Hello everyone.
In this video, we're going to talk about complex nth roots.
In particular, complex third roots, or complex cube roots,
because that's the problem that I picked.
So find the cube roots of 1 minus 2i.
You may leave part of your answer in terms of arctan. So this -
the answer to this question is going to be ugly and use
a lot of
notation. It's just the way it's going to be.
Because I don’t…
I didn't want to try to fudge the numbers so that
they were correct. I just decided hey let's just leave it and
let's just talk about the ideas because again the key ideas are what are really important
in these examples, as opposed to actually getting a complete…
well not a complete solution but you know what I mean, right,
in terms of making the numbers perfect.
So...
what are we going to do in this question? So we're trying to find
for the cube roots of 1 minus 2i, so we're trying to solve for z cubed equals 1 minus 2i,
and when we're solving for z cubed equals 1 minus 2i, we usually like -
well it's not usually, it's easier to deal with polar coordinates
than it is to deal with standard form. So we're going to let z equal to r
e to the i theta.
Then what do we have? Well if we
plug in this z value into z cubed equals 1 minus 2i,
we're going to get r cubed e to the i 3 theta is equal to 1 minus 2i,
and from here we have options. What I'm going to do
is I'm going to determine r first. r is always easy to determine
because you can determine the lengths of both sides.
The length of e to the i 3 theta is always going to be -
is always going to be 1,
because this is just cosine plus i sine
of our argument 3 theta,
and cosine of 3 theta squared plus sine of 3 theta squared, that's always 1.
So that's something you should double-check for yourselves, but taking lengths here is actually
going to simplify this a lot.
So it's going to be r…
okay so let's take lengths, what are we going to get? We're going to get…
taking lengths gives
this is equal to square root of…
it's equal to the following.
I guess I should say by…
okay, so let's take a look at everything that's just happened here.
So we're going to take lengths.
So by PM, what can we do? Well by PM, we can split this to a product,
the other term drops out,
so maybe I'll even include that step as well.
This term is going to go away
So let me do this.
So r cubed e to the i 3 theta, that's the -
if I split the modulus - or if I…
the length of the product of two complex numbers is equal to the
product of their lengths.
The second term is 1, so that's going to go away,
so then I'm left with - and I can move the cube outside
because again it splits across products,
and this value the length of this was equal to the length of 1 minus 2i,
and that's going to be the square root of 5.
So hence,
r is equal to the
sixth root of 5,
and the ugly numbers have already begun.
Okay?
Again, because the modulus - so remember in this notation,
r is equal to the modulus of z, which has to be a positive number, so
r is positive and it's real so we - well it’s, okay, not negative.
r isn’t always positive, it's always not negative, it could be 0.
I mean clearly it's not in this case, but it could be 0, so it's non-negative
and it's definitely real number, so the length is just going to be the
positive cube root of
the square root of 5, and that's going to be this value.
Now we want to solve in terms of 3 theta.
So next…
next we…
this back into - so we substitute this back into the
first displayed equation
to get that
e i to the 3 theta is equal to
r inverse of 1 minus 2i.
So e to the i 3 theta is equal to r inverse of 1 minus 2i.
So again, I could…
I mean, I can - I want to solve for 3 theta.
You pick your way that you'd like to solve this, I'm going to do -
I don't even know what am I going to do. What am I going to do?
I'm going to solve for the argument by taking -
what can I do? I could try to convert this to polar coordinates, I guess.
Is that the best option? Probably.
So to convert this into polar coordinates, it's going to be…
what do I want to say?
How do I want to do this?
I can write this as e to the power of
pick your favorite Greek letter, I'm going to pick alpha.
I'm going to say where alpha is equal to arctan
of...
what am I going to say? 2r…
2r negative 1,
divided by r negative 1.
Let's see.
Oh negative 2, oops.
Negative 2, which is equal to arctan of minus 2.
Now something to keep in mind, is this the actual correct
one? Do we want
arctan of negative 2 or do we want arctan of negative 2 plus pi?
And again we can answer this,
arctan of negative 2, I don't know the exact number of this,
but that doesn't matter. I know that arctan of minus 2
is going to be a negative radian value.
So that's going to put us in the fourth quadrant, and 1 minus 2i
scaled, so 1 minus 2i lives
in the fourth quadrant, so life is okay. So I don't need to add pi here, I can just
take alpha to be arctan of minus 2.
So then what does this give? So if we want to solve for
theta, so hence,
3 theta is equal to
alpha plus 2 pi k
for any integer k.
So if I compare these two things - oh I should have mentioned.
This should be e to the i alpha, let me
typo that in there.
I want e to the i alpha, where alpha is equal to the angle, right?
So this is going to give us 3 theta is equal to alpha plus
2 pi k for any integer k, right, because again, up to…
these values, e to the i...
e to the i 2 pi is always going to be equal to 1.
So if I take any multiple here, it's going to give me the same value for e to the i alpha.
I mean these two angles must be equal up some multiples of 2 pi, right,
because I can keep rotating around the complex plane, it's going to give me the same angle.
So doing this, and then just solving for theta,
so solving for theta gives
theta is equal to alpha over 3
plus 2 pi k divided by 3.
So these are our possible solutions.
Again, as in class,
right, we've already seen that two of these angles are equal if
the k values are congruent mod 3.
We don't want to keep doing that. We only want a unique set because we want -
because they correspond to complex numbers and
for complex numbers, eventually you just start to cycle on to yourself.
And when does that happen? Every 2 pi multiple,
right? So when I hit 3, I hit a 2 pi multiple,
and so I’m going to go back to when k was 0, and so on and so forth.
So, great. So solving for theta gives us this,
and as in class, we have that k must be either 0, 1, 2, or 3…
or 0, 1, or 2 to give us the 3 possible solutions.
So hence,
how do I want to say this? So hence,
z must be an element of…
which values?
The sixth…
the sixth root of 5 times e to the power of
i arc-
-tan of minus 2 over 3,
comma,
this is going to look nasty but that's okay.
Alright.
Let's see if I did this correctly. So hence z is an element of - oh my God. Okay.
The sixth root of 5 times e to the i arctan of
negative 2 divided by 3,
and then if I add a 2 pi over 3 multiple, and if I add a 4 pi over 3 multiple.
Again if you wanted to, if you're a lot
more patient than I'm going to be, you could convert these back
to standard form.
It's going to take a little bit of effort since we don't know what arctan of minus 2 is.
If you have a calculator, this is very easy. You can convert back and forth no problem.
As it is though, I'll leave the answer like this.
Okay, so what else do I want to say? Is there anything else I need to say about this?
Notice that the angles do differ by 2 pi over 3, and that's what's given by the
Complex nth Roots Theorem, it says once you have a root, all the other ones you just need to
rotate by 2 pi divided by n where n was the z to the n part,
and here n is 3.
So that's fantastic. Oh I forgot a little
bracket over here, that's okay.
Yeah that's basically it. I mean, that's how you solve something like this.
You need to be a little bit careful when you're doing these, but
again, the only real part where you could get a little bit tripped up is this
e to the i alpha part, right, just the same thing as in the previous video
about changing to polar coordinates, you might have to worry about adding pi or not adding pi.
Here again, we know that two of these
theta values are equal if they differ by a multiple of 2 pi,
and so we only need to consider three values for k because
after three values of k, so after 0, 1, and 2, they start to repeat themselves up to
2 pi multiples. You could think of this as like modulo 2 pi,
but that would need a lot of explanation that we haven't talked about,
but that's the way you can sort of think of it in your head.
And otherwise, I think that's
all I want to say. So
like cube roots of 1 minus 2i, not pretty.
If you plug these values in they will work,
ideally, if everything - if I didn't make any typos or anything.
But yeah, hopefully this gives you an idea of how to solve these types of problems,
how to set them up and how to attack them.
So that's great. Thank you very much for your time and good luck.