Math Girls: Chapter 3.4 Orbits and Roots Off the Unit Circle
I was in the library after school, as usual, a couple days after Miruka showed me the omega waltz. I was thinking about how I kept going to straight lines for the orbits, rather than circles, and I wondered if the concept could be generalized further. Why was it always equal to 1? Why not some other integer? For that matter, did it even have to be an integer? I decided to write up a problem for Miruka to try.
Problem 3-4
What does the orbit of omega^3 = 8 look like?
A slight rustling of fabric alerted me to Miruka's presence behind me, but I still jumped when I felt her hand on my shoulder as she sat down beside me.
"What's this?" she asked, snatching up the card with my problem on it.
And, again, she skips any attempt at small talk. Sigh.
"It's a problem for you to try. You've given me homework, so I decided to return the favor."
"I see," she mused. "And have you worked this problem out for yourself yet?"
"Well," I hedged. "Not yet. I was hoping we could look at it together."
"So it's not really 'a problem for me to try,' is it?" She smirked, then continued. "But I like your initiative. So, what were you thinking when you came up with this?"
"What, too tough for you?" My smirk hardly lasted a second after her response.
"Don't get a big head, now." She shook her head. "No; I just wanted to make sure what I thought the focus of the problem is matches what you intended."
"Right." I smiled, reminded of my discussion of Tetra about who's behind the math. "So, I was thinking about generalizations. What if we have some constant other than 1 in the omega waltz?"
"Okay, that's what I figured," she said with a nod. "But why'd you write 8 as the constant?"
"Well, it's the next perfect cube after 1," I replied with a half shrug. "I wasn't sure how changing the constant would change the overall outcome, and I hoped this choice would keep the numbers simple to work with."
"I see." Miruka tilted her head to the side. "But if you wanted a generalization, shouldn't you have used an arbitrary constant instead?"
"Fine," I huffed. "So omega^3 = kappa. What's its orbit look like?"
"I don't remember off the top of my head, so let's explore it together." She turned to a new page in the notebook and started drawing a table. "So, we need columns for omega, kappa, theta, cos(theta), i*sin(theta), anything else...?"
"How about a polar grid?" I suggested. "Ms. Mizutani might have some we could use."
"Good idea."
When I returned with a pad of polar graph paper, Miruka had a tense frown on her face. I sat down and asked, "What's wrong?"
"This whole problem..." she answered. "That's what's wrong."
I was bewildered. "What do you mean?"
Miruka closed her eyes and took a deep breath before answering. "We were confusing roots of complex numbers with their orbits. We could have 'what are the roots of omega^3 = kappa?' or 'what is the orbit of omega = some number?'" She opened her eyes, their intensity revealing that her confident side was back, and asked, "So, which do you want to try first?"
I made a snap decision. "Let's start with the roots, since we already have the equation."
"Let's get to it, then," she said with a nod.
"Well, looking at examples can help with understanding a general situation, so let's work with my original equation:
Problem 3-4, version 2
What are the roots of omega^3 = 8?
"The cube root of 8 is 2-" I began, but she cut me off.
"You're not wrong," she said, to which I raised an eyebrow. "One of the cube roots of 8 is 2, but how can we show this using trigonometric functions?" Before I could even start to say anything, she answered her own question. "Start by writing 8 in complex form."
"You mean 8 + 0i?"
"Again, not wrong. I meant the trigonometric complex form, 8(cos(2pi) + i*sin(2pi))."
"And we take the cube root of this, right?" I hazarded. I wasn't sure why she said 2pi, rather than 0, but I figured it would be clearer later. "So we have 2 times the cube root of cis(2pi)?"
"Oh, so you do know that abbreviation; I wasn't using it myself because I wasn't sure. But anyway-"
"Not wrong?" I guessed, going by her facial expression.
She nodded. "There's an equivalent, easier way to take the cube root of sine and cosine functions. Remember earlier when we talked about a single rotation by 2*theta being identical to two rotations by 1*theta?"
"I see," I said, catching on. "So in place of taking the cube root, which is equivalent to raising to the one-third power, we'll multiply the arguments by one-third, or divide by three! That does make it easier."
Taking my pencil again and writing the math while she spoke, Miruka said, "Exactly. So we have 2*cis(2pi/3) for one root. For the next root, we leave the cube root of 8 as-is, and take the cube root of cis(4pi), giving us 2*cis(4pi/3)."
"And the third root is 2*cis(6pi/3), or 2*cis(2pi)." I paused. "But wait, cos(2pi) is 1 and sin(2pi) is 0... Oh! That's what gives us 2 as a cube root of 8!"
"Right, and the others simplify to...?" Miruka prompted.
"I'll write it out as I go to make sure I don't make a simple mistake."
2*cis(2pi/3) = 2(cos(2pi/3) + sin(2pi/3)) = 2(-1/2 + sqrt(3)/2) = -1 + sqrt(3)
"And then the other one, graphically, is the reflection of this one over the real axis, so the real part is the same, and the imaginary part has the opposite sign, hence -1 - sqrt(3)."
"Right. So, in general, the solutions of omega^3 = kappa are:"
[Author's Note: By cbrt, I mean cube root.]
cbrt(kappa)*(cis(theta))
cbrt(kappa)*(cis(2*theta))
cbrt(kappa)*(cis(3*theta))
for given kappa, and theta the angle between the positive real axis and the vector representing kappa.
"Got it?"
"Yup. So, orbits now?"
"Sure. Give me an angle and a number."
"Um, pi/4 and 2."
"Alright. Now, look at this:
omega = r*cis(theta)
omega = 2*cis(pi/4)
omega^2 = 2^2*cis(2*pi/4) = 4*cis(pi/2)
omega^3 = 2^3*cis(3*pi/4) = 8*cis(3pi/4)
omega^4 = 2^4*cis(4*pi/4) = 16*cis(pi)
omega^5 = 2^5*cis(5*pi/4) = 32*cis(5pi/4)
omega^6 = 2^6*cis(6*pi/4) = 64*cis(3pi/2)
omega^7 = 2^7*cis(7*pi/4) = 128*cis(7pi/4)
omega^8 = 2^8*cis(8*pi/4) = 256*cis(2pi)
"With me so far?"
"I think so. Each successive point in the orbit of omega adds pi/4 to the argument and multiplies the r value by 2. Now we simplify those expressions, giving us:"
omega = 2*sqrt(2)/2 + 2sqrt(2)i/2 = sqrt(2) + sqrt(2)i
omega^2 = 4*(0 + i) = 4i
omega^3 = 8*(-sqrt(2)/2 + sqrt(2)i/2) = -4sqrt(2) + 4sqrt(2)i
omega^4 = 16*(-1 + 0i) = -16
omega^5 = 32*(-sqrt(2)/2 - sqrt(2)i/2) = -16sqrt(2) - 16sqrt(2)i
omega^6 = 64*(0 - i) = -64i
omega^7 = 128*(sqrt(2)/2 - sqrt(2)i/2) = 64sqrt(2) - 64sqrt(2)i
omega^8 = 256*(1 + 0i) = 256
I pulled out a sheet of square-grid graph paper and graphed the points, highly scaled down, considering how large the numbers got. "Can this still technically be called an orbit, even though it doesn't return to where it starts?" I mused.
"Good question. I don't know. I suppose so, since it would still spiral out around the full 2pi radians over and over, if we graphed more points." Miruka shrugged.
"Do you think it would spiral in if we used an r value less than one?" I conjectured.
"Why do you say that?" Miruka asked.
"Well, an r value of 1 puts us on the unit circle, and a value greater than 1 spiraled outward, so, by process of elimination, less than 1 spirals in."
"Hm. It's possible, I suppose. Let's try it!"
Before I could voice my agreement, another voice joined the conversation. "Not right here and now, you won't. The library's closed," Ms. Mizutani told us.
"Alright," I said, standing up to bow. "Oh, Ms. Mizutani, we didn't actually need that polar graph paper after all, so where would you like me to put it?"
"I'll take care of it," she said. "Now out!"
"Yes, ma'am," Miruka and I chorused.