# A Delightful Solution (77) - Manipulating Polynomials and Roots of Unity

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**Published on Jan 31, 2019**- We obliterate the problem using the art of clever manipulations - what all mathematicians enjoy.

Congratulations to Rishav Gupta, Kartik Sharma, Xu Chen Tan, Francesco Debortoli, Simon Armstrong, and The vast universe for successfully solving this math challenge question! Rishav Gupta was the first person to solve the question.

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Nasir Khan2 months agoThis just brilliant. Your videos always fascinated me and helped me to develop my ability to put on logical exertions in such a beautifully constructed problem.

SamRaxh' o_O4 months agoHow can 2020-(1010×2019) equal -170!?

Aarya Chaumal4 months agoHow do I reach to u

Like a message

Aarya Chaumal4 months ago@LetsSolveMathProblems f(x)=sin(sin(sin...70 sines... (x)

Find seventh derivative of f(x) at x=0

😋

Also if possible give a representation for n nested sines of such a function

All the best🤗

LetsSolveMathProblems4 months agoPlease feel free to share the problem here on the comment section. =)

Aarya Chaumal4 months agoI have a nice que I would like u to make a video on😁

Pls

Miyuki Umeki4 months agohow can I solve these insane questions by myself?

SeongKyun Jung4 months agoI failed this time, got to the polynomials but didn't think of comparing with x^2019+x^2018+...+1. I'll solve it next time tho! Great problem..

max power4 months ago^{+1}what tablet / input device do you use ?

Rahul Patil4 months agoSir can u plz guide me how to integrate e^sinx

Zelong Xiong4 months agoRahul Patil 😂 glad to help u

Rahul Patil4 months ago@Zelong Xiong hoa

Rahul Patil4 months agoHow

Zelong Xiong4 months agoRahul Patil chain rule?

Debopam Sil4 months ago^{+2}Oh goodness. I thought that omega was the complex cube root of 1 while solving the question. Just as I clicked on the video to see if my solution was correct or not, I discovered to my horror that omega is not what I thought. However, for those interested, my answer came out to be 1.

liam4 months ago^{+1}What software did u use to make this video?

Ben Burdick4 months agoYeah, this one went completely over my head...

LetsSolveMathProblems4 months agoIf there are any specific parts of the explanation you wish me to further elaborate on, please feel free to comment it here. I would love to help you fill in the missing pieces. =)

William Debdoot4 months ago^{+2}Sir, please solve some [JEE Advanced] mathematics problems (Indian Institute of Technology entrance exam).

G Dunken4 months agoYes, we are done.

SUPRATIM SANTRA4 months ago^{+1}Sir @ beautiful... Where can i get such beautiful collection of problem?

fengsheng Qin4 months ago^{+5}I can't understand the "P(x)" rule ,why it equal to X^2020 -1 ? like P(x)=(x-1)(x-2)(x-4)=x*3-7x^2+14x-8 not x^3-1 ,pls explain why ,thanks!

PDV Cubing2 months agoThere’s a theorem in algebra where, if a polynomial p(x) has leading coefficient a and roots r1, r2, r3, ..., rN, then p(x) can be expressed as:

p(x) = a(x-r1)(x-r2)•••(x-rN)

The nth roots of unity are all the complex numbers that, when raised to the nth power, are equal to 1. Algebraically, they’re the solutions to the equation:

x^n = 1

By rearranging this equation, we can clearly see that they are also the roots of the polynomial

p(x) = x^n - 1

When n = 2020,

p(x) = x^2020 - 1

1 is a trivial solution, and all the complex solutions are powers of ω = e^(2iπ/2020), which can be written in the form ω^k, k = 1,2,3,...,2019. Using the rule from the first paragraph, we can rewrite x^2020 - 1 in terms of its roots,

p(x) = (x-1)(x-ω)(x-ω^2)•••(x-ω^2019)

This is sort of simplified so please let me

know if I got anything wrong!

antonofka90184 months ago@LetsSolveMathProblems I don't understand what you did at 8:25

antonofka90184 months agoWatch this video usclip.net/video/8GPy_UMV-08/video.html

LetsSolveMathProblems4 months ago^{+4}I believe you may not have learned about the roots of unity, which is a prerequisite to thoroughly understanding this video and, in fact, the problem. Here goes a brief introduction.

It can be proven that, given any complex number z and positive integer n, there are exactly n different complex numbers such that each is an nth root of z. That is, there are 3 complex cube roots of 1; there are 2 complex square roots of 2019+2020i; etc.

The complex numbers that are roots of 1 (the unity) are named roots of unity. It can be proven that nth roots of unity are 1, w, w^2, ... , w^(n-1) where w = e^(2pi*i/n). In our case, n = 2020.

Consider a polynomial x^3-1. The zeroes of this polynomial are complex numbers such that x^3-1 = 0, or x^3 = 1, that is, the cube roots of unity. Thus, x^3-1 = (x-1)(x-w)(x-w^2), where w = e^(2pi*i/3). Note that 2 and 4 are NOT roots of unity, so the example you provided does not work.

I hope this helps. If you wish me to further elaborate on some of the above points or provide you with resources for learning more about roots of unity, I am willing to. =)

Omar4 months agoCorrect. and as you see it was given:

ω=e^(2πi/2020) which corresponds with the solution of the equaution x^2020=1⇒ x=e^(2kπi/2020) (in polar form) ∀k 0≤k≤2019 which implies

x=(e^(πi/1010))^k which is equivalent to (as you see) ω,ω²,ω³...,ω^2019.

UbuntuLinux4 months ago^{+10}I understand

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nothing...

LetsSolveMathProblems4 months ago^{+2}If there are any specific parts of the explanation you wish me to further elaborate on, please feel free to comment it here. I would love to help you fill in the missing pieces. =)

Supanat Sukpiriyakul4 months ago^{+1}What is that big pi

Johannes H4 months ago^{+1}If you watch the video, you see it

Zυвi4 months ago^{+2}Same as sigma notation but you multiply instead of adding

Joe Potillor4 months ago^{+2}Product notation

nizar aberqi4 months agoGreat