3:26 "Two things happen when you multiply by a complex number: [...] you scale, that comes from the real component, and you rotate, which comes from the imaginary component." No, just no. Multiplying by 2i (which has no real component) is still scaling by 2 (besides the rotation by pi/2). It's the modulus (or absolute value) which is scaling, and the argument which is rotating.
12:40 You could also use the trigonometric identity cos^2(x) + sin^2(x) = 1 as cos^2(x) - i^2 * sin^2(x) = = cos^2(x) - (-1) * sin^2(x) = = cos^2(x) - (-sin^2(x)) = = cos^2(x) + sin^2(x) = = 1 No matter which angle is "x" (in our case it was PI/4), the result is equal to 1. P.S. The identity cos^2(x) + sin^2(x) = 1 in the imaginary form: cos^2(x) - i^2 * sin^2(x) = 1 or [cos(x) + i * sin(x)] * [cos(x) - i * sin(x)] = 1
@@matemaatika-math It's a small slip up that can tend to happen at the end of a long piece of working. Presumably, he had the word "minus" on his mind because he just was just referencing the pi/3 *minus* pi/4. Commentating on what he had just written possibly made him write minus instead of plus. I hope this answers your wondering of why. If you're asking why in terms of mathematics - because the final line purports that cos(pi/12) - i sin (pi/12)= cos(pi/12) + i sin (pi/12), which is not true. It doesn't have a effect on the answer of the final question because the the real part of the expression is still correct. I'm certain that had been live in a classroom, one of the students would have pointed the small copying error out.
@@matemaatika-math It's errant because the sign of the sine function should not change. sin (pi/3 - pi/4) is still positive sin(pi/12), but like he said, since we're dealing with just the real part, it didn't matter, and probably why Eddie did not notice it either. He surely would have if that part were important.
12:29 This is just cos^2(x) + sin^2(x) = 1 , the pythagorean identity no need for exact values; although it is reassuring we get 1 anyway. Let conj(z) be the conjugate of z z*conj(z) = |z|^2 if z = cos(x) + isin(x) Then z*conj(z) = |z|^2 = (sqrt(cos^2(x) + sin^2(x)))^2 = (sqrt(1))^2 = 1
Question: to realise the denominator, instead of multiplying the top and bottom of the fraction by the denominators conjugate, couldn’t you just multiply the top and bottom by ‘i’ ?
@@matemaatika-math Its related to the Mod-arg form used in the video since e^iθ = cosθ + isinθ. Using it is pretty much doing the same thing as he mentions at 15:45 where you can just divide the modulo and substract the angles.
That would be the cosine of 30 degrees. Its sine is 1/2, so the pythagorean theorem gives you half of the square root of 3 as the answer. I take it you're doing it the hard way.
3:26 "Two things happen when you multiply by a complex number: [...] you scale, that comes from the real component, and you rotate, which comes from the imaginary component."
No, just no. Multiplying by 2i (which has no real component) is still scaling by 2 (besides the rotation by pi/2).
It's the modulus (or absolute value) which is scaling, and the argument which is rotating.
I never get tired of this kind of teaching. I plan to work a lot of problems to master these skills.
12:40 You could also use the trigonometric identity
cos^2(x) + sin^2(x) = 1
as
cos^2(x) - i^2 * sin^2(x) =
= cos^2(x) - (-1) * sin^2(x) =
= cos^2(x) - (-sin^2(x)) =
= cos^2(x) + sin^2(x) =
= 1
No matter which angle is "x"
(in our case it was PI/4),
the result is equal to 1.
P.S.
The identity
cos^2(x) + sin^2(x) = 1
in the imaginary form:
cos^2(x) - i^2 * sin^2(x) = 1
or
[cos(x) + i * sin(x)] * [cos(x) - i * sin(x)] = 1
ahaaaa That's pretty nice.
15:42 tiny correction: *plus* i sin pi/12
Why?
@@matemaatika-math It's a small slip up that can tend to happen at the end of a long piece of working. Presumably, he had the word "minus" on his mind because he just was just referencing the pi/3 *minus* pi/4. Commentating on what he had just written possibly made him write minus instead of plus. I hope this answers your wondering of why. If you're asking why in terms of mathematics - because the final line purports that cos(pi/12) - i sin (pi/12)= cos(pi/12) + i sin (pi/12), which is not true. It doesn't have a effect on the answer of the final question because the the real part of the expression is still correct. I'm certain that had been live in a classroom, one of the students would have pointed the small copying error out.
@@MrBennedy You're right. Thank you!
About denominator with cosine squared of pi over 4 plus sine squared of pi over 4, is possible to use the Fundamental Pythagorean Identity...
An errant "-" instead of a "+" creeps in at 15:38 , but it doesn't affect the answer.
Why do you count it as an errant? He explained what formula he used.
@@matemaatika-math It's errant because the sign of the sine function should not change. sin (pi/3 - pi/4) is still positive sin(pi/12), but like he said, since we're dealing with just the real part, it didn't matter, and probably why Eddie did not notice it either. He surely would have if that part were important.
12:29
This is just cos^2(x) + sin^2(x) = 1 , the pythagorean identity
no need for exact values; although it is reassuring we get 1 anyway.
Let conj(z) be the conjugate of z
z*conj(z) = |z|^2
if z = cos(x) + isin(x)
Then z*conj(z) = |z|^2 = (sqrt(cos^2(x) + sin^2(x)))^2 = (sqrt(1))^2 = 1
eddie woo one of the best asian youtube teacher 😄
18:15 "(1 + sqrt(3)) / (2*sqrt(2))" - now you could multiply both nominator and denominator by sqrt(2) to get
cos(PI/12) = (sqrt(2) + sqrt(6)) / 4
Lovely work. The 1 in the denominator was a delight 🤗
Question: to realise the denominator, instead of multiplying the top and bottom of the fraction by the denominators conjugate, couldn’t you just multiply the top and bottom by ‘i’ ?
You can use Cis notation
Did this topic recently, a bit of revision won't do any harm :)
Question B would be made so much easier with exponential form XD
What form is that?
@@matemaatika-math The form z=re^(iθ)
@@Seppevh I'm not aware of that and pretty sure that this form isn't in the curriculum.
@@matemaatika-math Its related to the Mod-arg form used in the video since e^iθ = cosθ + isinθ.
Using it is pretty much doing the same thing as he mentions at 15:45 where you can just divide the modulo and substract the angles.
if you know r1cisA/r2cisB=r1/r2cis(A-B) it should take 5 sec
Astonishing...
That would be the cosine of 30 degrees. Its sine is 1/2, so the pythagorean theorem gives you half of the square root of 3 as the answer. I take it you're doing it the hard way.
What exactly?
π/12 = 15°
The exact value of cos(п/12) is cos(п/12)
Sir how can I become a great mathsmatician like you.
We learn this in class 10😅😅
Where Eddie comes from, we learn this in Year 11/12.
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