Transcript for the Math Jam "AoPS Intermediate Trig/Complex Numbers" on Sep 13.

Math Jam hosted by nsato (Naoki Sato ).

nsato (19:30:33)
Welcome to this Math Jam. Today we will be dealing with complex numbers.

nsato (19:30:45)
Before we get started, I'd like to remind you all of a few classroom procedures.

nsato (19:30:49)
First of all, this classroom is moderated. This means that the messages you type will come to the instructors rather than going directly into the room. The instructors will choose some of the messages to share with all of the students.

nsato (19:30:58)
Also, sometimes after asking a question, the instructor will wait a couple minutes before proceeding, giving many students in the class a chance to respond. Take the time to make sure your responses are clear, and include your reasoning in your responses.

nsato (19:31:04)
Next, only the instructors have the ability to send private messages in the classroom. Sometimes they will use these to respond to your comments. These notes will appear in your window with the note 'Single click on this message to reply'. You do not need to click on the message to reply -- just type your reply if you have one into the reply box as usual and the instructors will receive it.

nsato (19:31:23)
Complex numbers are tricky things. Sometimes we can use complex numbers to solve problems that appear to have nothing to do with complex numbers. In particular, trigonometry problems are very susceptible to solutions with complex numbers.

nsato (19:31:29)
We start by recalling a few things about complex numbers.

nsato (19:31:33)
A complex number is a number of the form a+bi, where

nsato (19:33:33)


nsato (19:34:30)
It is often convenient to write complex numbers in exponential form. A nonzero complex number can be written in the form

nsato (19:34:56)


nsato (19:35:04)
Here?s a geometrical interpretation of exponential form:

nsato (19:35:14)


nsato (19:35:15)
http://www.artofproblemsolving.com/Classes/Complex/Images/retotheitheta.gif

nsato (19:35:27)
This representation is not unique: if we replace theta with theta+2pi, we get the same complex number. We also want to make r>0.

nsato (19:35:36)
It would make sense to explain what e^(i theta) is. It is equivalent to cos (theta)+i*sin(theta), so we sometimes also write r cis(theta) instead of re^(i theta). (cis is an abbreviation for cos+i*sin.)

nsato (19:35:42)
One of the most important sets of complex numbers is the set of the roots of unity. A complex number z is a root of unity if, for some positive integer n, z^n=1. Then z is said to be an nth root of unity. If n is the smallest number so that z^n=1, then z is said to be a primitive nth root of unity.

nsato (19:35:54)
All the nth roots of unity are of the form e^(2pi*i*k/n), where k = 0, 1, 2, ?, n-1. In other words, the roots of unity form a regular n-gon inscribed in the unit circle. Thinking geometrically like this is extremely useful in solving problems, especially about complex numbers.

nsato (19:36:14)
There are a few more properties of complex numbers that can be useful. If we have a complex number z=a+bi=re^(i theta), then it has a magnitude,

nsato (19:36:20)


nsato (19:36:33)
It also has a real part, Re{z}=a, and an imaginary part, Im{z}=b. It also has a conjugate called z-bar,

nsato (19:36:43)


nsato (19:36:50)
Any questions?

mdk (19:37:03)
so far so good

boostage (19:37:33)
yeah

nsato (19:37:56)
Great.

nsato (19:38:00)
Let?s start doing some problems!

nsato (19:38:11)


nsato (19:38:20)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-6829972.gif

nsato (19:38:26)
At a first glance, this problem appears to have nothing to do with complex numbers. But given what we just did, can we turn this problem into one that uses complex numbers? And how?

nsato (19:39:05)
We see the cosine and wonder if we can make it into the real part of a complex number. How do we do that?

.cpp (19:38:59)
Should we change cos to cis?

nsato (19:39:11)
Right!

nsato (19:39:16)


nsato (19:39:22)
Does this help?

nsato (19:39:40)
What kind of sum is the second sum?

boostage (19:39:49)
geometric series

nsato (19:40:09)
So what do we do with a geometric series?

nsato (19:40:29)
Sum it!

nsato (19:40:32)
We get

nsato (19:40:43)


nsato (19:40:53)
But we?re trying to divide by a complex number! Are we in trouble?

nsato (19:41:16)
No; we have a trick that allows us to divide by complex numbers:

nsato (19:41:22)


.cpp (19:41:21)
Multiply by z-bar?

nsato (19:41:47)
That's exactly it. It's an example of rationalizing the denominator.

nsato (19:42:01)
So now we know how to divide by complex numbers in the a+bi form. But here we have a complex number in the form 1-e^(i theta). So what do we need to do?

nsato (19:42:39)
We need to turn 1-e^(i theta) into a+bi (which is known as rectangular form). How do we do that?

nsato (19:43:36)
We have that e^(i theta) = cos theta + i sin theta. What is cos theta?

boostage (19:43:53)
well from the beginning, cos(theta) = 1/5, so b=1/5

nsato (19:44:20)
We haven?t used the bit of information at the beginning that says that cos(theta)=1/5! The time to use it is right now. But we need the sine of theta as well, don?t we? So what is it?

boostage (19:45:09)
from sin^2(theta)+cos^2(theta)=1

nsato (19:45:26)
We can use the relation cos^2 theta + sin^2 theta = 1.

boostage (19:44:49)
sqrt(1-1/25)

nsato (19:45:47)


nsato (19:46:01)
So now, what is 1-(e^(i theta))/2?

.cpp (19:46:45)
1/5 + sqrt(24)i/5

nsato (19:47:09)
That's cis (theta).

nsato (19:47:24)
1 - (e^(i theta))/2 =

nsato (19:47:29)


nsato (19:47:42)
So now what do we do?

.cpp (19:48:29)
plug it in and multiply by z-bar/z-bar

boostage (19:48:28)
multiply by the conjugate

nsato (19:48:45)


nsato (19:48:53)
So what is the answer to our problem?

.cpp (19:49:23)
6/7 + i4sqrt(6)/21

nsato (19:49:47)
We simply take the real part to get (9/10)/(21/20)=6/7.

nsato (19:50:02)
Here's the next problem.

nsato (19:50:08)


nsato (19:50:18)
http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-177638878.gif

nsato (19:50:30)
How much do we know already about raising i to various powers?

nsato (19:50:50)
We know how to raise i to integer powers. But we don?t even know how to take the square root of i yet, so how can we possibly hope to raise it to the ith power without being at least a little bit clever?

.cpp (19:50:49)
1, i, -1, -i repeated?

nsato (19:51:09)
Fortunately, we can be clever. How?

nsato (19:51:16)
We can write i in a different form. Which one?

boostage (19:51:09)
put i in exponential form

nsato (19:51:35)
We can write

nsato (19:51:40)


nsato (19:51:55)
How does that help us?

.cpp (19:52:09)
we can multiply exponents

nsato (19:52:39)
Now we can raise it to the power of i to get

mdk (19:51:55)
e^(-pi/2)

nsato (19:52:52)


nsato (19:53:00)
which I find to be a really beautiful result.

nsato (19:53:03)
Are we done?

nsato (19:53:24)
No, actually we aren?t. We picked i=e^(i*pi/2), but recall that we can replace theta by theta+2pi and get the same number in exponential form.

nsato (19:53:33)
So what do we get if we plug in 5i*pi/2 instead of i*pi/2, we might get a different answer. Do we?

nsato (19:54:24)
Yes, we do.

nsato (19:54:27)


nsato (19:54:42)
What is the general form that encompasses all the solutions?

.cpp (19:55:28)
e^(k*pi/2) where k is odd?...

nsato (19:55:43)
Close!

nsato (19:56:08)
We write i=e^(i*pi/2+2k*i*pi) for any integer k.

nsato (19:56:12)
What is i^i in that case?

boostage (19:56:25)
e^-(1+2k(pi))/2

nsato (19:57:36)
Almost.

nsato (19:57:41)


.cpp (19:57:45)
e^(-pi/2-2kpi)

nsato (19:57:57)
Isn?t in amazing that an imaginary number to an imaginary power can be real?

nsato (19:58:07)
It may seem strange that the number i^i can be all these values. It's all due to the fact that the exponential form of a complex number is not unique. Complex exponents put a whole new spin on things.

nsato (19:58:16)
There are many beautiful results in the theory of complex numbers, and I encourage anyone who knows multivariable calculus to study the theory of complex variables.

nsato (19:58:21)
However, complex numbers also lend themselves to a lot of traps. Here is my favorite one, which is due to one of the pioneers of the field of complex analysis. We will analyze what is wrong with this sort of reasoning. If you have any trouble following the steps, feel free to stop me and ask why it works.

nsato (19:58:35)
For any integer (positive, negative, or zero),

nsato (19:59:29)


nsato (19:59:46)
Now multiply by e to get

nsato (19:59:51)


nsato (20:00:15)
Now we raise both sides to the power of 1+2i*pi*n to get

nsato (20:00:19)


nsato (20:00:27)
Now we use the property (a^b)^c=a^(bc) to expand the left side out to get

nsato (20:00:43)


nsato (20:01:08)
Now we divide both sides by e to get

nsato (20:01:13)


nsato (20:01:33)
Now e^(4i*pi*n)= 1 since e^(2i*pi*n)=1. Therefore we can get rid of that term to get

nsato (20:01:40)


nsato (20:01:49)
But that is only true when n=0!

nsato (20:02:05)
Now we will discuss what went wrong. The normal thing that goes wrong in these sorts of tricks is that we use different values of the same expression. For example, a standard method of ?proving? that 1=0 involves taking the positive and negative square roots of a number and pretending that they are the same.

nsato (20:02:31)
Did we do anything like that here?

.cpp (20:03:10)
different e^x values

nsato (20:03:32)
Yes, we did. When we raised both sides to the power of 1+2i*pi*n, that wasn?t very kosher. In general, raising a number to a complex power will give you not one, not two, but infinitely many different possible results, just like when we calculated i^i.

nsato (20:03:36)
Here we chose two results that suited us. After that point, we were doing operations on false equality, so they were meaningless.

nsato (20:04:21)
Are there any questions about the class?

Jashper (20:05:07)
If I have no clue of what has been going on thus far, but I can do the prereq. problems, will I be able to understand this class?

nsato (20:06:05)
If you mean the pre-test, yes. The material in the class will be easier, and be explained in more detail than what we've done tonight.

.cpp (20:05:12)
Will the course spend more time on complex numbers and trigonometry combined or those separate subjects?

nsato (20:07:05)
About half and half. The first half will be on trig, and the second half will be on complex numbers.

xadowlugia (20:06:15)
e^(2i *pi *n) = 1

xadowlugia (20:06:24)
why is that true?

nsato (20:07:32)
It's because of something called Euler's Formula.

nsato (20:07:45)


nsato (20:08:21)
The proof is somewhat advanced, because it requires calculus.

mdk (20:07:56)
where is the pre-test?

nsato (20:08:46)
On the Enroll page.

nsato (20:08:47)
http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php

nsato (20:09:00)
Scroll down to the trig class, and click on ""Are You Ready?"".

.cpp (20:09:24)
Are the questions we have done tonight standard for the problem sets, and if so, at what point in the class will we reach this type of problem?

nsato (20:10:37)
I would say yes, at least for the final. We reach this point in about the last third of the course. For example, class 9 is on the exponential form.

mdk (20:09:47)
thanks

mdk (20:10:38)
can we do one more problem?

nsato (20:10:55)
I'm afraid that's all I have.

mdk (20:11:07)
that's cool

nsato (20:11:16)
Thanks for attending today?s Math Jam. I hope you enjoyed it.