Polar coordinates, the last thing we talked about this week was transferring
things in standard form to polar coordinates.
You can also go the other way too, it's not too difficult,
and we'll see that in a minute.
But let's just go through and talk a little bit about polar coordinates. So
a point in the plane can correspond to a length and an angle.
So the length - so here's a little picture on the first quadrant.
My length here is that r value, the length from the origin to the point.
My theta is the angle that it makes from the positive x-axis to the line,
and from here we can see a little…
what do you call it… just trigonometry, right?
Our r sine theta is going to be the length of the
altitude here, and then the length
along the x-axis is going to be r cos theta.
As an example, if we have r theta is equal to 3 and pi over 4,
this corresponds to the complex point, let's say, 3 times cosine of pi over 4,
plus i times 3 sine
pi over 4, according to this little picture, and if we simplify, we're going to get
3 over root 2, plus 3 over root 2i.
We can see that by the picture, right? So if we draw a little picture here, we have
3, and then pi over 4,
and we know that, again, for pi over 4, we have a pi over 4, pi over 4, pi over 2 triangle.
So 1, 1, root 2 are the side ratio lengths,
but if the root 2 is equal to 3, then each of these side lengths should be 3 root 2.
That's what we get.
This correspondence…
okay so there's the 0, 0 point that we have to deal with,
but we're going to ignore that issue for now.
0, 0 doesn't really correspond - well,
I mean, I guess the r value is 0 and the theta is 0, I guess.
I mean you could argue that theta value can be anything.
So let’s exclude 0, 0 from the argument. We can see that we can correspond every…
pair, r theta, to a standard form point and every standard form point
to a pair, r theta.
So how do we see that? So given z equals x plus y i,
we see that r is equal to the mod of z, which is the length of z, so
square root of x squared plus y squared,
and theta here is equal to…
well there's lots of ways to see it. It's the arccos of x over r,
the [arcsine] of y over r, or the arctan of y over x,
but something we have to watch out for here, okay?
So pictorially, it makes sense, right, tan of theta is y over x,
that's true.
But to say theta is arctan of y over x,
it's actually being a little bit incorrect, there's a couple of problems here.
What are some of the problems that come up? So tan theta
is y over x,
but we have some issues. So for example, like
what is tan of 90 degrees, right? That's a problem.
So the things on the imaginary axis, we can just do by eyeballing.
We don't need to compute arctan of something, right,
we just know that on the imaginary axis
that we're either 90 or 270 degrees,
so pi over 2, or 3 pi over 2.
Now when you compute theta to be the arc tan of y over x,
you have to be a little bit careful because arctan…
what's the
range of arctan? The range of arctan is negative
pi over 2 to pi over 2, not including [negative] pi over 2 and pi over 2, but we already
talked about [negative] pi over 2 and pi over 2.
So the range is the interval from negative pi over 2 to pi over 2,
but however, we have points that could live, you know, let's say
pi degrees away from the x-axis.
So we have to be a little bit careful, and how we
account for this is we just have to double check at the end, when we do this computation,
that it's either…we either want theta or we want -
we want theta to be either arctan of y over x, or pi plus arctan of y over x,
depending on what quadrant we’re in.
That's the thing we'll see in the next example, and
hopefully that'll clarify what I mean here, okay?
So the modulus, the length,
of our complex number, that's very easy to determine.
Just the square root of x squared plus y squared.
The thing that changes here though is
possibly the theta, right, the theta is arctan of y over x,
or pi plus arctan of y over x depending on where we are.
Let’s see an example:
write z equals the square root of 6 plus the square root of 2 times i
using polar coordinates.
What about negative z? So write
negative z as well. What's the difference?
Spend a couple minutes, try it. Use the formulas on the previous page, pause the video.
Now that you're back, let's give this a shot.
So the length of z is always easy to compute, just
6 plus - the square root of 6 plus 2, which is 8.
So square root of 8 is 2 root 2,
and the angle that we're computing is
the arctan of y over x, the arctan of root 2 over root 6,
and if we do that, we get arctan of 1 over root 3
and arctan of 1 over root 3 we know is pi over 6.
That again, you can see these things, so if these things are unfamiliar to you,
these trig identities, I mean
draw a picture, of course, and if you don't remember how to do that, I would advise checking out
the CEMC courseware. There's some good material on there.
Okay.
So thus, what do we know from this? So,
the perceived angle is pi over 6 and if we think about where this z lives,
it has a positive x value and a positive y value, so it's in the first quadrant.
pi over 6 corresponds to a point in the first quadrant, so we're okay.
So thus, r theta is equal to 2 root 2 and pi over 6 is the
polar coordinate for z.
Now what about for minus z? If you think about minus z,
you're going to get the same answers here. You're going to
get that r is also equal to 2 root 2, and if I take the arctan of
minus 2 divided by minus 6 - or minus root 2 divided by minus root 6,
that's pi over 6. So what happened here? What are we doing? What's really going on?
So we have to be careful, and what's happening here is that
instead of being in the first quadrant, we're all the way around in the third quadrant.
We've looped around by 180.
So well now to get your final answer,
instead of saying it’s pi over 6, add pi to it.
That'll give us 7 pi over 6, and that's where our minus z point lives.
So that's it for converting standard to polar form,
that's it for modulus and
Properties of Conjugates and complex numbers. Hopefully…
it’s a brief overview…in this video, but I think
the ideas are fairly clear. A lot of you have seen this before as well, so I'm not too, too
worried about spending a lot of time with the gory details.
If anything here is unclear, or if you don't know anything that I've said,
it's a good idea to work it out.
Try to pencil and paper out some of the proofs,
try introducing coordinates, try doing some
of these problems without coordinates if you can.
Some of the proofs,
especially in the Properties of Modulus, there’s a couple of those proofs that you can
do without going to coordinates.
Just remember that when you're dealing with like a length, a modulus,
you can square both sides and that usually makes the computation easier.
That's all I have to say, so thank you very much for listening and good luck.