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Polar coordinates, the last thing we
talked about this week was transferring

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things in standard form to polar coordinates.

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You can also go the other
way too, it's not too difficult,

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and we'll see that in a minute.

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But let's just go through and talk a
little bit about polar coordinates. So

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a point in the plane can correspond 
to a length and an angle.

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So the length - so here's a little 
picture on the first quadrant.

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My length here is that r value, the
length from the origin to the point.

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My theta is the angle that it makes 
from the positive x-axis to the line,

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and from here we can see a little…

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what do you call it… 
just trigonometry, right?

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Our r sine theta is going 
to be the length of the

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altitude here, and then the length

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along the x-axis is going to be r cos theta.

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As an example, if we have r 
theta is equal to 3 and pi over 4,

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this corresponds to the complex point, 
let's say, 3 times cosine of pi over 4,

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plus i times 3 sine

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pi over 4, according to this little picture, 
and if we simplify, we're going to get

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3 over root 2, plus 3 over root 2i.

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We can see that by the picture, right? So 
if we draw a little picture here, we have

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3, and then pi over 4,

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and we know that, again, for pi over 4, we 
have a pi over 4, pi over 4, pi over 2 triangle.

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So 1, 1, root 2 are the side ratio lengths,

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but if the root 2 is equal to 3, then each 
of these side lengths should be 3 root 2.

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That's what we get.

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This correspondence…

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okay so there's the 0, 0 point 
that we have to deal with,

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but we're going to
ignore that issue for now.

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0, 0 doesn't really correspond - well,

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I mean, I guess the r value is 
0 and the theta is 0, I guess.

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I mean you could argue
that theta value can be anything.

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So let’s exclude 0, 0 from the argument. We
can see that we can correspond every…

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pair, r theta, to a standard form
point and every standard form point

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to a pair, r theta.

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So how do we see that?
So given z equals x plus y i,

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we see that r is equal to the mod 
of z, which is the length of z, so

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square root of x squared plus y squared,

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and theta here is equal to…

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well there's lots of ways to see 
it. It's the arccos of x over r,

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the [arcsine] of y over r, 
or the arctan of y over x,

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but something we have to 
watch out for here, okay?

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So pictorially, it makes sense, 
right, tan of theta is y over x,

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that's true.

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But to say theta is arctan of y over x,

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it's actually being a little bit incorrect, 
there's a couple of problems here.

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What are some of the problems 
that come up? So tan theta

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is y over x,

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but we have some issues. 
So for example, like

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what is tan of 90 degrees,
right? That's a problem.

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So the things on the imaginary 
axis, we can just do by eyeballing.

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We don't need to compute 
arctan of something, right,

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we just know that on the imaginary axis

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that we're either 90 or 270 degrees,

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so pi over 2, or 3 pi over 2.

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Now when you compute theta 
to be the arc tan of y over x,

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you have to be a little bit 
careful because arctan…

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what's the

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range of arctan? The range 
of arctan is negative

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pi over 2 to pi over 2, not including [negative] 
pi over 2 and pi over 2, but we already

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talked about [negative] 
pi over 2 and pi over 2.

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So the range is the interval from 
negative pi over 2 to pi over 2,

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but however, we have points that 
could live, you know, let's say

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pi degrees away from the x-axis.

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So we have to be a little
bit careful, and how we

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account for this is we just have to double 
check at the end, when we do this computation,

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that it's either…we either 
want theta or we want - 

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we want theta to be either arctan of
y over x, or pi plus arctan of y over x,

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depending on what quadrant we’re in.

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That's the thing we'll see 
in the next example, and

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hopefully that'll clarify 
what I mean here, okay?

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So the modulus, the length,

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of our complex number, that's 
very easy to determine.

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Just the square root of 
x squared plus y squared.

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The thing that changes here though is

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possibly the theta, right, the 
theta is arctan of y over x,

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or pi plus arctan of y over x 
depending on where we are. 

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Let’s see an example:

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write z equals the square root of 
6 plus the square root of 2 times i

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using polar coordinates.

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What about negative z? So write

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negative z as well. What's the difference?

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Spend a couple minutes, try it. Use the formulas 
on the previous page, pause the video.

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Now that you're back, 
let's give this a shot.

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So the length of z is always 
easy to compute, just

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6 plus - the square root 
of 6 plus 2, which is 8.

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So square root of 8 is 2 root 2,

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and the angle that we're computing is

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the arctan of y over x, the 
arctan of root 2 over root 6,

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and if we do that, we get 
arctan of 1 over root 3

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and arctan of 1 over root 
3 we know is pi over 6.

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That again, you can see these things,
so if these things are unfamiliar to you,

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these trig identities, I mean

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draw a picture, of course, and if you don't remember 
how to do that, I would advise checking out

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the CEMC courseware. There's 
some good material on there.

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Okay.

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So thus, what do we know from this? So,

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the perceived angle is pi over 6 and 
if we think about where this z lives,

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it has a positive x value and a positive 
y value, so it's in the first quadrant.

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pi over 6 corresponds to a point in 
the first quadrant, so we're okay.

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So thus, r theta is equal to 2 
root 2 and pi over 6 is the

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polar coordinate for z.

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Now what about for minus z?
If you think about minus z,

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you're going to get the same 
answers here. You're going to

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get that r is also equal to 2 root 
2, and if I take the arctan of

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minus 2 divided by minus 6 - or 
minus root 2 divided by minus root 6,

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that's pi over 6. So what happened here?
What are we doing? What's really going on?

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So we have to be careful, and 
what's happening here is that

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instead of being in the first quadrant, we're 
all the way around in the third quadrant.

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We've looped around by 180.

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So well now to get your final answer,

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instead of saying it’s 
pi over 6, add pi to it.

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That'll give us 7 pi over 6, and
that's where our minus z point lives.

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So that's it for converting standard to polar form,

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that's it for modulus and

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Properties of Conjugates and
complex numbers. Hopefully…

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it’s a brief overview…in 
this video, but I think

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the ideas are fairly clear. A lot of you have 
seen this before as well, so I'm not too, too

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worried about spending a
lot of time with the gory details.

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If anything here is unclear, or if you 
don't know anything that I've said,

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it's a good idea to work it out.

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Try to pencil and paper 
out some of the proofs,

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try introducing coordinates, try doing some

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of these problems without 
coordinates if you can.

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Some of the proofs,

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especially in the Properties of Modulus, 
there’s a couple of those proofs that you can

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do without going to coordinates.

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Just remember that when you're 
dealing with like a length, a modulus,

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you can square both sides and that 
usually makes the computation easier.

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That's all I have to say, so thank you 
very much for listening and good luck.