00:01
So over here we're given a question which is asking us to find the complex qubits of 4 plus i.
00:08
So it's just something else to find the qubit of 4 plus i.
00:13
Okay.
00:15
So to do this, the first thing we're going to do is remember the moire's short picture in that name, but that is the moirems.
00:23
So the moiref's formula.
00:30
So that's one way and the mars formula says that if you have, perhaps i should write that in red.
00:40
It's not that to write that in red.
00:43
So the mars formula just says if you have some complex number and you try to find the entry to that, but forget this is going to be equivalent to rate to write to this, right? well, this is going to be, you'll find the modulus of that number.
01:06
Okay, remember what the modulus is.
01:08
But it's just the square root of the real part square plus the square of the coefficient of the imaginary part.
01:18
Okay, something like that.
01:20
So this is exactly, we're going to do this, but i just wanted to do that, i guess, an advanced district of modulus, right? it's kind of like a magnitude.
01:27
To think of factors.
01:30
So not equals it, sorry, this is the equal to the part and what's going to do.
01:36
This is going to be square of 17, that's 16 plus 1, okay? and this is square of 16 plus 1, which is not this, yeah, this is going to be the, this is when you find that's the marginal, right? it's going to be a square root to you, okay? now, this now would be common, well that would be that to the power 1 over n.
02:03
Okay? and this is going to be times the cosine of some angle theta, right? the angle theta plus 2k pi over n plus i.
02:27
I times the sign of the same thing here to the theta plus two pi two k so 2k over and what is k well k k is just the whole number right k is an element of zero sorry is it integers, so yeah, two, three, so on and so forth to n minus one.
03:11
Okay? so it could be any of those values, i'm just going to shorten it here because it is, it's just, you know, this is how it's usually seen.
03:26
But again, it could be any of any, any integer to n minus one.
03:34
Alright? so just to say this again, i'm going to just get rid of this might have any confusion here.
03:46
So we're just going to first of all multiply the modulus, okay, the modulus, raise that to some 1 over n which is the n root, we try and therefore, we'll multiply that by the cosine of theta, and theta is the argument.
04:15
Argument, alright, they spell that right, over 2k pi over n plus i times sign, same thing, theta plus 2k pi over n.
04:33
So let's go ahead and do what we're doing again.
04:36
Let's find the first the modulus.
04:38
The modulus is just this, as we saw earlier, would get the real part, square that, plus the coefficient of the imaginary part.
04:50
And the coefficient here is positive 1, so because it's plus i, so that's positive 1, it will not really matter ultimately because we're doing squares here, right? and then you find a square root of that.
05:03
What we're going to get here is the square root of 16 plus 1, and this is going to be the square root of 17.
05:12
So we found modules.
05:15
Now what is n? we know from here that again we're looking for this.
05:22
We're looking for remember this.
05:28
We're looking for the end root of some number.
05:35
What we're really looking for here is that number raised to 1 over n, correct? so we can just say this is 4 plus 1, plus i to the power 1 third.
05:48
So this is the whole number here, okay? so this is what we're really looking for.
05:53
Now that means n here is 3.
05:58
Therefore, k would range from, again, k is just the whole numbers from 0, which positive infinity, integers, by the way.
06:09
Right? zero under natural numbers.
06:13
So k would be from 0 to n minus 1.
06:18
So it's going to be n minus 1 there's 3 minus 1 which is 2.
06:22
So it's 0 1 and 2.
06:25
So we'll get stuck.
06:27
So the so our task here will be to add and to use each of these numbers.
06:34
Replace n k with each of these numbers here.
06:39
So let's get started.
06:40
The first root, the first root that we're going to have would be square root of 17, go into this, all right, raised to one third, okay? this is going to be divided by, sorry, that's multiplied by the cosine, okay, of what, it's a cosine of what's the argument? we've got to find an argument.
07:09
And what is the argument? i forgot to do that part.
07:13
So let me just quickly do that here.
07:18
If this is what the modulus is, we will know what n is, we therefore know what k is from that.
07:26
So let's see what the argument theta is.
07:29
But how do you find the argument theta? just remember that you have a...
07:35
This is the real part and this is the imaginary part.
07:42
We can just do trace that line there.
07:46
And if the imaginary part is the, what's that the coefficient is 1, the real part coefficient is 3, it means that this argument theta is going to be given by the tangent of theta would be, well, it's going to be opposite.
08:06
Or other adjacent, which is 1 over 3.
08:12
That means theta would be the arc tangent of 1 over 3.
08:21
Okay? alright, so we could go ahead and find what the arc tangent of 1 over 3 is.
08:39
We could put that value here.
08:46
I'm sorry, it's not 1 over 3.
08:47
I'm not sure.
08:49
I guess i don't know why you got one over three.
08:50
It's supposed to be one of a four.
08:53
Okay, this is one of a four.
08:58
It's the arc tangent of one of a four.
09:01
So we could figure whatever the arc tangent of one for four is...