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Welcome to lecture four in our course analysis of a Complex Kind.

Today we'll talk about roots of complex numbers.

You all know that the square root of 9 is 3,

or the square root of 4 is 2, or the cubetrid of 27 is 3.

But how would you take a square root of 3+4i, for

example, or the fifth root of -i.

That's what we're going to talk about today.

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Suppose w is a complex number.

Then we say an nth root of w is another

complex number z such that z to the n = w.

So we'll see that if the number w is

a root we're taking, is not = 0, and

there are exactly n distinct nth roots.

So there are two distinct square roots of any number.

There are three distinct cube roots and so forth.

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So we'll write w as rho e to the i phi, and z is re to the i theta.

So you noticed we just introduced yet another Greek symbol.

Namely, the simply rho, here for the radius.

Because r is already used up for the radius of z.

Now, if we use these notations.

Then z to the n equals w.

That equation, that defining equation for being an nth root.

That equation becomes, if we write instead of z re to the i theta.

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Then we get r to the n e to the i n theta.

By the rules of how to manipulate this e to the i theta

is equal to w, which is a rho to the i theta.

So to compare these two complex numbers and polar notation one or

two numbers the same.

They have the same distance from the origin and were angles,

degree modular to pi.

So r to the n has to be equal to rho.

And this angle and

beta has to be talking about the same angle that z is talking about.

They can differ by 2 pi or by 4 pi, but they can't differ much other than that.

So m theta is pretty much the same as z,

expect maybe for At 2k pi.

So r to the n = rho, that means r itself is the nth root of rho and

since rho is just a positive real number by nth root.

When I write the symbol nth root I mean the regular nth root that we all know.

The regular nth root, not a complex root.

Just the regular nth root of rho.

And n theta has to be equal to phi + 2k pi where k is some integer.

So I can divide this equation by n and

I get theta a has to be phi over n plus 2k pi over n and it turns out.

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If I look at the values of K from 0,1 and so

forth to n-1 I have exhausted all possibilities.

After that, things start repeating.

So these are actually distinct roots, and

if I look at k equals n for example when k = m.

What would I get?

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I would get theta = fi

over n plus 2, in our case = n.

2 n pi divided by n that N would cancel out so

I would have phi over n plus 2 pi.

But I already looked at the angle theta equals phi over n 1k was equal to 0.

So I would get kind of the same angle as the one I got when k was equal to 0.

And so, I don't get any new angle.

I'm getting different angles here for k equals 0 through n minus 1 and

then the angles start repeating.

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And so, I get all my different solutions as w to the 1 over n.

This is how I denote my solutions is = to the nth root

of ro times etha i, fi over n plus 2 k pi over n.

Where k runs through the values 0, 1 all the way to n minus 1.

So these are the nth roots of w and there all these things from each of.

One special root is the principal nth root of w.

It's the one that we obtain when we choose the principal argument of w for

pe and k equals zero.

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So here again I wrote down this formula we just found.

If W is of the form rho e to the I Phi, then the nth roots of

W can be found with this formula, for these n values of k.

So let's find the square root of 4i as an example.

So first of all 4i,

we need to bring that into this form, in the form rho e to the I phi.

So we have to figure out what's rho, what's phi.

4i is 4 times instead of i we can write e to the i pi

over 2 we discovered that during the last lecture.

And so in other words, this row is 4.

And phi and a possibility for the angle phi pi over 2, and

we wanted to take this square root of the second root.

So n is 2.

So now we have everything we need to plug in this formula.

So we find 4i to the one-half is equal to

the square root of 4, times e to the i.

Now we need to take phi and divide it by n.

Phi is pi over 2, divided by n.

That gives me pi over 4.

Plus, 2 k pi over n.

We'll just write that down.

2 k pi over 2 because my n is 2.

So here's the complete formula, and we need to plug in k = 0, and then k = 1.

So we get two separate solutions.

For k = 0, this whole second term 2 k pi over 2 which is not there.

And so, we find square root of 4 which is 2 times e to the i, pi over 4.

That's one of my two square roots, and when k's equal to 1,

2k pi over 2 becomes 2 pi over 2 which is pi.

So I get 2 times either the i pi over 4 plus pi, but

remember what e to the i pi equals?

Pi is a 180 degree angle.

E to the i pi, is this number right here.

That forms a 180 degree angle.

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We also looked at that during the last lecture e to the i of pi

over four was this number on the unit circle at a 45 degree angle.

And we found that was root over two over two plus i times root 2 over 2.

If I multiply that number by two it becomes root 2 plus i over 2.

So root two plus i over two.

I see that right here, and with a plus or minus.

Plus for the k equals zero solution, minus for the k equals one solution.

So in the picture, I'm drawing a lot of pictures here all over the place.

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The square roots of four i, where are they?

Four i is kind of far up on

the Y axis, i is here.

The square root of 4i are at distance 2.

So on the circle radius 2.

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And at a 45 degree angle.

That's one of the square roots, and the other one is over here.

So these are the two square roots of 4i.

And in coordinates, they are plus or minus root two plus i root two.

Let's do the same thing for the cubed roots of -8.

First step again, we have to write -8 in polar form.

For it's just not minus eight because the polar form has to

have a positive radius right here.

While the radius of minus eight, the norm of minus eight is eight.

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And phi the angle is pi.

And you want to take the cubed roots, so my n is 3.

Again now, I have everything I need to plug in to this equation and

I can go ahead and do that.

So the cubed roots of 8 can be found by taking

the regular cubed root of 8 This is just one cubed over eight,

the one for eight being a positive number, so that's actually two.

This number right here, that's simply two,

because two cubed is eight times e to the i, phi over n.

Phi is pi, n is three.

So I get a pi over three right there.

Plus two k phi over answer, two k phi over three.

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And now I need to consider k equals zero, one and

two because I'm taking the cubed root.

So let's start with k equals zero.

So this whole second term is in there when k is equal to zero and

assumed to get two times e of the i phi over three.

When k's equal to 1, the second term becomes 2 pi over 3.

So I have pi over 3 plus 2 pi over 3.

Well, that's 3 pi over 3, or simply pi.

k = 1 my root becomes 2 to the e ipa 1,

but remember e to the i pi is negative 1.

So the second solution is negative 2, and indeed if you check that negative 2 cubed.

What's negative 2 cubed?

Well negative 2 cubed is negative 8 indeed.

And the third solution is 2 times e to the i pi over 3 and

now we have to plugin k verse 2.

So 4 pi over 3 but pi over 3 is 4 pi over 3 that's pi pi over 3.

So this is the third solution that we find.

Let's turn to our picture where we find where this solutions are.

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So there's four and there's two.

So here is the circle of radius two all cubed roots of radius two and

one of them is at at angle pi over three.

One of them is an angle pi and then we said angle five pi over three.

So here's the angle pi, and there is phi over three and

there is phi, phi over three.

So these are the cubed roots of -8.

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Just a quick other definition, the nth roots of the number 1.

Those are called the nth roots of unity, because 1 is going to be unity.

So what do those look like?

Well the number 1 is in polar form, 1 times e the i times 0, so angle 0.

So if we plot that into our formula.

We get the nth root of 1 is the regular nth root of 1,

which is just 1, times e to the i phi over n.

Phi is 0, so 0 over n plus 2k pi over n, and k runs from 0 through n minus 1.

So replacing the nth root of 1 here.

Just with a number 1, I get e to the i to k pi over n.

So these are the nth roots of unity.

And again, let's draw a picture to understand what they are.

So this time 1 I'm going to

draw that right here and let me draw the circle of radius one right here.

So this is the circle of radius one and I want to find the nth roots of one.

And they're all of radius one and their angles are 2 pi k over n.

So for example, for the eighth roots of one I

need to find the angle 2 pi over 8.

This is 2 pi over 2, coz that's pi 2 pi over 4.

So 2 pi over 8 will be here.

2 times 2 pi over 8, 3 times 2 pi over 8, 4 times 2 pi over 8,

5 times 2 pi over 8, 7 times 2 pi over 8 And this is 0 times 2 pi over 8.

So I have these equally spaced points on unit circle that

correspond to the 8th roots of unity.