Why is this math community so much better than others that i have been involved in? Here, on this channel, nobody judges, nobody brags,.... its a utopia of curiosity and good vibes.
@@PrimeNewtons 😀 thank you! Its an honor to share this very small moment of time with you. The universe is so very old and here we are. We are all very lucky.
Sir I really understood the basics of this topic. At first I was really scared of this topic. But now after I got to know this topic explained with simplicity, I became a fan of your teaching. Thank you very much!!!!!!!!!!
but the way you were so excited to teach, omg. i smiled everytime you smiled and looked into the camera😭. first time watching your video, gonna continue for sure. love from India!
What's also interesting is that it doesn't matter which complex root is called omega. If you square it, the result is the other root which can be called omega-squared. The results are fascinating indeed.
This was fabulous, sir! I had trouble understanding from my own textbooks what unity meant in a case full of examples pertaining to it. The video helped marvelously!
You're such a great teacher! I had so much difficulty in understanding this topic but you explained it so well and so patiently. Truly helpful! Please keep posting videos and maybe start some courses on TH-cam as well because you're really a great teacher!
Nice - and when I taught this to chemistry graduate students who had forgotten (or never really learned) about complex numbers in a review session, at this point they were perfectly prepared to love the polar form of complex numbers. Because, nothing is better than going through the Cartesian version and seeing how much simpler it is to just write 1, exp(2πi/3), exp(2πi/3) for 1, ω, ω² - and performing the multiplications is trivial.
The only thing missing here is that (ω²)² = ω⁴ = ω³·ω = 1·ω = ω, or in other words, each of ω and ω² is the square of the other. Repeatedly squaring flips back and forth from one to the other. There's also that -ω is a sixth root of unity, but that's out of the scope of this video.
Truly, there IS an infinite number of solutions. Number 1 Can be expressed in euler's form which is periodic by 2kPI. Then using Moivre's formula we get those three solutions and each of them is also periodic by 2kPI, k is the whole number.
well .. i^4 = ω*ω² = ω^4/ω => (i/ω)^4 = 1/ω => ω = (ω/i)^4 = err ω^4 (and ω^5 = ω² and ω^6 = 1) the magics of 3 solutions on a circle in complex plan. (can also even guess some angles for R*(cos(phi)+i.sin(phi)) variations .. R = 1 here)
Yes, if ω² is one of the cube roots of 1, then (ω²)³ should be equal to 1. By laws of exponents this means (ω³)² = 1 and ω6 is also 1. On the complex plane, each factor of ω is a rotation of 120 degrees, so repeated multiplication just spins us around and around, reaching (1, 0i) every third term.
Man.... What it ended...oh god 12 min gone so quickly ....man your teaching style is just mesmerizing ...i just loved it and also it cleared my doubt ❤
Sir, you are really great. The way you make things understand is awesome. You explained the concept in a very simple and easy way.... love you Sir, keep posting these types of videos.
we can get 'n' th root of any number. Just take a circle with radius of the number in argand plane and divide 2pi by n and every {2zpi/n} z belongs to integer 0
We can generalise, that the sum of consecutive 3 powers of omega is always equal to zero. Btw, your views should be in millions. The way you teach is just awesome, even a 5th std kid would understand a 10th std concept by seeing your videos!! Hats off sir 🙏🙏
The magnitude of omega is 1 and argument of 120 degrees (2pi/3 radians), so repeated mulitplication by omega 'spins' us around the origin, reaching (1, 0i) every third step.
Thank you for the nice video. 💥 The grapical representation of 1, ω, ω² on the complex plane, as it seems, will be good thing to provide visual proof of sum 1 + ω + ω² = 0 and product 1 ∙ ω ∙ ω² = 1.
vieta's formula says it all. great video anyway! (i've been thinking about this since i started watching your videos: if there's an ASMR maths channel, you're definitely the one!)
Yes, I did pass the math exams while studying for an electronic engineer...and yes, I did learn how to use imaginary number. However, my colleagues and I are forever puzzled with how deep the whole is once humankind mathematicians accepted the convenient truths (like imaginary numbers or negative numbers etc.). What you accept because it is convenient may and will bite you in the as* down the road. No wonder we have so many time-space-quantum scientists who disagree on basic principles of anything around us.
We can generalise those to all nth root of unity. I am not a math teacher so I cant explain every thing but let me give you some tools. Imagine the unity cercle in complex plan. center or the cercle is 0.0 and it goes through 1,0 if you want the all the nth root of 1 you just need to divide the cercle into n equal parts. to add all of those you just need to add the vectors from 0.0 to those roots, and since they are by construction balanced around the circle, sum is 0. For the multiplication I dont have a visual proof to mind.
For the visual multiplication proof, you need to first establish that multiplying numbers on the unit circle is always equivalent to a rotation ; multiplication by _i_ is a 90° rotation, but other values result in rotations by different angles.
The multiplication property only works for odd roots of unity because you can always pair ω^k with ω^(n-k) to get 1. With even roots of unity you will be left with ω^(n/2)=-1 as a solitary factor. So they always give you -1. e.g.: 1*i*-1*-i=-1
With (u+v)^3=u^3+v^3+3uv(u+v) and Vieta formulas for quadratic x_{1}+x_{2} = -b/a x_{1}x_{2} = c/a we can solve cubic equation Cube roots of unity help us to find all roots of cubic equation not just one of them
Great video as usual, I have a question: is Omega the only number (besides 1) that equals the fourth power of itself (I noticed that omega^2 * omega^2 = omega)? This also implies that (omega^(2^k))^3 = 1 for any positive integer k, right? And the cool thing about this formula is that it also encompasses the basic case 1^3=1 by picking k=0!
in my opinion, representing the numbers on complex plan, as vectors would more show why the sum = 0 but formally all the video is great. polynom of degree n will always have n solutions either in R or C (and at times even more if k*2pi involved) Ps : your accent is so crystal clear to foreigners... it's a no accent or from anywhere ?)
I desagree. If I'm wrong, I would like someone with more knowledge to correct me. Thank you. In the video, in my understanding, the calculation of a root is being confused with the solutions or roots of an equation. The root symbol is reserved only for the principal root of a number or, if desired, the root of the principal argument complex. Thus, if x is equal to the cube root of 1 then x=1. There are no more values. Cube root is a function. And a function only returns one value. A example about I'm telling. Nobody disputes that √4 = 2. No one who knows anything about mathematics can claim that √4=-2. This is false. Since the symbol √ is reserved for the main root or positive branch (if real). According to the chain of reasoning in the video, the following happens. √4 = x we square (correct) (√4)² = x² It is correct to cancel the root and the square since 4 is positive 4 = x² x² - 4 = 0 (x + 2) (x - 2) = 0 Therefore, x + 2 = 0 → x = -2 x - 2 = 0 → x = 2 Therefore x = -2 is a solution of the equation x² - 4 = 0, but it is not a solution of √4 = x, because √4 ≠ -2, as everybody knows. In the original equation in the video, cube root of 1 only has one value that satisfies it, that is, x=1. The problem appears in the second step, when it is rised to 3 power snd transform in another different equation x³ = 1. In this step two extra (imaginary numbers) solutions have been added. They must be wasted.
You are correct that ∛1 = "the principal cube root of 1" = 1. Thus, the equation 𝑥 = ∛1 only has one solution, 𝑥 = 1. However, in "assuming that we don't know anything about the cube root of 1", our only option is to solve 𝑥³ = 1, which has three solutions.
It's just a case of poor notation. Like how sometimes ∞ means the element +∞ (the opposite of -∞) and sometimes ∞ means the generic unsigned, possibly complex, ∞. It can mean either the principal root or simply any number which cubes to 1 in a multivalued function sort of way, and you just should specify which you mean.
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Why is this math community so much better than others that i have been involved in?
Here, on this channel, nobody judges, nobody brags,.... its a utopia of curiosity and good vibes.
I think it's because you're here.
@@PrimeNewtons 😀 thank you! Its an honor to share this very small moment of time with you. The universe is so very old and here we are. We are all very lucky.
The answer is simple. Prime Newton sets the tone in such a way that that everyone is friendly. Now that is a sign of a good teacher!
@@nothingbutmathproofs7150 💯
I am a math teacher in Greece. You explain all these very . Congratulations.
Sir I really understood the basics of this topic. At first I was really scared of this topic. But now after I got to know this topic explained with simplicity, I became a fan of your teaching. Thank you very much!!!!!!!!!!
but the way you were so excited to teach, omg. i smiled everytime you smiled and looked into the camera😭. first time watching your video, gonna continue for sure. love from India!
You explained it very well....I'm the guy who is afraid of Maths but I have to study for my exams and I just find your video, it is amazing
"Those who stop learning stop living!" Excellent explanation. Thank you
Again... unbelievable. Especially when you put these solutions in a vector form. When you add these 3 vectors you end up at the origin.
What's also interesting is that it doesn't matter which complex root is called omega. If you square it, the result is the other root which can be called omega-squared. The results are fascinating indeed.
This was fabulous, sir! I had trouble understanding from my own textbooks what unity meant in a case full of examples pertaining to it. The video helped marvelously!
Sir🙏 , i am from India thank you for this amazing video which solves my all problems those are roaming in my mind.❤❤
You're such a great teacher! I had so much difficulty in understanding this topic but you explained it so well and so patiently. Truly helpful!
Please keep posting videos and maybe start some courses on TH-cam as well because you're really a great teacher!
Nice - and when I taught this to chemistry graduate students who had forgotten (or never really learned) about complex numbers in a review session, at this point they were perfectly prepared to love the polar form of complex numbers. Because, nothing is better than going through the Cartesian version and seeing how much simpler it is to just write 1, exp(2πi/3), exp(2πi/3) for 1, ω, ω² - and performing the multiplications is trivial.
The only thing missing here is that (ω²)² = ω⁴ = ω³·ω = 1·ω = ω, or in other words, each of ω and ω² is the square of the other. Repeatedly squaring flips back and forth from one to the other.
There's also that -ω is a sixth root of unity, but that's out of the scope of this video.
Truly, there IS an infinite number of solutions. Number 1 Can be expressed in euler's form which is periodic by 2kPI. Then using Moivre's formula we get those three solutions and each of them is also periodic by 2kPI, k is the whole number.
well .. i^4 = ω*ω² = ω^4/ω => (i/ω)^4 = 1/ω => ω = (ω/i)^4 = err ω^4
(and ω^5 = ω² and ω^6 = 1)
the magics of 3 solutions on a circle in complex plan. (can also even guess some angles for R*(cos(phi)+i.sin(phi)) variations .. R = 1 here)
Yes, if ω² is one of the cube roots of 1, then (ω²)³ should be equal to 1. By laws of exponents this means (ω³)² = 1 and ω6 is also 1. On the complex plane, each factor of ω is a rotation of 120 degrees, so repeated multiplication just spins us around and around, reaching (1, 0i) every third term.
@@mikefochtman7164 let's be serious ... 2*pi/3 radians.
Man.... What it ended...oh god 12 min gone so quickly ....man your teaching style is just mesmerizing ...i just loved it and also it cleared my doubt ❤
Sir, you are really great. The way you make things understand is awesome. You explained the concept in a very simple and easy way.... love you Sir, keep posting these types of videos.
Thank you SO MUCH for verifying a complex solution involving an odd power!
we can get 'n' th root of any number. Just take a circle with radius of the number in argand plane and divide 2pi by n and every {2zpi/n} z belongs to integer 0
nods, it's the geometrical way to consider this problem.
We can generalise, that the sum of consecutive 3 powers of omega is always equal to zero.
Btw, your views should be in millions. The way you teach is just awesome, even a 5th std kid would understand a 10th std concept by seeing your videos!! Hats off sir 🙏🙏
thats just common sense
Does this guy just know everything?
No. He is still learning.
@@robertpearce8394 he never stops
so simple yet so intriguing. Man mind blown
What's also amazing is when you look at these on the argand diagram and see how they relate when they are written in polar form :D
Why did i understand this so easilyyy
You're the best!!
Fascinating stuff. Entertaining mathematics.
This is elegant and beautiful
The magnitude of omega is 1 and argument of 120 degrees (2pi/3 radians), so repeated mulitplication by omega 'spins' us around the origin, reaching (1, 0i) every third step.
These last properties of the cube roots (they multiply to 1 and sum to 0) is a property of all nth roots of 1, where n is any natural number.
I wonder if it extends to irrational numbers. For example, 1^π has infinite possible values, e^(2π²ni) for integers n.
Thank you for the nice video. 💥
The grapical representation of 1, ω, ω² on the complex plane,
as it seems, will be good thing to provide visual proof of
sum 1 + ω + ω² = 0 and product 1 ∙ ω ∙ ω² = 1.
z^3 +0*x^2+0*x-1=0
The coefficient of x^2 is equal to 0, therefore according to Vieta's formulas, the sum of the roots is 0
vieta's formula says it all. great video anyway!
(i've been thinking about this since i started watching your videos: if there's an ASMR maths channel, you're definitely the one!)
You should have finished the video with a graphic representation of the 3 roots on a complex plane
Thank you for the video, hope you discuss the curvature of a curve in a simplistic way.
Oh. That would be a good video. Thanks for the suggestion.
Yes, I did pass the math exams while studying for an electronic engineer...and yes, I did learn how to use imaginary number. However, my colleagues and I are forever puzzled with how deep the whole is once humankind mathematicians accepted the convenient truths (like imaginary numbers or negative numbers etc.). What you accept because it is convenient may and will bite you in the as* down the road. No wonder we have so many time-space-quantum scientists who disagree on basic principles of anything around us.
No this is really insane wow!!
Bravo Prime Newtons! That's serious stuff. Power 3
(a+bi)³ = 1
(a+bi)(a²-b²+2abi) = 1
a³-3ab²+3a²bi-b³i = 1
a³-3ab²+i(3a²b-b³) = 1
a³-3ab² = 1 and 3a²b-b³ = 0
3a² = b² and a³-9a³ = 1
a³ = -1/8 and a = -1/2
3(1/4) = b² and b = ±sqrt(3)/2
The complex answers are (-1±isqrt(3))/2
We can generalise those to all nth root of unity.
I am not a math teacher so I cant explain every thing but let me give you some tools.
Imagine the unity cercle in complex plan. center or the cercle is 0.0 and it goes through 1,0
if you want the all the nth root of 1 you just need to divide the cercle into n equal parts.
to add all of those you just need to add the vectors from 0.0 to those roots, and since they are by construction balanced around the circle, sum is 0.
For the multiplication I dont have a visual proof to mind.
For the visual multiplication proof, you need to first establish that multiplying numbers on the unit circle is always equivalent to a rotation ; multiplication by _i_ is a 90° rotation, but other values result in rotations by different angles.
The multiplication property only works for odd roots of unity because you can always pair ω^k with ω^(n-k) to get 1.
With even roots of unity you will be left with ω^(n/2)=-1 as a solitary factor. So they always give you -1.
e.g.: 1*i*-1*-i=-1
With
(u+v)^3=u^3+v^3+3uv(u+v)
and Vieta formulas for quadratic
x_{1}+x_{2} = -b/a
x_{1}x_{2} = c/a
we can solve cubic equation
Cube roots of unity help us to find all roots of cubic equation not just one of them
this was a beautiful video
Thank u soo much sir, it really helped a lot ❤
Great video as usual, I have a question: is Omega the only number (besides 1) that equals the fourth power of itself (I noticed that omega^2 * omega^2 = omega)? This also implies that (omega^(2^k))^3 = 1 for any positive integer k, right? And the cool thing about this formula is that it also encompasses the basic case 1^3=1 by picking k=0!
Bravo from Greece.
Can you prove algebraically that (1^3) +(2^3)+(3^3) = (1 + 2+ 3) ^ 2. ??? Thanks
thank you sir
There are also nth roots of unity, which are all possible complex nth roots of 1
Each of the complex cube roots of unity is both the square and the square root of the other .
Excellent.....
Which class omega is studied in
Guess complex analysis. Omega is also used in maths history by Lagrange to get fun symmetries in 3rd order equations.
11 th grade in india
in my opinion, representing the numbers on complex plan, as vectors would more show why the sum = 0
but formally all the video is great.
polynom of degree n will always have n solutions either in R or C
(and at times even more if k*2pi involved)
Ps : your accent is so crystal clear to foreigners... it's a no accent or from anywhere ?)
Thanks 🙏🙏🙏🙏
Cool video
Power System Protection Engineers use this daily. We just don't call it "omega" and the square root of negative 1 is j.
Thanx
Do you know Probability and Statistics
Amazing...
very well, I wanted to write.
Which grade omega is studied in
11th in india
@@KRO_VLOGSin cbse? I just finished gr12 and we never did this
You forgot to mention, that (ω²)² = ω, which leads to (ω²)³ = ω * ω² = ω³, which you already showed, is 1.
I really thought of that but I wanted those who were curious to continue their quest privately
Now divide them 😈
My teacher just gave value of omega and that’s it :(
I desagree.
If I'm wrong, I would like someone with more knowledge to correct me. Thank you.
In the video, in my understanding, the calculation of a root is being confused with the solutions or roots of an equation.
The root symbol is reserved only for the principal root of a number or, if desired, the root of the principal argument complex.
Thus, if x is equal to the cube root of 1 then x=1. There are no more values. Cube root is a function. And a function only returns one value.
A example about I'm telling. Nobody disputes that √4 = 2. No one who knows anything about mathematics can claim that √4=-2. This is false. Since the symbol √ is reserved for the main root or positive branch (if real). According to the chain of reasoning in the video, the following happens.
√4 = x
we square (correct)
(√4)² = x²
It is correct to cancel the root and the square since 4 is positive
4 = x²
x² - 4 = 0
(x + 2) (x - 2) = 0
Therefore,
x + 2 = 0 → x = -2
x - 2 = 0 → x = 2
Therefore x = -2 is a solution of the equation x² - 4 = 0, but it is not a solution of √4 = x, because √4 ≠ -2, as everybody knows.
In the original equation in the video, cube root of 1 only has one value that satisfies it, that is, x=1. The problem appears in the second step, when it is rised to 3 power snd transform in another different equation x³ = 1.
In this step two extra (imaginary numbers) solutions have been added. They must be wasted.
You are correct that ∛1 = "the principal cube root of 1" = 1.
Thus, the equation 𝑥 = ∛1 only has one solution, 𝑥 = 1.
However, in "assuming that we don't know anything about the cube root of 1", our only option is to solve 𝑥³ = 1, which has three solutions.
It's just a case of poor notation. Like how sometimes ∞ means the element +∞ (the opposite of -∞) and sometimes ∞ means the generic unsigned, possibly complex, ∞. It can mean either the principal root or simply any number which cubes to 1 in a multivalued function sort of way, and you just should specify which you mean.
Hi brother
But, I thougt that n sqrt could use at real number.
I have a unity Shirt lol
😯😮😶🙂😀😃😍
What a great teacher!!
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