A Complex System of Equations | Putnam & Beyond
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- čas přidán 20. 05. 2020
- This question is from the book "Putnam and Beyond" Q156: amzn.to/3e1RomQ
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math for fun
Make the chain chomp teach us the chain rule writing with his chain
U mean Chen Lu ryt?
@@akshataggarwal4002 yeah I'm confused about that naming convention
ItzARubix hiiiii
i solved it in my head geometrically! I know for numbers on unit circle:
- adding them ---> is just like adding vectors tip to tail
- multiplying them ---> add their angles, stay on unit circle
so for x + y + z:
- you go one unit up, one unit down, and one unit right to get to one (i, -i, 1)
and for xyz:
- you go 90 clockwise, 90 counterclockwise, and then 0 degrees so you are at one (i, -i, 1)
it works!!
I did it the exact same way, I was trying to find this four sided shape and suddenly a square popped in my head, it was so obvious
I came to the same solution, but I still watched the video for the more thorough, algebraic proof. It's a pretty neat question
Fidi the same too trig form is faster
Or in terms of geometry of mass
Did the same, it was quite simple once you know that the imaginary parts need to cancel out to get the sum equal to 1 meaning the conjugate is there too. Then you get the hint from the product that their angles should add up to 0 as cis0 = 1 and finally they form a right triangle so if you set x,y,z as fixed solutions their permutations should also be solutions, so 3! gets the 6 of them for (1,i,-i).
|x| = |y| = |z| = 1 --> They are all on the unit circle.
xyz = 1 --> Their angles sum up to 360.
x + y + z = 1 --> Their imaginary parts vanish when you add them up, or two of them are complex conjugate.
The solution is pretty much guessable.
One can easily guess the set of solution as {1, -i, +i}.
Lol this is exactly what I came up with too! Too bad it's not only asking for just 1 solution. xD
That's how I got to the same answer pretty much within a minute too
Yep, I really expected the proof to go down a geometric path.
The full set would be all combinations of those three numbers, so there should be 3! solutions.
Its not satisfying when you guess though
I was guessing the third roots of unity!!
Surprised to see it was the forth roots of unity excluding -1!! Very exciting!
Thank you BPRP!
You're very welcome!
Actually, equation 1 can be written as:
x+y+z-1=0
There you have the four forth-roots of one. :-)
@@sergiokorochinsky49 Bravo!!
@@sergiokorochinsky49 yeah was about to say the same thing, if they were cube roots of 1, then their sum would have been 0, not 1 😊
I used another way : just look and guess the obvious solution. Not really rigorous, isn't it 🤭. And no proof it was the only solution.
I see blackpenredpen remained on stage 5 in Desert Land in "Super Mario Bros. 3" until the Chain Chomp broke free and followed him home.
loll
Too funny!!!
your enthusiasm is just so infective! Watching your videos just tops up my love for maths from time to time- thank you! :)
Like = love for maths = love for BPRP
Dislike = hate for maths = hate being killed by Chomps in Super Mario 64
There’s a nice geometric solution on the unit circle interpreting x+y+z=1 as the statement that you have a triangle on the unit circle with centroid at (1/3,0)!
The vector geometry perspective with Euler's formula provides quick intuition for solving a problem like this.
I always have a go before looking at how you did it and it's surprising how often you and I come up with different solutions. In this case the second condition says x = exp(i a), y = exp(i b), z = exp (i c) and the last says that a + b + c = 2 pi (or some other multiple). Then you can put that into the first condition and equate real and imaginary parts to get cos(a) + cos(b) + cos(a+b) =1 (eqn 1) and sin(a) + sin(b) = sin(a+b) (eqn 2). Using sin(a) + sin(b) = 2 sin( (a+b)/2 ) cos( (a-b)/2 ) and sin(a+b) = 2 sin( (a+b)/2 ) cos( (a+b)/2 ) in 2, you get either sin( (a+b)/2) =0 (which doesn't lead to a solution) or cos( (a+b)/2 ) = cos( (a-b)/2 ) which is true if a or b =0. Wlog, take b=0 and go back to eqn (1) to get 2 cos(a) = 0 or a = pi/2. By the condition a+b+c = 2 pi, then c =3 pi/2. So the solutions are {exp(0), exp(pi/2), exp(3 pi/2)} = {1, i , -i}.
Wow! This is very cool!!
Helps in my IIT-JEE prep
So true, i was studying complex numbers forJee 2020 and this video showed up on CZcams, so cool 😎
Lmaooo Indian tests are THAT difficult
You can think about x+y+z-1=0 as vectors and since all of them has length of 1 it make a rhombus where the sides are additive inverse 2 by 2
Now we can say that one of our variables has to be equal to 1 (the opposite side to the -1) and the others has to be x and -x.
Put them in the xyz=1 and you get a system of 2 equations (b^2-a^2=1) and (ab=0) where x=a+bi
After solve it you get the final answer (1, i, -i) and its permutations
I solved it pretty much the same way you did. Make it into a four vector problem and use symmetry to our advantage.
@@bsmith6276 Yeah it was so funny to think it, I love the problems that looks like hard or complex and u solve it like if you was 10 yo xD
This man deserves more subs
I had guessed those solutions right away just looking at it, but yours is much better to prove that those are the only solutions.
I felt good remembering this and being able to solve it easily
Just loved how you did that without mentioning Vieta!!!. Showing that conjugate z equals 1/z when modulus z=1 saved the day. Your demonstration also explains beautifully why the signs of ESPs alternate.
Also to avoid so many symbol clashes, maybe (w-x)(w-y)(w-z)=0 Letting w be the "indeterminate" makes it crystal clear imho. Thanks for what you do! you are the savior for so many struggling math students around the world.
This was awesome!
I solved this using polar form, you can show that the arguments of the exponentials have to sum to zero (or some multiple of 2*pi) for the product xyz=1 to hold true and you can also show that because all x,y,z are unit length that one of the complex numbers has to be 1. That gets you a solution of exp(i*pi/2)=i, exp(-i*pi/2)=-i, exp(i*2pi)=1 and you can prove it’s the only possible solution set by requiring the angles sum to zero or some multiple of 2*pi, which is needed to be consistent with xyz=1.
Oh man. I feel so big brain after watching this man's videos. Huge respecc
I came up with an answer pretty quickly, but I would not have known how to show it algebraically. Good one!
Your mics are so cool!!!!
Thanks!!
A very good exercise in the morning. Mental exercise, thanks. I love the smile of your pet. Hahaha
Before the vid, looking at just the thumbnail I saw a solution wherr
X=1
Y=i
Z=-i
Edit:
After the vid, it seems like I did have an answer, but not all off them
Im only 11th grade and dont know if my solution is mathematically right or valid, but I solved it this way:
1)
Because of |x|=|y|=|z|=1 they lie on the unit circle
2)
Write x, y, z as e^(ia), e^(ib), e^(ic) with a, b, c in [0, 2pi[
3)
Because of xyz=1, a+b+c=2k(pi)
4)
x+y+z=1 is like having a chain of 3 arrows (vectors) with the same length of 1 (because of |...|) in the complex plane. The "chain" starts at 0 and ends at 1. The distance between these two points is also one, so the vectors together with the distance have to form a rhombus. A rhombus is only possible if two opposite sites are parallel, so the second vector has to be parallel to the real axis so the first number, let it be x, is 1. That is because one is the only number in the unit circle which, seen as a vector, is parallel to the real axis.
Cant draw a sketch at this point sadly.
5)
The other two sites of the rhombus have to be parallel too, so (wlog) c = b + pi
6)
x = 1, so a =0
7)
Solving the equations in 3) and 5) with a =0 (7) you get the three complex numbers
1
i
-i
They can be ordered in any way.
Yes that's fine reasoning. Put differently, once you have x=1 you have yz=1 too. And with y+z=0, you get to the same answer.
Brilliant
This is beautiful! Keep on exploring math- one day, you may do great things.
Im in 7th grade but I’m taking AP Calculus BC and that looks right
Brilliant!
Love ur videos
Excellent ! Couldnt have the intuition about cubic equation ! Now i must watch the other cases where you used the same idea
LOLOL! A Chain Chomp?!? That's the funniest microphone-substitute yet!
I guess the funniest part about it is that, throughout the WHOLE video, he's holding it like it is his microphone, though his REAL microphone is clipped to his shirt. So the Chain Chomp is COMPLETELY unnecessary (except, I guess, to terrorize him if he got the answer wrong)!
alkankondo89 lol. I tried to put the mic on the chain chomp but it didn’t work so well.
WONDERFUL
Hi bprp can you please make a video on visualising complex roots of a cubic equation graphically??
Thanks
Thanks..its gonna help in iit jee prep
Big plus for the chain chomp :D
I found the solution 1, ±i intuitively and used a geometrical method to prove it was the only solution (up to permutations). As x+y+z+(-1)=0 the numbers x, y, z, -1 as vectors in the complex plane add up to 0 and so form a quadrilateral with sides of length 1. So it is a rhombus and one of the vectors is 1 (side opposite -1). The other two are simultaneously opposites (they add up to zero) and conjugates (their product is 1), so their real part is zero and so they are ±i. Solved.
Wow, this is so nice!!!
I really like this solution. Very well done!
It is interesting how the word "Note" on the right hand whiteboard is right in the right spot as to be also reflecting the light from the screen. Makes it hightlighted.
LoL, I’ve seen your videos for a long time ago and this is the first time that I realize you look so closely to Gendo Ikari from Evangelion. Awesome videos btw. :D
First I figured 1 for x, and two complex rotations of 1 for y and z that cancel each other out. (x, y and z can be in any combination.) Then it becomes obvious that only i and -i work for the two necessarily complex values for the sum and the product to each cancel. In angular terms this is e^{0,(π/2)i,(-π/2)i} interchangeable.
Love from philippines!
Solved this math geometrically. It was way more fun and easy.
This video is very cool. Technically, I solved it by mere guessing at first, because I kind of know how the system of equation works. But, this explanation is really what I need.
Come for the microphone, stay for the maths.
Yesterday some mental trigonometric limit integral to solve. Didn't figure it out.
Today, inspect for ten seconds. Solved!
Dammit, I solved in my head for x+y+z = 0, not 1.
If that’s the case, what’s the answer?
@@blackpenredpen the three cube roots of unity.
Think of x+y+z = b
And vary the value of b.
Then think how changing the values of b changes solution geometrically. On the C plane. If b = 0 they are cube roots of unity, if b = 1 then {1,i,-i}, if b = 2 then its { 1, e^(iπ/3), e^(-iπ/3) }. Thus from pure geometrical interpretation, we can conclude that for x,y,z to be distinct with the condition |x|=|y|=|z|=1 b < 3
Same on the other but we know that xyz = 1 so in the end their arguments should add upto zero.
Therefore, b ε (-1,3)
6:45 ...multiply eq3 by x
=> y=-z
Replace in eq1
=> x=1
Replace both in eq2
=> -z^2=1 => z=i => y=-i
6:09 From Vieta formulas x,y,z are roots of polynomial
u^3-u^2+u-1
and this polynomial is easy to factor
I love complex numbers!
There was this same question in the Australian NSW HSC (SAT/GCSE Equivalent in Australia) for the 2022 Mathematics Extension 2 Examination, watching this video would have granted the student full marks for the question (It was the last question also)
I solved it like this: Notice that x,y,z are on the unit circle. (2) tells us that the sum of their angles adds up to 0. (1)Tells us the complex parts cancel out and result in one. WLOG set x=1, and y=z* (conjugate). We want to find a conjugate pair (y,z) on the unit circle that cancels out and leaves 0 for the real part and it's easy to see i.-i fit the bill. Now to show that this along with all its permutations are the only solutions, we have to do some trig to show that one of them must be 1
Yeah draw the vectors on the unit circle. Adding them head to tail it's fairly obvious the solution is like x,1,-x. The product condition is -x^2=1 so only i,1,-i (or permutations) will work.
I just tried out x=1. That lead to y and z being opposite points on the unit circle because y+z=0 and the only opposite points on the unit circle that have a product of 1 are -i and i. Since I didn't have any paper to write stuff down on in reach, I was happy with this one set of 6 solutions and didn't try other values for x. But apparently there aren't any other solutions, cool!
In Link's Awakening, you can take a chain chomp for a walk and he happily eats enemies for you.
Chain Chomp is a good boi
Yea he is! Or in super Mario odyssey, you can become the chain chomp and hit enemies!
Hey Blackpenredpen,would you like to make a video about integrating with a floor function tho? I think it's gonna be interesting! Big Thanks!!
Just by looking at it: all 3 variables are on the unit circle. and for Eqn (1) to be true the complex parts of two of the variables must be conjugates, and the third variable must be real. So 1+0i, 0+i, 0-i. XYZ=1 isn’t needed to solve for X,Y,Z. but it’s true.
You should start a playlist for jee advanced questions
Wow very smart to make the polynomial appears
Chain chomp says all permutations of (1,-i, i)
Funny, I came up with the solution in my head quite easily: x+y+z=1 means you have three plane vectors of length 1 which together take you from origo to 1 so for example 1 step up, another right and finally one step down and luckily these steps multiplied together give 1 (1 * i * -i = 1). However, I wasn't able to solve this on paper.
1, A complex number is a vector. 2, if |x|=1, we call x is a unit vector. So x, y and z are all unit vectors. If x+y+z=1, according to the parallelogram law of vector addition, there should be either a 2π/3 angel or a π/2 angle. All the other angels can not cancel the imaginary parts by addting together. If the angle between x and y+z is 2π/3, let's say x=1/2+√ 3i/2, then y+z must be 1/2-√ 3i/2, again, there should be another 2π/3 angle between y and z, so let's say y=1 and z =-1/2-√ 3i/2, but this solution does not fit xyz=1, so the angle should not be 2π/3, so the only remain possibility is π/2. So let's set x=1, and y and z have π/2 with x, so y and z is one i and one -i. The position of x,y and z in this equation are all the same, so just replace x to y and to z, we got the other solutions.
BlackPenRedPen do you have a website or something?
I really miss math...
I solved the problem in my head, noting in passing that the i, -i, and 1 could be swapped between variables. Getting one answer, it would seem that the others naturally follow by cyclic permutation.
I used polar form and got a trignometric equation; where sum of the arguments was 0, sum of their sines was 0 and sum of their coses was 1. Kinda long but interesting :)
I didn't know that the absolute value can be extended to complex numbers or at least I'm not familiar with this idea excellent video 👌
For complex numbers it means the distance from the origin.
I solved it by inspection, but then was really unsure if it was the only solution, I was trying to use polar form and ended up with:
sinA+sinB=sin(A+B)
cosA+cosB=1-cos(A+B) where A=arg(x) and B=arg(y) (the third one will be defined by the other two)
A+B+C=0 C=arg(z)
I didn't got anywhere but the plot looks cool
That’s a cool chen chomp I wonder if it’s differentiable
I have a somewhat interesting integral for you: Integral{0, 1} [dx/sqrt(ln(1/x))]. Has quite an interesting solution imo.
Let y = sqrt[ln(1/x)], hence x = e^(-y^2), (0, 1) -> (♾, 0), and dx = -2y·e^(-y^2)dy, such that the integral you propose is equal to Integral{0, ♾}[2y·e^(-y^2)/y·dy], which is equal to Integral{-♾, ♾}[e^(-y^2)·dy] because the integrand is even. This is equal to Γ(1/2) = sqrt(π).
impressive
隙なく解こうとすると結構難しいんですね
はー 地力がないとだめだなーー
First time getting a solution in my head which makes me think there's more
1,i,-i any permutation thereof
This made me feel really stupid. I saw the thumbnail and tried to solve it before watching the video.
Since the magnitude of all three numbers is 1, I decided to work it out with trig. I wrote polar forms for the numbers (ie x = cos(X) + isin(X)) and rewrote the 2 equations as:
1=cos(X)+cos(Y)+cos(Z)
0=sin(X)+sin(Y)+sin(Z)
2pi = X+Y+Z
By the end of this process I managed to distract myself from the simple answer for almost an hour.
I finally solved it when I just thought, "What if you set x = 1? The other two would have to be complex conjugates, and when you add them, they need to cancel out entirely, not just the imaginary value, so they would need to be the negatives of each other."
A great puzzle: Simply stated, looks difficult, simple solution.
I though x, y and z were real
How innocent I was
If they are real the system is impossible....
@@mattethebest1 YESSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS
Nice mike bprp 👍
First, I tried out real solutions, but it does not work… then, I thought of conjugate pairs. Tat that moment, I believe that there is one real and two conjugate complex among x, y, z. So I tried x=+1/-1
Great
Consider x+y+z+w=0 where x,y,z,w all live in the unit circle.
I claim the solutions (up to permutation obviously) are z= -x and w=-y.
May assume wlog that the angle between x and y is
老師，我在點開影片之前，我覺得可以用畫圖的
因為老師有說到複數，所以根據第二式，x,y,z都在圓心為原點，半徑為1的圓上
根據第一式，一定有兩個為共軛複數，然後再用湊答案的方式
只是這樣比較沒有計算基礎，還在思考
獻醜了^_^
李祥數學,堪稱一絕
李老師好！ 好多人都直接想到集合的方法呢 我一般都喜歡直接計算 這就是為什麼我很少有幾何方面的影片 哈哈哈
還有 我建議一般書 叫做Putnam and beyond. 最近剛開始看 裡面有很多厲害題 你可以去找找來看
@@blackpenredpen 謝謝老師，我會去找書來看的^_^
i solve it by noticing that while xyz = 1, then at least 1 of them is real, and as all the numbers' abs val is 1 so they all ly on the unit circle, so you just put 1 or -1 into 1 of xyz, and rejecting -1 because if -1 is in the solutions, then the variables will not all have an abs val of 1, so sub x=1 then you get y+z =0,yz=1, go ahead and solve them and you get the ans set (1,i,-i). I think this is a fast and viable way to solve it.
0:46 Chain Chomp boyo!
The man who does not talk too much
Oh that's clever. I found the solution through guess-and-check but I couldn't come up with an argument that it was unique
Oh yeah
First instinct is that there are 5 equations here and 6 variables, so there's probably going to be more than one solution. Equations are symmetric so I expect each solution to generate up to 6 (permutations of size 3) solution, assuming x =/= y =/= z. x,y,z are all on the unit circle and their angles sum to 2pi, sum of imaginary parts vanishes and sum of real parts is 1. Easiest way to get that is (1,i,-i). Question is, are there any solutions not permutations of this? I don't think so but it's not immediately obvious.
How many poki balls are there at you ?
Wait a minute, Are you from "Team Rocket"..
What is the best math analysis book for u
Your "friend" can go back to the universe of Super Mario.
Thanks Michael
Super easy. x = 1, then y and z are 120 and 240 degrees around complex plain with radius = 1, and when multiplied will be r^2 * ( angle1 + angle2 ) = 1* (120 + 240) = 1* 360 = 1. So xyz = 1, their radius is 1 so all magnitudes |x| = |y| = |z| = 1, and since y and z are 120, 240 with same radius they make a conjugate pair and thus y + z = 0, and x + y + z = 1 + 0 = 1. Done.
argh, messed up x + y + z in my system = 0, not 1. The solution is y , z = +/- i respectively. My bad.
In my defense I did that in my head and not on paper.
Where did you buy your mic? ;) I love Chomp plushies!
Amazon!
@@blackpenredpen wow imma grab one once the traffic gets back up!
yes! more complex numbers! ^u^
You can see this geometrically. The constraints say x, y and z lie on the unit circle, that the sum of their real parts is 1 and of their imaginary parts 0, and that their arguments sum to a multiple of 2 pi.
12:16 it:I’m hungry...
why he has that ball? btw great videos!
Good proof. But I saw the answer pretty well straight away. It had to be symmetric about the real axis, and have one solution only on the positive real side (else xyz is negative). All 3 -ve won't work. So we start with X=1 and the rest follows.
Nice chain chomp!
Woof woof!
Nice Chain Chomp.
The chain chomp is going to eat you blackpenredpen, be careful
Imagine if the 🎾 just said “ but I’ll still eat u “
I didn't watch the video, but I think x + y + z is the location of the orthocenter for a triangle with vertices at x,y,z, in the complex plane. So if the orthocenter is on the circumcircle, then it must be a right angled triangle. I found the solution 1,-i,i from there. If it's right angled, and we have both xyz=1 and x+y+z = 1, then maybe that's the only solution. #Olympiad geo info.
Ah, to finish, you can note that if x,y,z, form a right triangle, then WLOG, x and z are diametrically opposite each other on the unit circle, so x + z = 0. Then, since x + y + z = 1, we have y = 1. Then, since arg(x) + arg(y) + arg(z) = 0, and arg(y) = 0, we have that arg(x) = -arg(z). So 1,i,-i and all permutations are the only solutions.
Btw bprp have u noticed that
Integral of best friend (1/(1+x)) = ln(best friend) +C
Integral of best friend of best friend (1/(1-x)) also = ln(best friend of best friend) + C
Please pin be bprp.
I am an 8th grader from India and totally love your videos
Could you please do an integral including partial fractions sir
I would be really thankful.
Good observation!