00:00Hi guys, okay, so in
00:02this video we are going
00:03to write complex numbers in
00:05modulus argument form and that
00:08is this form this or
00:10cos theta or cos theta
00:12plus i times sine theta
00:14and this is actually exactly
00:16I don't know why they've
00:16put this one in the
00:17middle of these two because
00:18this and this are the
00:20exact they're exactly the same
00:21thing. Sis theta is just
00:23a fancy way of writing
00:25cos theta plus i times
00:26sine theta hence the c
00:28the I S cis. So
00:31these two exactly the same,
00:32these are polar form modules
00:34are going to form. This
00:34is what we are doing
00:35today. This one here is
00:38Euler's form. We'll do that
00:39in the next lesson. Euler's
00:41form. Okay, so look, you
00:46will find out that there
00:47are many uses and very
00:50interesting applications of complex numbers
00:53when they're written in this
00:54form. It actually makes your
00:55life
00:56a lot easier for solving
00:58complex equations and things like
01:00that. But I'll get to
01:02that in the future lessons.
01:04First, all I want to
01:05do here is I want
01:05to write a complex number
01:07that you know in Cartesian
01:09form, like 3 plus 2i.
01:11I want to turn it
01:12from this into this. That's
01:13our goal for this lesson.
01:17OK, so the introduction. How
01:18can we write this in
01:19modulus argument form? So this
01:21is just basically an example.
01:23What's this?
01:24So we have our complex
01:27number. Now I want you
01:29to always remember this, um,
01:32this, uh, argon, diagram. This
01:34is an argon diagram. And
01:36we have real numbers and
01:37imaginary numbers. Now this three
01:40plus two i is going
01:41to be somewhere here. Three
01:43plus two i here. Now
01:46if you remember, I said
01:47we often write these using
01:50little vectors, right?
01:52here. So this is my
01:533 plus, let's write z
01:58equals 3 plus 2 i.
02:03Okay, now what is the,
02:06what do you think? What
02:07is the length of this
02:10vector? Well, if you think
02:12of this is 3, this
02:14length here is 3 over
02:17here. And this length here,
02:20is 2. This is where
02:22you go over 3 and
02:24up 2. So what's this
02:29length? Well, it's just going
02:31to be using Pythagoras. It's
02:32going to be the square
02:33root of 3 squared plus
02:352 squared. So what we
02:38can actually do is, if
02:41I get this angle theta,
02:43this angle here, what I
02:46can then say is, let's
02:48get
02:48this or first. So the
02:51size of z, the modulus,
02:53and this is what it's
02:53called modulus, the size of
02:57z or the modulus of
02:58z is equal to the
03:02square root, because this is
03:04straight from Pythagoras, the square
03:07root of 3 squared plus
03:092 squared, which is the
03:11square root of 9 plus
03:134, which is 13.
03:16and let's just write it
03:17like this. So the modulus
03:19of z, this is the
03:21modulus, is the square root
03:23of 13. So I'll actually
03:24draw this here just to
03:26show you what's happening. This
03:27is the square root of
03:2813. So now I can
03:32write, but look, if I
03:34can say the cos of
03:35theta is three over root
03:3713. Let me just do
03:38this over here, because you
03:39don't need this. Just want
03:40to just make sure you
03:41understand what's happening. Cos of
03:43theta is
03:443 over root 13. So
03:483 is actually equal to
03:533, sorry, root 13 cos
03:57theta. So this length is
04:02going to be equal to
04:03root 13 cos theta or
04:06or cos theta. So what
04:09I need to find now
04:10is theta. And theta
04:12is the argument. So when
04:14we say mod argument form,
04:18the modulus is the length
04:21of the complex number, the
04:22length, this length here, length
04:24of this vector, and the
04:25argument is the angle. And
04:27it's the angle from the
04:30positive real axis. So I'll
04:32get to some more when
04:33we have come with some
04:34over here and here. But
04:35it's always from here. And
04:37generally we go between negative
04:40pi and pi. Sometimes you
04:43may see arguments that are
04:46bigger angle, like an angle,
04:48this angle here, but generally
04:50they're between negative pi and
04:51pi. Okay, so how do
04:54we find theta? Well, theta,
04:57this is just, this is
04:58the right angle triangle. So
04:59theta is just the inverse
05:01tan, the inverse tan of
05:03two over three. And the
05:06inverse tan of two over
05:07three, let's do this, inverse
05:08of 2 over 3 gives
05:14a 0 .5888. Let's do
05:17a proximal equal to 0
05:19.5888. So this is my
05:23modulus. So I can actually
05:28now say that 3 plus
05:322i is equal to, instead
05:35of 3,
05:36I'm going to write this.
05:39I'm going to write root
05:3913, I'm going to say
05:42root 13, cos of 0
05:47.588 plus, and instead of
05:532, so I'm going to
05:54leave the i there, but
05:55instead of 2, I'm going
05:56to write, instead of 2,
06:01same kind of thing, I'm
06:02going to write, but with
06:03sine, root 13,
06:04and it's outside the bracket,
06:06i times sine of 0
06:09.58. So I definitely want
06:12you to understand why this
06:16equals this, this is just
06:18trigonometry. But once you have
06:22the modulus and the argument,
06:23you can just write it
06:24like that. And actually, you're
06:29able to do all this
06:30form now already. I'm going
06:31to explain it more in
06:32the next video, but
06:33You have OR and you
06:33have theta, and you've got
06:35an artist form. So this
06:36is how we write that
06:39in mod argument form, modulus
06:42argument form. And we can
06:45make extra fancy. We can
06:47write it as root 13,
06:50cis 0 .5888. Okay, done.
06:59I'll just show you how
06:59to do that one on
07:00the calculator.
07:01see I can do M3
07:04plus 2i and then we
07:10go menu number complex number
07:142's there's a lot of
07:15things here you can actually
07:16find you can find the
07:18real part imagine part easy
07:20you can find the angle
07:21you find the magnitude that's
07:23the the modulus the magnitude
07:25and convert to polar converter
07:29rectangular. So rectangular is the
07:32Cartesian form. That's rectangular form.
07:33So we could go back
07:34to that. But we just
07:35want to convert to polar.
07:37So for some reason, this
07:41calculator puts it into order
07:44form. But I guess order
07:45form is a type of
07:46polar form because the polar
07:48form just means that you
07:49have this angle. So because
07:56polar form is
07:57is to do with polar
07:58coordinates. I won't get into
08:00that. Anyway, this is the
08:01angle, and this is the
08:04modulus. And this is just
08:06root 13. Let me show
08:07you just to prove it.
08:09Root 13 is this. So
08:11we did get that. OK.
08:15So if you're in a
08:16paper too, and you have
08:17to write that in modern
08:18argument form, you can just
08:20go straight to this. But
08:22obviously write it as, or
08:25or cause this plus i
08:27times this. Okay, done, that's
08:31my introduction. Now, what happens
08:35when the angle, let's say
08:38the complex numbers over here,
08:42well, you need to get
08:43this angle there. That's your
08:47argument. And when the, let's
08:50say when it's down here
08:51somewhere,
08:53Then this is your argument.
08:59So we go from negative
09:00pi to pi. I have
09:04seen many times arguments that
09:06actually go beyond 180 or
09:08beyond pi and go down.
09:10This is the argument. It
09:13obviously gives you the exact
09:14same. It's the exact same
09:15complex number. This angle, if
09:18you put like the cost
09:20of this angle and the
09:21sign of this
09:21the same as the cos
09:24or the sine of the
09:26sine of the sine of.
09:29Okay, let's go and do
09:33some examples. So we are
09:35going to write in the
09:38form Z equals or cos
09:39theta plus i sine theta,
09:41just to not know confusion
09:42about what form they want.
09:44These form, we're going to
09:45do them without a calculator
09:46because obviously with a calculator
09:48it's very easy.
09:49You can just do it
09:50straight away. So for each
09:53of them, I always want
09:55you to draw, I always
09:57want you to draw an
09:58argon diagram. So let's do
10:01the first one. Let's do
10:03it here. So we are
10:04going to draw an argon
10:05diagram. Doesn't take long to
10:08do. Just this, this, this,
10:14this. Now the reason I
10:15like to do this is
10:16because you, it
10:17our phi is what's going
10:18on. So two minus two
10:19root three. So two minus
10:22two root three. I'm gonna
10:23go over two and over
10:25two and down two root
10:26three. So somewhere here. This
10:29is my point where my
10:31complex number is z equals
10:32two minus two root three.
10:35I. Now what angle am
10:38I trying to find? Well,
10:39I'm trying to find this
10:41angle. And what's the modulus?
10:45Let's get to that in
10:46a second. So this length
10:49here is two, and this
10:51length is two root three,
10:53yes, it's minus, but the,
10:55and it's minus two root
10:56three, I, but the length
10:57of it is two root
10:58three. So to get the
10:59modulus of z, the modulus
11:01of z, again, I'm gonna
11:02use Pythagoras, it's the square
11:04root, the square root of
11:07two squared plus two root
11:10three, all squared.
11:13which is equal to the
11:16square root of 4 plus
11:19this is 4 plus 4
11:23times 3 which is 12.
11:25This is a nice one.
11:27That's just root 16 which
11:28is 4. So this guy
11:31here is for the modulus.
11:33The modulus is 4. What's
11:35the argument? Now I just
11:39like what I like to
11:40do is I say
11:41Just use socatoa. Yes, it's
11:45going to be negative, but
11:46I don't say I prefer
11:48not to think of, um,
11:51don't say get the tan
11:52of negative two root three
11:55over two. I prefer to
11:56just just think of it
11:56as a normal triangle, positive
11:58lengths. And we can say
12:00the inverse tan inverse tan
12:03of two root three over
12:06two. Because it's, this is
12:09So opposite, tan is opposite
12:15over adjacent to 2, root
12:163 over 2. So this
12:18is equal to the cancel.
12:20What's the tan of root
12:223? That's one of the
12:22ones you need to know.
12:24I'm going to close my
12:24eyes, visualize my triangle, and
12:27it is pi over 3.
12:32So this angle is pi
12:37over 3.
12:38but the argument, therefore the
12:41argument is, and that's why
12:45I didn't write theta equals
12:46this, I'm gonna say theta
12:47is equal to negative pi
12:51over three. So, yeah, as
12:55I say, I find that
12:56less confusing just to find
12:58the angle and obviously if
12:59I'm going this way, it's
13:00gonna be negative. Then I'm
13:02gonna say, z is equal
13:04to four,
13:06Open brackets, costs negative pi
13:10over three plus i times
13:14sine of negative pi over
13:18three. Done. Second one. Let's
13:23bring it down here. Part
13:26B. Minus four minus four
13:29i. Okay. Again, first thing
13:32we do, draw.
13:34And argon diagram minus 4,
13:37minus 4i, where is it?
13:40Well, and if you're good
13:41at the, by the way,
13:42this is real and imaginary.
13:43If you're good at the
13:45unit circle, if you like
13:47the unit circle, let's face
13:48it, we all do. You're
13:50gonna like this, you're gonna
13:51be good at this. So
13:52minus 4 minus 4i. So
13:54this is a perfect diagonal
13:56here, minus 4, minus 4i.
14:02So this is this length
14:07here is for, this length
14:09here is for, this angle
14:12here. Now, I like to
14:16get this angle first and
14:18then, but note, we're trying
14:21to, this is the argument.
14:23The argument is always from
14:25the real, the positive real
14:27axis. This is going to
14:28be the argument, but if
14:29we find this angle
14:30obviously it's just pi minus
14:31that. So, well let's do
14:34the mod first. So the
14:37modulus of z is equal
14:39to square root of 4
14:43squared plus 4 squared, which
14:46is equal to root 32.
14:51Or, that's actually 4 root
14:532. So that's the mod
14:55z. So 4 root 2,
14:58And the argument, so let's
15:02get, now you might actually
15:03wanna call this alpha, and
15:06then call this theta. That's
15:09what I'd like to do.
15:10So, let's do 10 of
15:14alpha is equal to 4
15:18over 4, which is 1.
15:20So alpha is equal to
15:22pi over 4. Therefore, theta
15:26is equal to, well it's
15:30negative pi, it's negative pi,
15:38negative, how about this? It's
15:41negative pi, it's actually negative
15:43pi plus pi over 4.
15:46But look, we just think
15:47of it intuitively. If that's
15:49pi over 4, then that's
15:523 pi over 4 and
15:53it's negative.
15:54So negative 3 pi over
15:574. So the complex number,
16:02z is r 4 root
16:052. Open bracket. I forgot
16:10to bracket here, look. Open
16:13brackets. And cos negative 3
16:18pi over 4 plus i
16:21times sine
16:22and negative 3 pi over
16:264. Okay, that's the second
16:28one done. Now, let me
16:32just show you that one
16:33on the calculator. So this
16:36was minus 4 minus 4i.
16:39So minus 4 minus 4i.
16:47And we want to write
16:48this
16:51complex numbers, tools, write this,
16:54convert to polar, and we
16:57get this. So just note,
16:59look, this is the, the
17:02modulus, and this is the
17:05argument. Now I can, promise
17:07you, this is minus three
17:09pi over four, and this
17:11is four root two, all
17:13to six significant figures. But,
17:15like, if they ask you
17:17to, to, to, do,
17:18to write your answer in
17:21exact form, then you're gonna
17:23be in trouble with the
17:23calculator, so certainly you need
17:25to know how to do
17:25it without the calculator. Okay,
17:27part C, let's bring these
17:29two down here. So these
17:34two are gonna be easier.
17:35I am gonna draw. They're
17:39easier, but until you see
17:42them, you may not know
17:44exactly what to do. So,
17:46real?
17:46Imagine, I'll do them both
17:49on the same argon diagram.
17:51So where is 3i? Well,
17:553i is here. It's 0
17:58in the real axis and
18:00up 3. So it's 3.
18:01What's the length of it?
18:03Well, the length of it
18:03is just 3. So I
18:05can say straight away z
18:06is equal to 3. Open
18:09brackets. Cause. What's the angle?
18:12What's this angle here? Well,
18:14Let's write angle, let's pi
18:16over two. So three costs
18:17of pi over two plus
18:19i times sine of pi
18:21over two. That's it, done.
18:25This is part C. And
18:28then part D, where is
18:29minus five? Well, it's over
18:32here. This is minus five,
18:34there's no imaginary part. And
18:36what's the length of this?
18:37What's the size of this
18:38from here? To here, what's
18:41the magnitude?
18:42So it's five. So dz
18:48is equal to five, cos
18:52five, cos, you could use
18:56pi or negative pi, but
18:57obviously what you're better off
18:59just using pi. Five, cos
19:00of pi plus i times
19:03sine of pi. And just
19:06to check that you're right,
19:08multiply it out, and this
19:09is actually how you turn
19:10back into
19:11Cartesian form just multiply that
19:12out. What's cost of pi?
19:14Cost of pi is negative
19:17one. So you five times
19:19negative one and what's sine
19:20of pi? Zero. So five
19:22times negative one is just
19:24negative five, which is this
19:26same here. Multiply this out.
19:28Cost of pi over two
19:29is zero. Sine of pi
19:30over two is one. So
19:31you've got three times one
19:34i, which is three i.
19:35And you can do the
19:36same for the other ones.
19:37Okay, that's it.
19:39And hopefully that made sense.
19:42Look, that's not easy. Of
19:44course it's not. Perhaps this
19:49is where it gets kind
19:50of tricky. Just make sure
19:52you're finding the angle from
19:55the real, the positive real
19:58axis. Yeah, as always, let
20:03me know if you have
20:04any doubts and I will
20:06see you
20:07in the next video.