00:00Hi guys, okay, in this
00:02lesson we are going to
00:03look at the morph... the
00:04morphs theorem. I was struggle
00:07to say that name. So
00:08the morphs theorem is this.
00:10If you have a complex
00:12number raised to a power
00:14of n, then it equals
00:17or to the n times
00:18cos of n theta as
00:19you multiply the theta by
00:20the n plus i times
00:21sine of n theta. I
00:24have a little bit of
00:25a... I've always had a
00:26bit of an issue with
00:26the morphs theorem because
00:28It gets this big name
00:32and it's a big thing
00:34in mathematics. Everyone knows Du
00:36Maur's theorem. But the truth
00:40is, once you know Euler's
00:42form like this, if you
00:44know a complex number 2E
00:48to the i times theta
00:51like pi over 3, then
00:53we know that if I
00:54want to raise this to
00:56the power of, by the
00:58way, I wanted this to
00:59be to the power of
01:00five. If I wanted to
01:02raise this to the power
01:04of five, then we know
01:06that it's just two to
01:08the power of five times
01:09e to the i times
01:11five pi over three, because
01:14of the rules of indices.
01:16So you can actually prove
01:17to more of this theorem
01:18pretty easily using order. However,
01:23it's
01:24from what I can gather,
01:25Dermorv came up with this
01:26before Euler did his thing
01:28with his identity and all
01:30this stuff. So we used
01:33Dermorv's theorem and we need
01:38to know how to prove
01:38it by induction. Now I
01:41will do that in the
01:44proof -band -duction section. So
01:45you need to be able
01:46to prove it for n
01:47is a positive integer. So
01:49here is a complex number
01:50to the power of n.
01:51You need to know how
01:52to prove
01:52prove it using induction for
01:55n is an element of
01:57the positive integers. But note,
02:01and again, this is also
02:03a little bit, this isn't
02:07as clear as I'm going
02:08to make it sound, but
02:11for the purposes of everything
02:13you need to know in
02:14the IB, this holds for
02:18rational numbers.
02:20n can be any rational
02:21number. And here I'm going
02:23to show you how to
02:25use it for 12 to
02:26the power of a half
02:27because it's square root. Okay,
02:29so how do I write
02:30this in hertesian form? So
02:33the first thing we need
02:34to do is write it
02:37in modulus argument form. So
02:40again, hopefully, and I see
02:41when I taught you how
02:42to do this, I said
02:43this comes up all the
02:44time. So hopefully you're getting
02:46good at it. So 1
02:47plus root 3.
02:481 plus root 3. This
02:54is root 3. This is
02:57theta. This length here has
03:00to be 2, because the
03:01square plus the square equals
03:02the square. And the angle,
03:07so let me write here
03:11or is equal to the
03:15square root. Well, it's 2.
03:16But there he was there
03:17and then tan of theta
03:19is root three so theta
03:22is pi over pi over
03:26three So now this thing
03:31this one plus a root
03:36three i to the power
03:39five is equal to and
03:44two, cis theta. Let's go
03:49with this. It's two. Well,
03:53it's this first. It's two
03:55times cos. So I'll leave
03:57the, we're going to leave
03:57the five outside. So it's,
03:59I'm writing this in modular
04:00circle and form. So it's
04:01two cos of pi over
04:04three plus i times sine
04:08of pi over three. But
04:11this to the power of
04:12five
04:12means it's this all to
04:15the power of five. So
04:16I get my two to
04:18the power of five and
04:20then this to the power
04:21of five becomes, this is
04:23de Morvius theorem, I multiply
04:24by the five here. So
04:26it's two to the power
04:27of five times cos, so
04:29I multiply the theta by
04:30five, cos five power of
04:32theta plus i times sine
04:37of five pi. Sorry, I
04:39say theta, I'm
04:40over three. There. So that's
04:44Demorvus theorem. To set it
04:46up like this. Now, obviously
04:47he said, write it in
04:49Cartesian form. So I need
04:51to now go backwards. So
04:52imagine, look, I could have
04:53used the binomial expansion there
04:55and multiply it out. But
04:58this is a much quicker
04:59way, especially if this was
05:00much bigger than five. Imagine
05:01if this was 25, then
05:03the last thing you want
05:04to be doing is the
05:04binomial expansion. This actually works
05:06out quite nicely.
05:08I just need to now
05:10find the well two to
05:13the power 5 is 32.
05:16Costs of 5 pi over
05:17three. So 5 pi over
05:18three would be down here.
05:21So it would actually be,
05:23um, and take your time
05:25to figure this out if
05:26you want, but it would
05:27be down here and the
05:29cost would be one over
05:32two. And it's because it's
05:33down here is positive. So
05:35this would actually
05:36be 1 over 2 plus
05:40i times sine. Actually, it's
05:43going to be negative, but
05:46let's do that in a
05:47second. So it's plus i
05:49times sine of 5 power
05:523 is actually negative root
05:553 over 2. So I'm
05:59left with 32 times 1
06:00.5 is 16, 16 minus
06:0416 minus 16 minus 16
06:11root 3 I and Okay,
06:19yeah, that's it 16 minus
06:2216 16 minus 16 root
06:243 I that is this
06:26written in Cartesian 4 okay,
06:29and
06:33Next one, this. So the
06:36only kind of difference with
06:38this is well that the
06:39exponent is a rational number.
06:40It's going to be a
06:41half and I've just written
06:42it in as a square
06:44root. So this, this is
06:46actually equal to this equals
06:48the square root of negative
06:49two plus two i equals
06:53negative two plus two i
06:56to the power of a
06:58half. And this equals, we're
07:01going to
07:01Let's do this down here.
07:03We want to write this
07:05in modulus argument form. So
07:10minus 2 plus 2i is
07:11actually up here. This is
07:162, this is 2, so
07:18this is going to be,
07:19this square plus this square
07:20root 8. This is going
07:22to be root 8. This
07:23is going to be alpha.
07:25This is going to be
07:26theta.
07:29So, or equals root eight,
07:3510 alpha equals two over
07:39two, which is one, alpha
07:40equals pi over four, then
07:43theta has to equal pi
07:45minus pi over four, which
07:46is three pi over four.
07:49So this thing here, minus
07:52two plus two i to
07:54the power of a half,
07:56equals
07:57thing we're going to do
07:58this, we're going to write
07:59this in modded desego and
08:00form which is root 8
08:02root 8 times, at this
08:07time I'm just going to
08:07say cyssi, theta is 3
08:11power 4, all to the
08:14power of a half. Okay
08:20so this gives me root
08:218 to the power of
08:22half which is actually 8
08:24to the power of a
08:25half
08:25half to the power of
08:26a half, eight to the
08:29power of half, which is
08:30actually eight to the power
08:31of a quarter. And then
08:34it's going to be cis,
08:36cis, three pi or four
08:38times a half, which is
08:40three pi over eight. This
08:47is then the eight to
08:50the power of a quarter,
08:52eight to the power of
08:53a quarter,
08:53order times sis 3 pi
08:57over 8. Now look, he
08:58didn't actually say, this time
09:02he said write it in,
09:04find this in Cartesian form.
09:05So I had to put
09:06it in Cartesian form. This
09:07time he just said find
09:07this. So there, that's it.
09:10That's the square root of
09:11this. I've written it in
09:13sis form, fair enough. So
09:15that's the morphs theorem. In
09:19the next lesson you're going
09:20to see that this is
09:21Well, it's very useful when
09:24it comes to solving our
09:28finding roots of complex equations.
09:31So that's what we're going
09:32to do in the next
09:33dozen. So I'll see you
09:34then.