BAM!!! Mr. Tarrou Let's take a look at DeMoivre's Theorem and learn how to expand complex numbers
in polar form. I am going to help explain where this pattern comes from, do not a proof
but start to show you how that pattern comes up so you can see where that comes from, and
do one example. In the next video linked above we are going to use DeMoivre's Theorem to
find the root of some complex numbers in polar form. But for now, how is that we can just
take a complex value of r times cosine theta plus i times sine theta, raise it to some
ridiculous power like the seventh power which will be our example I do, and just take the
radius and raise it to the n power and then do the cosine of n times theta plus i sine
of n theta? You can see this is a pretty simple looking formula. Even if I take this complex
number to the seventh power, I can pretty much just write the answer. This is why. Here
we have a complex number in polar form of r times cosine theta plus i times sine theta.
If you don't want to see this, this is not going to be a full proof, but if you want
to skip the next few minutes and get to the example, go right ahead. We are going to take
this complex number in polar form and square both sides and I am going to show you the
amount of work it takes to raise this complex number only to the second power and then hopefully
point out how you can see the pattern in this formula and just apply it in the next example.
Ok. So, I am going to take this and square it. Now this is going kind of like power to
power and I am going to distribute this power of two. I kind of missed there. It is going
to be equal to r squared, that is where I missed that line, times the cosine of theta
plus i times the sine of theta squared. Well, you know, my students like to short cut this
squaring of binomials all the time and get the wrong answer. When you square a binomial
you should be getting three terms. So, I am going to go ahead and drop that power of two
and just write this twice. Now I am going to start the distributive property multiplying
those binomials together. We have r squared. Now cosine times cosine is cosine squared
theta, and cosine times isine(theta) is positive isine(theta)cosine(theta). That is these two
multiplications and now let's continue the distributive process. We have i sine(theta)cosine(theta)
again, then plus i squared, because i times i is i squared, sine squared theta. Normally
if it were just regular binomials we would end up with three terms. This time we are
going to end up with two and this is why... r squared. And what do we have? Let's not
forget that i squared, i is the square root of negative one, so i squared is going to
be replaced with negative one. So, we have cosine squared theta and then with the i squared
minus sine squared theta. Here we have i sine(theta)cosine(theta) and i sine(theta)cosine(theta), they are like
terms so we are just going to add those together. Ok, so I said that we were only going to be
left with two terms but we have three...1,2,3. But within these three terms we have some
double angle identities. The cosine theta squared minus sine squared theta is the cosine
of 2theta, this is plus i times 2 sine(theta) cosine(theta)... 2 times sine times cosine
is the right hand side of the double angle identity for sine...so 2sine(theta)cosine(theta)
is the sine of 2theta. Z squared is equal to r squared times cosine (2theta) plus i
sine(2theta). Now if I wanted to raise this complex number to a third power, it would
be z squared times z and I could take this and multiply it again by r(cosine(theta)+isine(theta))
But that would even be more work and you can already see the pattern show up. I took this
complex number in polar form and raised it the second power and indeed we do have r squared,
I took this complex number in polar form and raised it to a second power and we have 2theta
and again 2theta. You can see how DeMoivre's Theorem is working here. We have r to the
n power times the cosine of n theta plus i sine n theta and indeed we have that pattern
show even with the power of two. If you go to the power of three and four and so on,
that pattern will continue. There is not a proof, this is incomplete for a proof, but
a demonstration of how this pattern shows up and a little bit about why DeMoivre's Theorem
works. Let's use that theorem to do an example. I am actually going to kind of do two examples at once. I would like to
raise negative one, or excuse me, that is not right, (-1+ i) to the seventh power. That
complex number is not even in polar form. We are going to get that complex number into
polar form which is going to be a little bit of work, and then we are going to raise it
to the seventh power. That is still going to be less work than me trying to do (-1+i)^7,
me writing that seven times and trying to do all of that distributive process over and
over and over again. Even though this example is going to be a little bit long, it is still
a lot less work than just trying to raise that to the second power with plain algebra.
We need to first review how to get a complex number into polar form. The first thing that
you going to want to do is graph it. This is left one and up one. Clearly I am not going
to make this a point that is super hard for me to work with. A complex number in polar
form is r(cosine(theta)+isine(theta)), so I need to find r which is the distance that
complex number is from the origin or the absolute value of z and then find theta. r is equal
to the square root of a squared plus b squared, or the square root of negative one squared
plus one squared. That is of course going to be the square root of two. Now let's find
theta. My x and y coordinates are the same, so hopefully...well, my real and my imaginary
coefficient of one for this complex number...they are basically the same at -1 and 1. Theta,
you could probably already know is going to be what...135 degrees or 3pi/4. But, if this
were an angle that we could not just do in our head we would be doing the tangent of
theta is equal to b over a or one over negative one. Theta is equal to the inverse tangent
of negative one. If you were using your calculator, the calculator would tell you that this is
equal to negative 45 degrees. I am going to do this in radians and again I have an answer
that is basically on the unit circle if you are studying trig, if you are watching this
lesson I would think so. You might get negative forty-five degrees if your calculator is in
degree mode because of the restrictions on then inverse trig functions. Negative 45 degrees
is quadrant four and my point is in quadrant two. Let's get to the deal here. Theta is
3pi/4, because I going to do this problem in radians. Just make sure that when you are
trying to find theta, when you are going from complex form into polar form of a complex
number, that your theta value that you end up using gets you in the appropriate quadrant.
Don't just take the number from your calculator if that is what you are using to come up with
the answer and think...oh that is it. This comes out to z equals the square root of two
times the cosine of 3pi/4 plus i sine of 3pi/4. Now I have my complex number in polar form.
I want to raise that to the seventh power. z to the seventh power is going to be this
complex number in polar form raised to the seventh power. The DeMoivre's Theorem says simply take the square root of
two, or the r, and raise it to the seventh power. Then take that power, that seven, that
n, and multiply it to all of the angle measures. It is going to be cosine of 7 times 3pi/4
plus i times sine of 7 times 3pi/4. We get the cosine of 21pi/4 plus i sine of 21pi/4.
What to do about the square root of two to the seventh power? You can just come out with
it, doing it off the top of your head. The square root of two times the square root of
two is two. So, two times two is four, times two is eight, and so it is eight square root
of two. You could, if you can't do it in your head, do a little scratch work. This is equal
to two to the power over root and that could be rewritten as 2 to the 6/2 power times 2
to the 1/2 power. Six divided by two is three. Two to the third power is eight, two times
two is four, four times two is eight. Power over root and we have eight times square root
of two. If you are a little bit weak on your radical notation and powers, you can write
your powers as fractional exponents and work it out that way. You can do this a multitude
of ways. I am not going to be happy with the square root of two raised to the seventh power,
so it is 8 square root of two. That is how you take a complex number and convert into
polar form, and then take that polar form, that complex number in polar form and raise
it to a very very high power using DeMoivre's Theorem. That is it, I am Mr. Tarrou. BAM!!!,
Go Do Your Homework. NO!!! That was not it:) This question started off in complex form
and i left my answer in polar form. That might be ok, however it might not. Let's take this
number that is in polar form, this complex number in polar form and put it back into
rectangular form. That is going to be eight square root of two. The cosine of 21pi/4.
There are a couple of ways that you can do this. You can just count it out which is what
I am going to do. 0, 1, 2, 3, 4pi/4, 5, 6, 7, 8pi/4. A full rotation is 8pi/4, so if I
go another full rotation it is going to be 16pi/4, 17pi/4, 18pi/4, 19pi/4, 20pi/4, 21pi/4.
We are in quadrant three, so the cosine of 21pi/4 is negative square root of two over
two plus i, and the sine of 21pi/4 that is below the x axis if you're are doing standard
position rotation, that will also be negative square root of two over two. Now when I multiply
through my 8 square root of two what is going to happen? Well, we have eight times negative
one and square root of two times square root of two is two. Square root of two times square
root is two and we have a denominator of two, so the two in the numerator is going to cancel
with the 2 in the denominator leaving you with negative eight. This is also negative,
so minus eight i. Now that is it, now I am done. BAM!!! Go Do Your Homework.