all solutions to x^3=1

[@chessandmathguy]

beautiful. just beautiful. won't help me get back with my ex, but that's a different story for a different day.

[@lucasoliveirasaintrain4298]

- Oh you guys were such a cute couple, why did you break up? - She said that 1 was the only solution to the cubic root of 1...

[@crazye7132]

Try a different approach, maybe you'll get then the solution

[@donmoore7785]

lol - that is the nth root of crazy you are trying to solve for.

[@guythat779]

That sounds like a challenge I love those

[@MauryaAcademy]

kzbin.info/www/bejne/m5LNoJamgd93iJY

[@adamhrankowski1298]

I prefer the polar form, because it's so easy to extend it to higher roots. Plus the regular polygons centred on the origin provide such an effective geometric illustration of the fundamental theorem of algebra.

[@blackpenredpen]

😮 I just noticed your comment from 3 years ago!

[@jithu8018]

@@blackpenredpen so to find nth root of 1 we just need to find the cordinates of n sided regular polygon right. wow thats cool

[@aryanjoshi3342]

@@jithu8018 and since these polygons have evenly spaced angles, the complex numbers have angles 0*2pi/n, 1*2pi/n, 2*2pi/n, 3*2pi/n ... (n-1)2pi/n so just plug that into euler's formula and youve found the numbers!

[@dataweaver]

​@@jithu8018 It is, isn't it? I can just see a 3B1B-style animation of what happens as you take the nth roots of 1, as n gradually increases: from a point (n=1) to a line (n=2) to a triangle (n=3) to a square (n=4) to a pentagon (n=5), and so on.

[@guitaristxcore]

Finding nth roots like this is just so much more fun in polar form.

[@Baltie3]

I am amazed by your fluency in utilizing 2 markers at the same time!

[@b5lovermore]

Right?! I was looking for this comment haha

[@richardgratton7557]

Black Pen, Red Pen….🫣

[@AexisRai]

onehanded dualwielding

[@BenjaminRoyes-n4c]

The chopsticks genes

[@TramNguyen-pk2ht]

@@richardgratton7557Where's blue pen?

[@flamingpaper7751]

These are all the fourth roots of 1: 1 -1 i -i

[@ottotutzauer2788]

A bit easier to calculate, you just take all the z values at the points which have angles 90, 180, 270 and 360

[@potatopassingby]

fifth roots!

[@KCD3000]

Jino Tzino 1. drive.google.com/file/d/1ZHtKfKJYHsp1RLSJH1JEnYicL6b00bin/view?usp=drivesdk

[@KCD3000]

Jino Tzino 1. Here are the solutions.

[@riadfikani3735]

the big bang theory

[@matthewzuelke6721]

I prefer the first solution when solving generically but the 2nd solution is a fun alternative that i wouldn't have thought of

[@davidrheault7896]

The second solution comes from Euler, e^(2*i*pi) = 1, then transform to polar form and use deMoivre theorem.

[@skilz8098]

1+1 = 2 is the algebraic method of arithmetic... The geometrical representation of this equation is the unit circle translated by 1 to the left... so that one of the 1's is at -1, the other 1 is at 0 and the 2 is now at +1. Yes, the unit circle with its origin defined at (0,0) as well as the Pythagorean Theorem are embedded in the very first mathematical expression/equation (1 + 1) = 2. Fun Facts!

[@dadoo6912]

@@davidrheault7896 it does not come from euler's formula. absolutely different things

[@donmoore7785]

I always liked it when professors showed multiple solutions. Well done.

[@chengkaigoh5101]

Go on,list all infinite solutions to this in exponential form

[@pbuj]

@@chengkaigoh5101only 3

[@saxbend]

Bring in the quaternions too, why not? :)

[@alexwang982]

saxbend What will that get us? Platonic solids?

[@chanderule605]

Wont that simply give you extra 4 numbers where you replace i by j or k?

[@juhaszelodmusic]

I graduated two years ago with a BSc in Electrical Engineering and now I simply enjoy your math videos more than ever before when I had to do maths at university 😂 Thank you ❤

[@jackeea_]

2:26 This is 1 Minus 4 that's -3 Quik mafs

[@obinnanwakwue5735]

Quick maths*

[@markorezic3131]

Obinna Nwakwue no its mafs*

[@gian2kk]

Quick maffs*

[@blackpenredpen]

Mafffs*👻

[@LondonModzHD]

Bcuz of the ayota the - is removed

[@jeromesnail]

Another method is just to think about 1 as e^(2kπi) for all integer k. Then ³√1 = (e^(2kπi))^(1/3) = e^(2kπi/3) If k=0 you get e^(0) = 1 If k=1 you get e^(2πi/3) = cos(2π/3) + isin(2π/3) = -1/2 + i√3/2 If k=2 you get e^(4πi/3) = cos(4π/3) + isin(4π/3) = -1/2 - i√3/2 For k2 you get the same results knowing that for any integer n and any real number x,, cos(x+2nπ) = cos(x) and sin(x+2nπ) = sin(x)

[@davidrheault7896]

Euler's identity in all its glory :)

[@thebrooksarmy2318]

do you think you're clever or something?

[@bensevin1931]

n-th roots of unity : Un={ exp(i2kπ/n) , k∈⟦0, n-1⟧ }

[@w0lf370]

@@thebrooksarmy2318 (even if this comment is sent like years ago) bruh they just wanted to point out another perspective, it’s a nice thing, nobody’s here to boost their own “smartness shit skill” and that’s the cool part :)

[@MallorcaDuck]

@@thebrooksarmy2318 do you feel offended?

[@JeanDAVID]

When you multiply by i, you rotate by 90°. Here we have three roots so three rotations with different angles. In fact, for x=1, rotation is 0°, for -1/2+i sqrt3/2, rotation is 120°, for -1/2-i sqrt3/2, rotation is -120° or 240°. Three following rotations from position (1,0i) with these angles will return to position (1,0i)

[@m_riatik]

“If you find this (algebraic) method easier, you are a devil worshipper” - Fematika 2017

[@wkingston1248]

MatrixWolf computationally it is harder conceptually most people doesn't even know what the complex plane is.

[@skilz8098]

They are both easy!

[@MauryaAcademy]

kzbin.info/www/bejne/m5LNoJamgd93iJY

[@skilz8098]

@@wkingston1248 That's only the complex number line against the real numbers. What about a complex volumetric plane with complex-axes (i,j,k) being perpendicular to the volumetric real plane (x,y,z). And they said the former was hard?

[@harleyspeedthrust4013]

@@skilz8098 11 month update: your comment was major cringe

[@holyshit922]

They will appear in formula for roots of cubic equation I think that the Moivre will be useful later in so called casus irreducibilis Way for solving cubics can be generalized to quartics For cubic equation we can use following thing (A+B)^3=A^3+3A^2B+3AB^2+B^3 (A+B)^3=A^3+B^3+3AB(A+B) (A+B)^3=3AB(A+B)+(A^3+B^3) which is in the same form as the equation y^3=3py+2q , (after shifting variable to eliminate x^2 term) For quartics we can use the same concept (A+B+C)^4=2(A^2+B^2+C^2)(A+B+C)^2+8ABC(A+B+C)+4(A^2B^2+A^2C^2+B^2C^2)-(A^2+B^2+C^2)^2 which is in the same form as the equation y^4=2py^2+8qy+r , (here we also shift variable to eliminate x^3 term) I got this identities after playing with symmetric functions After comparing coefficients we will get system of equations which can be transformed to Vieta formulas for sextic polynomial reducible to lower degree polynomial (for cubics sextic resolvent is reducible to quadratic and for quartic sextic resolvent isreducible to cubic) Vieta's formulas involving elementary symmetric functions of the roots and coefficients of polynomial can also be useful for solving cubics and quartics Symmetric function does not change after permutation of its variables Another way for solving cubics and quartics is x^3+px+q=A(x+n)^3-B(x+m)^3 x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}=(x^2+px+q)^2-(rx+s)^2

[@Fematika]

Fun fact - An Eisenstein integer is a + b * w for integers a and b and w = e^(2pi/3), a cube root of unity. The units in this ring (the numbers that have a multiplicative inverse) are simply just plus or minus the cube roots of unity. Then, you can extend this to what is known as an Eisenstein prime, which is an Eisenstein integer who's only divisors are the units, the cube roots of unity, and themselves.

[@pbondin]

Thank you for your clear derivation of the cube roots of unity. I also remember my high school teacher (more than 50 years ago!) demonstrating that the cube roots of unity are: 1, w and w^2. From your derivation, it was easy to see that, if you let w = (-1/2 + i*sqrt(3)/2), then w^2 = (1/2 + i*sqrt(3)/2)*(1/2 + i*sqrt(3)/2) = (-1/2 - i*sqrt(3)/2). I know this is obvious to most of your subscribers, but for me it’s another reminder of how much I enjoyed (and still do) mathematics.

[@bobh6728]

I know it’s been a year, but when you expanded w^2, I think you left the negative off of the 1/2. The answer you gave is the correct one if you include the negative. So out of 100 points, I would give you a 98%.

[@General12th]

Using degrees? On a BPRP video? I guess we're not grown-ups anymore!

[@jehoiakimpillay342]

What's not grown up about degrees

[@hailnijo7099]

Cos real men use radians

[@hello_2632]

@@hailnijo7099 Real Men use Gons.

[@ccuuttww]

add radian to your watch clock instead of 1 -12 ?

[@amaxingmusic9334]

@@hailnijo7099 *cos*

[@aeuhgv139]

you could also do: x^3=1 x^3=e^(k * 2pi * i) with k an integer x=e^(k * 2pi * i)^1/3 x=e^(k * 1/3 * 2pi * i) x=e^(k * 2/3pi * i) k=0, k=1 and k=2 give respectively x=e^0 x=e^(2/3pi * i) x=e^(4/3pi * i) you know that 1/3pi rad = 30° so 2/3pi rad and 4/3pi rad are just mutiples of 30° Every found number on the unit circle forms a special right angle triangle with one 30° angle With the unit circle and special right angle triangles you find that x=1 x= -1/2 + 1/2 * sqrt(3)i x= -1/2 - 1/2 *sqrt(3)i

[@blackpenredpen]

AeuHG V yes!

[@biohead66]

Love how he uses 2 different color markers and alternating them really fast.

[@jacoblund8289]

Hey, great video! When I was first learning about the roots of unity there was an exercise which asked me to prove that they sum to 0, and Id like to share my answer with you: Where w is an nth root of unity and I'm going to call the series: 1 + w + w^2 +... w^n-1= S because w^n = 1: 1 + w + w^2 +... w^n-1= w^n+ w + w^2 +... w^n-1 therefore: S = wS If assume S is not 0 and dicode both sides by S, we get w = 1 which is a contradiction for all n other than 1 therefore S = 0 Thanks for reading!

[@blackpenredpen]

Jacob Lund Hi Jacob! Thanks for sharing your proof. This gives me an idea for making my future video. Can I take a screenshot of your comment and use in my video? Thanks

[@NotYourAverageNothing]

Jacob Lund Where is the contradiction for w = 1?

[@aahpandasrun]

Thanks for explaining so many details that other videos tend to just skip over, such as "roots of unity means roots of 1"

[@PM13501]

Really liked how you handled the two colors for the markers without making a mistake even once!

[@Ali_Pxll]

I was studying calculus, & I studied a lot of ur content, then I faced x^3 - 1.. searched on YT and found u. you're a real math finisher

[@GeekTommy]

Your videos about series and integrals helped me a lot to pass my calculus exam, thank you so much! I'm Italian and we call that "Analisi Matematica 1", it's the exam about limits, derivatives, series and integrals. Even if I'm not English, your explanations are really clear and understandable, I wish all the teachers were like you! Keep up the great work! A question, will you ever make videos about linear algebra (matrices and so on) and probability stuff?

[@contaminatedduck4959]

Your english is great

[@Medesogato]

​@@contaminatedduck4959Your english is good to.

[@jey_lhnq68]

Hai ragione uomo analisi uno colpisce tutti

[@jeromesnail]

We're all grown ups now, we use radians :p

[@skilz8098]

Sometimes with simple and common angles degrees are easier to memorize and work with. As for me, it doesn't really matter as I can use either! But in truth, the amount of separation between two vectors by a given angle is defined as degrees. Radian is the actual arc length that is generated by that angle with regards to the radius of the circle of that arc. They are in a way interchangeable and easy enough to convert from one to the other, however, they are not the same. If we look at the unit circle we know the radius is 1 so we don't have to use it within the equation or definition to define arc length, we can just omit it. If we look at the unit vectors along the x and y-axis that point at the value 1 on their respective axes they will both be points that lie on the unit circle. The actual (angle) that separates these two vectors is 90 degrees. The length of the arc that is generated by the 90-degree angle from the points (1,0) & (0,1) is PI/2. So there is a direct relationship between Degrees and Radians as the following set of pairs are interchangeable: (0 && 360 degrees, 0 && 2PI radians), (30 degrees, PI/6), (45 degrees, PI/4), (60 degrees, PI/3), (90 degrees, PI/2), (180 degrees, PI radians). The value on the left is the actual value of the angle, the value on the right is the arc length. Now, this only works directly in the unit circle because the "r" or radius value was omitted. If you have a circle that doesn't have a unit distance, then you would have to factor the radius into the equation as these arc lengths would change by the factor of that radius, however, the corresponding relationships will remain. I use both, but it depends on what is being referred to. If you are asking me for the angle (then it is implied as degrees). That is the rotational distance between two vectors! Now if you want the arc length that is generated by that rotation in degrees, then you are talking radians! There is a difference! Most overlook this distinction! I know, because I use both as I write many programs that deal with angles and some of then need to be in degrees while others need to be in angles. Now, in the context of writing programs, they are in a sense interchangeable as long as you provide the appropriate conversion from one to another! There is a 3rd form or notation of angular displacement which many don't refer to, talk about or use is grads! Not too many people use gradians anymore or even know what they are, mostly the french, or those who do land surveying! Just something to think about! Is knowing when and where to each convention, staying consistent with the convention and using the appropriate conversions when translating between one and another!

[@jeromesnail]

@@skilz8098 1) it was an inside joke 2) It's the first time in my life that that I hear about the fact that degrees should used to mesure angles between vectors, and I studied maths for years. I can't even find one reference about this fact anywhere...

[@skilz8098]

​@@jeromesnail The rotational separation between two vectors is the measure of degrees or gradians where there are 360 equally divided sectors of a full circle for degrees, and for gradians, it is 400. These are the amount of rotational separation between vectors. It is a linear distance but in 2D. When you talk about radians, this is the arc length that is generated by those angles. Here's what I mean: Draw 3 concentric circles where they all have the same midpoint. Now draw a vertical and a horizontal line that passes through all 3 circles and intersects at the center. Label the center (0,0). Make the innermost circle have a radius of 1, the next circle a radius of 2, and the outer circle a radius of 3. We know that the angle between the vertical and horizontal lines is 90 degrees! This amount of separation between these two lines will never change. The arc length of the inner circle that has intersecting points on both of these vectors will have a length of PI/2 radians. This is why 90 degrees and PI/2 are interchangeable since Arc length is defined by r * theta. However, Arc length (s) is usually in terms of radians. So we have to convert theta from 90 degrees to PI/2. Since r = 1 we end up with PI/2 radians. When we go out to the next circle, the angle is still 90 degrees or PI/2, but the generated arc length for the second circle will be PI since r now equals two instead of one. And for the third circle, the generated arc length will be 3PI/2. The observation here is that PI/2 is the common factor of a 90-degree angle for all circles with a given radius of r. This is what I was getting at! The concept of degrees being the separation of two vectors is in the definition of a degree! Take a circle and divide it into 360 equal parts. The separation or distance between each of these vectors is a uniform and equal distance. Just look at any information on any given navigational system that is modeled, built and based on the compass! There is a subtle distinction between degree and radian. A degree is the actual rotational distance, the amount of turning at a fixed point, where radians is the actual arc length that is generated by that rotation based on a unit vector. It's never really taught this way directly, but one can easily come to this conclusion. When you begin to work with Vector Calculus, Transformations, Physics, etc. You will easily see the relationship between the two and their differences. They are slightly different approaches of measurement, but there is a corresponding relationship to them. For example; an Arc Length of PI radians will be generated by a 180-degree angle when the radius is 1. However, an Arc Length of PI radians can also be generated by an Angle of 90-degrees when the radius is 2! This is why you have the half and double angle properties for the trigonometric functions! I hope this gives you a little bit of clarity about what I was getting at!

[@imperialrecker7111]

@@jeromesnail I have been studying with radians for 2 years now, and I feel uncomfortable with degrees now. Radians is so useful in maths.

[@skilz8098]

@@harleyspeedthrust4013 I can't help it if this younger generation is too lazy to read a few paragraphs...

[@loich.9133]

the formula for the unity root is exp(i*k*pi/n), with n the highest exponent of the equation, and k∈[[1;n]]

[@ok64234]

{e^(i theta)}^3 = 1 ... then ... e^(3 i theta)} = 1 then ... cos(3 theta) + i sin (3 theta) = 1 This happens when ... 3 theta = 0°, 180°, 360° then .. theta = 0°, 60°, 120° //// Replacing value of theta on original number e^(i theta) ... it means any polar number: 1, e^(i 60°) or e^(i 120°) is a cubic root of 1 Thanks //

[@skylardeslypere9909]

but as we know e^(pi i) is -1. So that quared is (e^(pi i))^2 = e^(2pi i) = (-1)^2 = 1 so when we take the cube root of that number, we get (e^(2pi i))^(1/3) = 1^(1/3) = cbrt(1) so another cube root of 1 is just e^((2pi i)/3) 2 * pi * i ------------ 3 e

[@seanl.5181]

Before I watch the video, are the answers 1(trivially), -0.5-sqrt(3)i/2, and -0.5+sqrt(3)i/2?

[@blackpenredpen]

Sean L. Yes! HEYYYYY SEAN!!!!!!

[@seanl.5181]

I figured that out by the second method you showed.

[@mosab643]

What does the cube root have to do with dividing the circle of 360 deg into 3?

[@mtd197]

Without a doubt the left one was easier, but the second one felt as if I understood more now. I'm 14 so learning this was really helpful. Thanks so much and you have earned a sub 😊

[@bismandeep5266]

why do u wanna learn abt cube roots of unity at age of 14 lol,, i learnt them when i was 16 ,,, the second method btw is de moivre's theorem basically

[@grantbell8143]

Not to sound like a smart ass but the second approach is actually easier if you have a good understanding of Euler’s formula. You get can away with not using the quadratic formula and the only thing you essentially need to know is how to divide 2pi by the n-th power of the equation. Plug those values into Euler’s formula and boom you have all solutions for that polynomial

[@blueberryjackson7539]

14 as well!! altho you're an adult now ig lol??

[@archangecamilien1879]

Roots of unity...rewriting e^{3 ln(x)} would help, etc, I mean, complex numbers...using de Moivre's formula, if that's what it's called, e^{i theta} = cos(theta) + i sin(theta), etc, get sin(theta) to be 0, so try to make 3 ln(x) = 0, pi or 2pi, if I didn't forget how the sine is zero, lol...which would mean x = 1, x= e^{1/3 pi} or e^{2/3 pi}...I mean, lol, if I didn't screw all of this up...

[@shilopkala9178]

​@@bismandeep5266because they want to know? I am also learning this at the age of 15...one of my friends at 14..

[@squared8290]

OK, first off, love your video and many of your others too. Secondly, I jumped so quickly to solving for x using x = a + ib I didn't even think of using the quadratic equation so thank you for that! My only comment is more like a question that I'd love your feedback on. When you explained the solution using w (instead of x for all future references in this comment) as a complex number, you said we know w = 1 and we need to rotate it three times so it must be at 120 degrees and also at 240 degrees, to then get back to itself if I understood you correctly. That's not how I would explain it. I feel like we are starting at the end state and skipping some steps perhaps when we explain it that way. I would say, w, whatever it is, is a vector expressed in the complex plane, that when you cube it, you arrive at w^3 = 1. Now usually, multiplying by a vector both changes the direction of the vector you started with AS WELL AS its magnitude. However, since we know that w^3 = 1, we know that the starting vector, w, must be NORMALIZED in its starting state or there's now way cubing w would yield 1. (Note that I therefore view the problem as having a starting vector and then applying just two operation to it, once multiplying by w to get the square, and once again to get the cube. I don't think of this therefore as three steps or three components but rather two operations applied to a starting vector). Therefore, since w must be normalized, we know it must lie somewhere on the unit circle. So the question is, where does that starting vector, w, lie on the unit circle? Well if we start at w and then apply two operations to it, first squaring it, and then multiplying again by w to get the cube, we know by symmetry that the starting angle of the vector w must be 360/3 degrees away from the Real axis at zero degrees, so either at 120 degrees or at -120 degrees/240 degrees. At 120 degrees, we would be rotating the vector twice, each time 120 degrees, to get to the Real line in the counterclockwise direction, whereas if we start at -120 degrees or 240 degrees, we could be rotating the starting vector twice in the counterclockwise direction to get to 1. That's my intuitive, vector-like explanation of what is actually happening of what w must be for w^3 to equal 1. Please let me know what you think! I'd very much appreciate your feedback, comments, and any corrections or refinements you could offer to my approach and explanation. And thanks again for all the great videos. Cheers!

[@sab1862]

I love second. Unit circle is very beautiful and most symmetrical!

[@wkingston1248]

Sa B i hate circles. Hyperbola is best shape.

[@sab1862]

WiSpKing Well, however, circle and hyperbola are friends. :) Circle is defined as x^2+y^2=1, and hyperbola is defined as x^2-y^2=1! Also, trigonometric functions and hyperbolic functions are friends, too. For example, cos(ix)=i cosh(x). Circ and hype are best friends!

[@wkingston1248]

Sa B i just don't like circles aesthetically. :>

[@sab1862]

WiSpKing Aha, okay. But in my case, I like circle aesthetically. ;)

[@mikyjet065]

I like this video. I just do not understand why did you ritate around the circle 3 times, is it because of the power or was it just random? If someone could answer my question please.

[@Music--ng8cd]

Polar form is nice because you can solve it graphically and see all the relationships. Also the formula works for any equation so it's more plug and play.

[@counterleo]

If you like the polar form he used in the video, wait till you see the exponential notation - DistanceFromOrigin + e^(i * AngleInRadians)

[@saranshsharma3147]

A more interesting thing, For € = - 1/2 +i(3^1/2)/2 €^2 = - 1/2 - i(3^1/2)/2 Where €^2 is also an answer for cube root of 1!!!!!! That means, the values for cube root of one are 1,€,€^2!!!!

[@Magic73805]

Today; You are looking very very very very very..................Smart Sir.👌👌👌👌👌👌👌👌👌👍👍👍👍

[@Mnemonic-X]

You're wrong because the imaginary unit is not a number. So we have only one cubic root of 1.

[@OptimusPhillip]

x^3=1 when x=e^(2nπi/3), where n is an integer. This gives us three distinct solutions on the complex plane: when n=0, x=e^0=1; when n=1, x=e^(2πi/3)=-1/2+sqrt(3)*i/2; and when n=2, x=e^(4πi/3)=-1/2-sqrt(3)*i/2. Every integer value of n gives you one of these three solutions.

[@yarooborkowski5999]

Could You show us in similar way why we can not solve any general equation of degree higher then 5? ax^5+bx^4+cx^3+dx^2+ex+f=0

[@mariomario-ih6mn]

That is more advanced than you think.

[@3marElsayed]

You're the best ever.. My teacher told me only that the cubic roots of 1 is just this numbers and didn't tell me why! while you explained in algebraic and trigonometric forms, thank you very much.:)

[@blackpenredpen]

Yay!!!

[@bismandeep5266]

i dont see no reason for ur teacher to not to teach u abt cube roots of unity unless u study in a lower grade

[@ScottKKern]

Great video and, as always, very enjoyable to watch. One thing that may be missing is showing that 240 degrees rotated three times also lands you at 0 degrees (mod 360 degrees). 120 degrees is sort of obvious because it’s 360 degrees, I guess. And 0 degrees x 3 = 0 degrees, sure. But the 240-degree case is a good one to demonstrate the modularity. And oddly enough omega 1 x omega 2 times omega 3 is also 1, right? But that’s not the point at all!! So it may be good to show that *each one* lands you at 1+0i.

[@brianblumberg7599]

Why do you have to rotate three times? What is the relationship between the angle of "rotation" and the formula?

[@apenasmeucanal5984]

Great, you just helped me derive an algorithm for finding the n complex roots of a real number!

[@MauryaAcademy]

kzbin.info/www/bejne/m5LNoJamgd93iJY

[@AndyGoth111]

Cubing each successive cubic root of unity indeed brings you back to unity, but "along the way" there are zero, one, or two full revolutions. This might be interesting to explore in another video utilizing de Moivre. General form: ª√1=exp(i2nπ/a), where n∈{0,...,a-1}. If n is any other integer, exp(i2nπ/a) will give the same results as for n∈{0,...,a-1} because it is a periodic function, so the number of a'th roots is a. See also: Ethan Jensen's comment about tripling angles to get 0° [more precisely, 2nπ].

[@jakehu8733]

Fun fact, you can outvert it into sin(1/3 * pi)i - 0.5, -sin(1/3 * pi)i - 0.5, respectively in the way blackpenredpen wrote the answer.

[@AbouTaim-Lille]

It is easier to use the polar coordinates of complex numbers to find the roots. If z=r.cis(t) is the root then 1= z³= r³.cis (3t) . so r= 1 and cis (3t) = cos 3t + i. Sin 3t = 1 . so cos (3t) = 1, sin 3t =0 . which has the solution 3t = 2πk . k=0,1,2,3,... Thus: t= 0, 2π/3 , 4π/3. Or more simply: z= 1, e^2π/3i, e^4π/3i

[@karstenmeinders4844]

My guess is that solution for higher roots n>=4 can more easily be determined using the polar form way - just like one the visitor presented for n=4. Howerver, for higher n the sin (i*2*pi/n) and cos(i*2*pi/2) values will be more difficult (impossible?) to be expressed in algebraic Terms. Anyway, great video as always!

[@gamerguy2369]

notice that with every 120 degree in unit circle three time its making an equilateral triangle with its vertices exactly the outline of unit circle ... another property of cube roots of unity

[@purim_sakamoto]

はー　複素平面ほんと便利すぎる 日常生活を実数だけで送ってるのほんとばからしくなる

[@usmasuda]

This can be easily generalized to x^n=1. x = cos(2pi/m) + i*sin(2pi/m), where m = 1, ..., n

[@owlsmath]

very nice video! I was expecting the 2nd method and not the first but both really good :)

[@proofofalifetime488]

I just did this last week with the polar form of a complex number, that is really cool you made a video of it.

[@MauryaAcademy]

kzbin.info/www/bejne/m5LNoJamgd93iJY

[@Jodabomb24]

Hey, I don't know if you ever take requests, but I remember this really interesting problem from years ago, and I wonder if you might like to take a crack at it, or if you've seen it before: Integral from 4 to 8 of ln(x-3)/(ln(x-3) + ln(9-x))

[@blackpenredpen]

Joseph McGowan Oh hi, I actually have a video that's very similar. kzbin.info/www/bejne/rJm2l4ufh7eXj5Y

[@Jodabomb24]

Ah, so you do, thanks for pointing me to it.

[@ulisespachecosanchez5058]

More than. True teacher in this case when You represent the equation from x3 to x2 You make the differents bewteen in two differents equations that You do. So very interesting to learn more in mathematics

[@shivajoshi9068]

Can u make a vid. On nth root of unity please

[@sunrazor2622]

On the complex plane, all solutions: x = e^( 2 i n pi / 3 ) n = all real integers In Quaternions, you get x=1 and x=infinite solutions on a sphere: x = 1, and... x = -1/2 + a.i + b.j + c.k a^2 + b^2 + c^2 = 3/4 a, b, c are real numbers

[@MathematicsForAll]

Its really helpful for all.... thanks

[@AnchoriteWolf]

Online classes, teacher's notes didn't have an explanation for cube roots of 1. Brilliant video

[@rupeshmishra-vnc]

Plz. Solve find all cube root unity of z^5+32

[@thiagodonascimento7926]

@blackpenredpen, how do you know that the third root of 1 means that you just have to find the three values in the complex numbers with radius 1, and each time rotated by 120°? You just start at this point without any explanation.

[@ryulesmana]

In 1999 while I was attempting math c class, I profound method of solving this by drawing XY real and i diagram with 0/0 as center then put 1 as 1 from 3 Root, Another 2 Roots found by rotate them, one 120°, last one - 120°. Congratulations, you've remind me of those stuff 😊🤝🧬

[@Ravenheart__]

Using the quadratic formula just uses a rule that students have learnt and helps them understand why there’s 3 roots however using the Argand diagram and using the polar forms of the complex numbers to actually show visually where everything comes from is better, even if it means that you’ll need to do a bit of research yourself, (like conjugates, polar forms of complex numbers, multiplying by i = 90 anti-clockwise rotation etc...), to understand everything being explained, great video though!!

[@mijmijrm]

i like using -1 as the basis rotation on the complex plane w^3 = (-1)^2n w = (-1)^(n 2/3) .. where n = 0 , (+,-)1 w = cos(pi n 2/3) + j sin(pi n 2/3)

[@Re-lx1md]

Can you derive the arc length formula?

[@awfuldynne]

The distance between two points on a curve f(x) is √{[f(x + Δx) - f(x)]^2 + Δx^2} where Δx is some nonzero number. I hate reading the in-line math, so feel free to write it down as you follow along if you know how it's supposed to be written. The mess of brackets above is the distance formula √(a^2 + b^2) where a is the vertical distance from f(x) to f(x + Δx) and b is the horizontal distance from x to x + Δx, which simplifies to just Δx. The sum of all the Δx across the curve becomes an integral as Δx becomes arbitrarily small, but there's a problem in writing it as in integral, since this formula is for the actual length of the short arcs, which is like a formula for the actual area of the rectangles under the curve in a normal integral; you need to factor out the Δx that becomes dx to turn it into a proper integral. Well, when you factor out a Δx from under a square root, you divide each term under the square root by Δx^2, which gives us √{[f(x + Δx) - f(x)]^2/Δx^2 + Δx^2/Δx^2} Δx which, if you took calculus, parts of that might look familiar, especially if you take the limit as Δx approaches 0 like you do when you make an integral. When you take this limit (I'm simplifying Δx^2/Δx^2 to 1 here because why would I leave it?), calculus notation turns all instances of Δx into dx. At this point, it simplifies down to √{[df(x)/dx]^2 + 1} dx, where df(x) represents the difference between f(x + dx) and f(x); If you aren't familiar, or didn't recognize it for whatever reason, df/dx is one of the available notations to indicate a derivative. I always see the arc length formula rearranged so the 1 is in front, so our integral, the arc length formula, is ∫√{1 + [df(x)/dx]^2} dx That's the integral with respect to x of the square root of the sum of 1 and the square of the function's derivative with respect to x. The bounds of the integral are your start and endpoints on the arc.

[@levistepanian5341]

I just had a thought, 2 can be factored into (1-i)(1+i)

[@ariadame102]

When i watch your videos, i always feel frustrated since all i see is you solving a problem. I think you are too capable, as a mathematician, to just solve problems alone. A physicist, chemist or anybody can do it, but it's up to a mathematician to find patterns and relationships and identifying meaning in solving the problems in the first place. I think solving a problem being a supplement to delivering a bigger lesson is much more worthwhile than being on its own. For example, you could expand the problem to root n of unity, and explain what root of unity really means, then you can relate them being on the unit circle centered at origin of complex plane to taking root of root of unity or something like that. I see a lot of unrealized potential here.

[@blackpenredpen]

Aria Dame I see what you mean. I should have leave a question or something at the end of the video to let the viewers think more. That's a great idea! I actually have something planned for my new videos but it's also good to do a "tease". Thanks for sharing ur ideas.

[@madhuranjandev7849]

Your writing style with marker is awesome

[@SensSword]

Fun video. Liked switching up the colors as you write. Reminds me of how much I rely on syntax highlighting to code. It's artificial synesthesia.

[@reesetyra902]

I like the 2nd method better. If my intuition is going in the right direction, it seems like solving for x^n = 1 would involve chopping the unit circle up into n equal pieces which is a pretty solution.

[@awfuldynne]

Your intuition is right. x^n = 1 has n distinct solutions of the form e^(iτk/n), where k is an integer and τ is equal to 2π by definition-it represents one full turn around the circle, measured in radians. Take any of those solutions to the nth power, you travel k times around the unit circle and land back at 1-e^ix is cyclical, with a period of τ Also confirming what you suspected, since the solutions are integer powers of e^(iτ/n) (the k=1 solution) from 0 to n-1, they're equally spaced around the unit circle because of the way multiplying complex numbers works-the final result is equivalent to adding the angles and multiplying the lengths. Here the lengths are all 1 and the angles are all integer multiples of a wedge spanning 1/n of the unit circle.

[@awfuldynne]

I just pictured multiplying the corners of a square-with the sides vertical instead of the first corner at (1, 0) The right corners multiply to +1 and the left corners multiply to -1, so the product of all four corners is -1. Generalizing to an n-gon, I notice you can only rotate the shape τ/n before it repeats the starting conditions, and the surplus angle each vertex contributes to the product is the same as the position of the first angle, so the product of n points evenly spaced around the unit circle is the same answer as "what's the first point to the nth power?". Alternatively, ask how far the first point is around the circle in terms of how far the points are from each other; since the distance between points is 1/n the distance around the circle, the ratio does our multiplying.

[@n0ita]

I discover something ( at least for me ), if you integrate a circle in the complex plane, you get 0. That maybe should be obvious for someone, but for me, i think it is pretty cool. o god, i don't want to be wrong.

[@gian2kk]

Did you also discover that sinx integrated from 0 to 2pi is zero too? That would be amazing

[@andywright8803]

r0ad288kma8x You are perfectly correct as long as you circle is centred on the x axis

[@georgepennington908]

Is it true that integrating a complex function, like the function which would produce a circle centred at 0, gives the centre of "mass" of the function, so integrating the circle gives 0 because its centre of "mass" is at 0

[@JJ_TheGreat]

1:58 What if instead of using the quadratic formula, you subtracted the +1 from both sides of the equation, so you followed with: x²+x+1=0 x²+x=-1 x(x+1)=-1 Therefore: x=-1 or x+1=-1 So: x=-1 or x=-1-1=-2 ?

[@lumina_]

Let's say we have the equation x² - x + 2 = 0. We can factor the left-hand-side: (x-2)(x+1) = 0 and then we can set each factor equal to zero. But why are we allowed to do that? Well, it's because of the zero product property: any number times zero is zero. So that means if we know that one of the factors is equal to zero, the entire expression is equal to zero. The value of the other factor doesn't matter because it'll be multiplied by zero. Now let's say that we have (x-2)(x+1) = 1. Are we allowed to set each factor equal to 1 to get our solutions? Well, just because one factor is equal to 1 doesn't necessarily mean that the entire left-hand-side is equal to 1. This is why we're not allowed to find solutions this way So to conclude, we can find solutions to a quadratic by setting each factor equal to zero ONLY when we have 0 on one side. This is because we know that x * 0 = 0 for all x, so the value of the other factor doesn't matter (because it's being multiplied by zero)

[@imsounak19]

I must honestly admit that your and mind your decisions' videos make my day

[@r0m4an21]

I don't know if I'm happy because I've discovered it by myself or sad because I found it was already a thing....

[@WildStar2002]

Be happy! 😁 It doesn't matter that you weren't first, you *still* discovered it on your own!

[@davidwright8432]

Thanks! It's good to see both ways - to show you can approach the same question - what are the cube roots of unity? - by two very different paths that don't even look related - and still get the same answer! (Relief!) And there will bee deeper reasons why this is so - which maybe you can investigate some time?

[@dheerendranagaria1032]

Why not use Euler's identity to get: Cos 2pi/3 + i sin 2pi/3 = e^(i 2pi/3) = (e^(i pi))^(2/3) = (-1)^(2/3). Similarly the last answers would be (-1)^(4/3). I know they would essentially become 1^(1/3). But would that not be enough for the viewers to get a surprise?

[@MrRyanroberson1]

all the nth roots of 1 are cos(tau* k/n)+i sin(tau* k/n) for k=[0...n-1] the algebra solution is a bit cooler though, since x^n -1 can be factored into (x-1)*(sum of x^k from k=0 to n-1)

[@ashishanand9629]

I think he is best teacher, love from India

[@cipherunity]

I never thought of the 2nd method of polar form to find the cube roots of unity. Thanks for that. I just want to add some thing interesting here. As you have shown that the cube roots of unity are: 1 & -1/2 ± i √3/2 Let W = -1/2 + i √3/2 then W^2 = ¼ -3/4 - i √3/2 = -1/2 - i √3/2 which is the other cube root of unity. Therefore the cube roots of unity are 1, W & W^2 Also 1 + W + W^2 = 0

[@DjVortex-w]

I have been wondering how hard would this hypothesis be to prove: "The nth root of any integer is either an integer or irrational" (for any positive integer n.) (This is, in a way, a generalization of "the square root of 2 is irrational.")

[@authenticknowledge9349]

Teacher: ANYWAY!!! Me: Leaves online class...

[@Magic_Unhinged]

Just a reminder because this fact goes unnoticed- The method used to solve the quadratic formula here(2:05) is called Shridharacharya Formula. Yep, He's Indian & this method is quite substantial for solving Quadratic equations, yet very few people know it...

[@WerewolfLord]

DEGREES!?!?!?!?!?😜

[@kidistzewdu9875]

Thank you Mr

[@collinskanyensha8169]

I really enjoy your tutorials and I need you to present also on this.find zeros z^3-1/ z^3+1

[@chengshengway]

i thought for the complex method you would use re^iθ.

[@ZoroarkLover98]

I did this in high school, and my calculations were about three times longer than these :(

[@sudhirdwivedi9319]

how to integrate 1/ln(x) ??

[@sahu6675]

Special integration lix + C

[@sahu6675]

Or Integrate I = (1/lnx).1 by parts method You will find a quadratic equation in terms of 'I'. Then solve for I to get the answers Which is pretty hard So i'd stick with li(x) + C lol

[@sudhirdwivedi9319]

your explanation is toatally wrong ,,because if we assume I =integral of [1/lnx] then the value of the integral of [1/lnx]^2 can never be I^2 .

[@dropdatabase8224]

I use sign language to remember which way round sin and cos are. "Why sign? 'Cos of X." For X,Y situation "I Sign, 'cos it's real" For complex situation.

[@carlocirillo8708]

7:22 i don't understand, why is r = 8 if it said omega = 8? shouldn't it be 8? what am i missing

[@brian554xx]

Now do a similar thing using quaternions!

[@Christian_Martel]

The infinite roots of unity are a circle. How poetic is that.

[@RigoVids]

My completely random guess would be something like the second answer is the point on the imaginary unit circle that is 1/3 of the way around, so -1/2 + (sqrt3)/2 i

[@dosu5064]

Do this has also p-adic roots

[@LightChu2.7183]

Quick questions: 1. Why w^3 = 1 could be a circle in the complex polar form? 2. Are the other points on the circle also the solution? For example, 90 degrees.

[@mikef5951]

The reason for the angles makes a bit more sense if you do out the problem in polar form. So in polar form, you have the magnitude, and the angle and it's written as something like 2∠90. Where 2 is the magnitude and 90 is the angle on the complex plane (in degrees) in rectangular form this would be written as 2i Some important context, the way multiplying polar coordinates works, is you multiply the magnitudes together, and then **add** the angles. Division similarly, you divide the magnitudes, and then **subtract** the angles. Extending on this, to take the root of a polar number, you take the real root of the magnitude, and then divide the angle by whatever the root is. So we want to solve for x^3 = 1. Well first let's write it in polar form. x^3 = 1∠0, straight out to the right side of the circle. If you were to go all the way around the circle 360 degrees you end up back at the same point, so we can say 1∠0 = 1∠360 (and 1∠720). If we then cube root each of these, we get ____ 3√1∠0 = 1∠0 ____ √1∠360 = 1∠120 ____ and 3√1∠720 = 1∠240 Which are the final answers, if you convert those back to rectangular they'll be the same as in the video. You could keep going with the earlier step beyond 1∠720 and get things like 1∠1080 or 1∠1440. However when you cube root these, you'll end up with 1∠360 and 1∠480 1∠360 is equal to 1∠0 which is already an answer so is a duplicate, and 1∠480 is equal to 1∠120 and so is another duplicate - and this pattern will go on forever. Why can't 90 degrees be a solution? Well if we work backward from 1∠90 and cube it, we get 1∠270 which in rectangular form is -i. This also means that if you cube any positive purely imaginary number, you'll always get a negative purely imaginary number. To drill this idea home, this can also be used to find the other 2 cube roots of -i -i = 1∠270. 1∠270 = 1∠630 = 1∠990 3√1∠270 = 1∠90 = i 3√1∠630 = 1∠210 = -0.866 -0.5i 3√1∠990 = 1∠330 = 0.885 - 0.5i

[@LightChu2.7183]

@@mikef5951 Understand! Thanks for your wonderful comment. Hope you are doing well!

[@Magic73805]

Sir, You are looking so smart in this dress. 😍😍👍👍👌👌👌👌👌👌👌👌👌👌👌👌👌👌

[@tsadiamabel]

Maths is beautiful. I love your explanations. Thanks for saving me this morning.

[@nahilachowdhury1785]

I've fallen in love with this man

[@MrSeezero]

1 @ 120 degrees cubed is 1 @ (120 + 120 + 120) or 1 @ 360 or 1.

[@gonzalotapia1250]

This means that the quartic roots of 1 are 1, - 1, i and -i?

[@AizenSosukesama]

Theres more

[@desrepeerc206]

Can we solve this by doing (a + bi)^3 = 1, expanding this to a^3 + a^2 bi - ab - b^3i = 1 + 0i and equating the real and imaginary parts? We get a^3 - ab - 1 = 0 and a^2b - b^3 = 0. Not sure how to proceed.

[@seja098]

isnt the square root of -3 the same as 3*i? I saw you wrote i*3aquared

[@FredDeSouza409]

The square root of -3 is i*√3.