a complex number so that $-\pi < \theta \le \theta$. Then we define

$\log z = \ln r + i \theta$.

Definition. Let $z$ and $w$ be complex numbers so that $z \ne 0$. Then we define

$z^w = e^{w \log z}$

Technical point: for the latter two definitions, these are the principal values of the functions. In complex analysis, these are usually considered multiply-defined functions. But I’m not going to worry about this technicality here and will only consider the principal values.

In the remaining posts in this series, I want to explore which properties of exponential functions remain true when complex numbers are used.

To begin, if $w$ is a real rational number, then there is an alternative definition of $z^w$ that matches De Moivre’s Theorem. Happily, the two definitions agree. Suppose that $z = r e^{i \theta}$ with $-\pi < \theta \le \pi$. Then

$z^w = e^{w \log z}$

$= e^{w [\ln r + i \theta]}$

$= e^{w \ln r + i w \theta}$

$= e^{w \ln r} e^{i w \theta}$

$= r^w (\cos w\theta + i \sin \theta)$

Next, one of the Laws of Exponents remains true even for complex numbers:

$z^{w_1} z^{w_2} = e^{w_1 \log z} e^{w_2 \log z}$

$= e^{w_1 \log z + w_2 \log z}$

$= e^{(w_1 + w_2) \log z}$

$= z^{w_1 + w_2}$.

However, in previous posts, we’ve seen that the rules $(x^y)^z = x^(yz)$ and $x^z y^z = (xy)^z$ may not be true if nonpositive bases, let alone complex bases, are used.

We can also derive the usual rules $z^0 = 1$ and $z^{-w} = \displaystyle \frac{1}{z^w}$. First,

$z^0 = e^{0 \log z} = e^0 = 1$.

Next, we think like an MIT freshman and use the above Law of Exponents to observe that

$z^w z^{-w} = z^{w-w} = z^0 = 1$.

Dividing, we see that $z^{-w} = \displaystyle \frac{1}{z^w}$.

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 22)

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

$z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

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.

Theorem. If $z = x + i y$, where $x$ and $y$ are real numbers, then

$e^z = e^x (\cos y + i \sin y)$

Definition. Let $z = r e^{i \theta}$ be a complex number so that $-\pi < \theta \le \theta$. Then we define

$\log z = \ln r + i \theta$.

At long last, we are now in position to explain the last surprising results from the calculator video below.

Definition. Suppose that $z$ and $w$ are complex numbers so that $z \ne 0$. Then we define

$z^w = e^{w \log z}$

Naturally, this definition makes sense if $z$ and $w$ are real numbers.

For example, let’s consider the computation of $i^i$. For the base of $i$, we note that

$\log i = \log e^{\pi i/ 2} = \displaystyle \frac{\pi i}{2}$.

Therefore,

$i^i = e^{i \log i} = e^{i \pi i/2} = e^{-\pi/2}$,

which is (surprisingly) a real number.

As a second example, let’s compute $(-8)^i$. To begin,

$\log(-8) = \log \left( 8 e^{\pi i} \right) = \ln 8 + \pi i$.

Therefore,

$(-8)^i = e^{i \log(-8)}$

$= e^{i (\ln 8 + \pi i)}$

$= e^{-\pi + i \ln 8}$

$= e^{-\pi} (\cos [\ln 8] + i \sin [ \ln 8 ] )$

$= e^{-\pi} \cos (\ln 8) + i e^{-\pi} \sin (\ln 8)$

In other words, a problem like this is a Precalculus teacher’s dream come true, as it contains $e, \ln, \pi, \cos, \sin$, and $i$ in a single problem.

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 21)

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

$z = r(\cos \theta + i \sin \theta) = r e^{i \theta}$

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.

Over the past few posts, we developed the following theorem for computing $e^z$ in the case that $z$ is a complex number.

Definition. Let $z = r e^{i \theta}$ be a complex number so that $-\pi < \theta \le \theta$. Then we define

$\log z = \ln r + i \theta$.

Of course, this looks like what the definition ought to be if one formally applies the Laws of Logarithms to $r e^{i \theta}$. However, this complex logarithm doesn’t always work the way you’d think it work. For example,

$\log \left(e^{2 \pi i} \right) = \log (\cos 2\pi + i \sin 2\pi) = \log 1 = \ln 1 = 0 \ne 2\pi i$.

This is analogous to another situation when an inverse function is defined using a restricted domain, like

$\sqrt{ (-3)^2 } = \sqrt{9} = 3 \ne -3$

or

$\sin^{-1} (\sin \pi) = \sin^{-1} 0 = 0 \ne \pi$.

The Laws of Logarithms also may not work when nonpositive numbers are used. For example,

$\log \left[ (-1) \cdot (-1) \right] = \log 1 = 0$,

but

$\log(-1) + \log(-1) = \log \left( e^{\pi i} \right) + \log \left( e^{\pi i} \right) = \pi i + \pi i = 2\pi i$.

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