The expression | z1+2 z2+3 z3| is not invariant under the rotation of z1->z2->z3, but the preconditions are. So there are actually 3 different solutions, corresponding to z1=z2=i z3, z2=z3=i z1, and z3=z1=i z2. They are respectively 3 sqrt(2), sqrt(26), and 2 sqrt(5).
Came to post this and found your comment here already. It's a curious and terrible blunder by Michael to completely miss the asymmetry of the final question.
@@zunaidparker Yes, it was Michael's mistake, but I find comment section to be more interesting because of these not deliberate mistakes. Michael is producing a lot of material and many times sacrifices rigorosity for output. This concept helps to grow his channel, since discussion and involvement bring people back.
@@mguzjebesku2591 agreed. Usually his mistakes are things like transcription errors, but in this case overlooking an entire branch of solutions is unlike mistakes he's made in the past.
If you write the product of lambdas in terms of z1, z2 and z3, you immidiately get that it's equal to one: λ1*λ2*λ3 = (z1*z2*z3) / (z2*z3*z1), and everything cancels out. A nice and elegant problem, thank you! :)
While the z's play symmetric roles in the setup of the problem, they do *not* do so in the quantity we want to calculate. If you set lambda_2 = 1 instead, you get z_2 = z_3 = a and z_1 = +/- ia, so our desired quantity is |a(5 +/- i)| = sqrt(26). And if you set lambda_3 = 1, you get z_3 = z_1 = a and z_2 = +/- ia, so our desired quantity is |a(4 +/- 2i)| = 2sqrt(5)
They don't actually play a symmetric role even in the given conditions. The first one is cyclically symmetric, but e.g. (z_1, z_2, z_3) |--> (z_3, z_2, z_1) does not preserve the equation.
6:16 seriously?! Just multiply the lambdas and you get 1. Plus your answer is not complete. The fact that the Zs are symmetric (and thus the lambdas also) doesn't mean that the target function is also. In fact it is not. For the othr cases you get 3|1+i| (edited because i forgot the 3) |1+5i| |1-5i| |4+2i| |4-2i|
an alternate approach for finding the "lambdas" that I haven't seen mentioned: all the z's having modulus 1 means all the lambdas have modulus 1. From a graphical perspective, adding up 3 complex numbers of length 1 to make the number 1 forms a rhombus - 3 sides of length 1 plus a line from 0 to 1. Rhombus' sides come in parallel pairs, so one of the lambdas must be 1 and the others must be a positive/negative pair, from which it's easy to find the rest.
1. Find all complex triplets (z1, z2, z3) where z1/z2+z2/z3+z3/z1=1 and |z1|=|z2|=|z3|=1 Write z1=e^a1•i etc, so that the 2nd condition is fulfilled automatically. Note that 1/z1=e^-a1i etc. So the 1st condition becomes e^(a1-a2)i + e^(a2-a3)i + e^(a3-a1)i = 1 Two equations and 3 variables leave one degree of freedom. Since only angle *differences* occur in the exponents, we could add the same angle to each a without changing the result. Which gives us the freedom to rotate solutions. A particular solution is a1=a2=0, a3=π/2, giving (z1, z2, z3) = (1,1,i). Then also its conjugate (1,1,-i) is a solution, and their permutational symmetries. General solutions then are (α,α,+/-iα), (α,+/-iα,α) and (+/-iα,α,α) for any complex α of norm 1. 2. |z1+2z2+3z3| = sqrt 18 = 3 sqrt(2)
I'm surprised at how cool it was to see an algebra problem on here, given how the channel is usually only more "advanced" topics-- but this feels like your number theory videos, except with one of my favorite fields (heh) of math!! What a great puzzle!
Am I missing something? Why conjugate @3:40 only to "multiply by 1" @4:50, just do your "multiply by 1" trick at @3:38? You still arrive at the linear coefficient's, with less work.
There is issue with this ques let z1=1 z2=1 z3=i it will satisfy all the given conditions. we have |z1+2z2+3z3|= 18^(1/2) but z1=i z2=1 z3=1 will also satisfy all the conditions but we will get different ans
@@MushookieMan I also checked and you're wrong. The first condition is z1/z2 + z2/z3 + z3/z1 = 1 and we can see that 1/1 + 1/i + i/1 = 1 - i + i = 1. So first condition satisfied. It's also obvious that |z1| = |z2| = |z3| = 1 because |i| = 1. So second condition is satisfied. Check that both { 1, 1, i } and {1, 1, -i } work for those conditions. Note that 18^(1/2) = √18 = √9√3 = 3√2 as found in the video when z3 = ±i. Cyclically permuting { 1, 1, i } for { z1, z2, z3 } makes |z1+2*z2+3*z3| equal to |4+2i| = √20 = 2√5 when z2 = i; and equal to |5+i| = √26 when z1= i.
6:22 Wouldn't it be much easier to calculate the triple product by noticing that all three z terms are both at top and at bottom so it will be also 1, instead of multiply it through the things out as you did?
without algebra there is also a trignometric solution: let Z1/Z2 = e^iA, Z2/Z3=e^iB, Z3=e^{-i(A+B)) now according to given conditions we get sinA+sinB=sin(A+B) and cosA+cosB=1-cos(A+B) now squaring and adding and then simplifying both equations we get cosAcosB=0 now from this we get different cases (5 in total) and we can easily eliminate 2 of them
It seems simpler to calculate z1/z2 * z2/z3 = z1/z3 = 1/lambda3 = conjugate of lambda3. Similar results for the other terms then the 2nd parenthesis = conj. of 1=1.
You can solve the problem by applying the algebra of complex numbers. Zk=exp (i*φk), k=1,2,3. exp(i*α)+ exp(i*β)+ exp[-i*(α+β)]=1, (1) α=φ1-φ2, β= φ2-φ3, -(α+β) = φ3-φ1. |Z1 +2*Z2+3*Z3|= |exp(i*φ1)+2*exp(i*φ2 )+3*exp(i*φ3)]|= |exp(i*φ2)*[exp(i*α)+2+ +3*exp(-i*β )|=|exp(i*α)+2+3*exp(-i*β )|. (2) From (1) follow the conditions for the angles α and β: 1) sin[(α+β)/2] =0 & cos[(α-β)/2]=0. 2) sin( α/2)=0 & tg(β/2)= ±1. 3) sin( β/2)=0 & tg(α/2)= ±1. Finding the corresponding pairs of values α and β from these conditions, we get the answers: 2√5 in case 1) ; 3√2 in case 2); √26 in case 3). This solution method turns out to be more cumbersome. But don't come up with a trick.
If there is a unique answer, then it must be independent of the choice of z1, z2, z3 that satisfy the conditions. One such choice is z1 = 1, z2 = 1, z3 = -i. That choice will make |z1 + 2 z2 + 3 z3| = |3 - 3i| = 3√2.
Spoiler alert, there ISN'T a unique answer. Other commenters have skyway mentioned there are 3 different answers because the final question isn't symmetric.
But unfortunately there isn't a unique answer. (Michael screwed up here! Perhaps deliberately, I don't know...) The answer *does* depend on the choice of z1, z2, and z3. It can be sqrt(18), sqrt(20), or sqrt(26). Michael only found the first.
@@zunaidparker it doesn't matter there isn't a unique answer, as long as the initial conditions are met, the final result has to be the same, as the final result does not depend on the choice of z1, z2, z3.
great video! but i have a question…what if z1=i, and z2=z3=1; then z1/z2+z2/z3+z3/z1= i/1+1/1+1/i=i+1-i=1, so the first condition is satisfied. as well, |z1|=|i|=1 and |z2|=|z3|=1, so the second condition is satisfied as well. so the modulus we want will be |i+2+3|=|i+5| which is 26^1/2.
The expression | z1+2 z2+3 z3| is not invariant under the rotation of z1->z2->z3, but the preconditions are. So there are actually 3 different solutions, corresponding to z1=z2=i z3, z2=z3=i z1, and z3=z1=i z2. They are respectively 3 sqrt(2), sqrt(26), and 2 sqrt(5).
Came to post this and found your comment here already. It's a curious and terrible blunder by Michael to completely miss the asymmetry of the final question.
@@zunaidparker Yes, it was Michael's mistake, but I find comment section to be more interesting because of these not deliberate mistakes. Michael is producing a lot of material and many times sacrifices rigorosity for output. This concept helps to grow his channel, since discussion and involvement bring people back.
@@mguzjebesku2591 agreed. Usually his mistakes are things like transcription errors, but in this case overlooking an entire branch of solutions is unlike mistakes he's made in the past.
If you write the product of lambdas in terms of z1, z2 and z3, you immidiately get that it's equal to one: λ1*λ2*λ3 = (z1*z2*z3) / (z2*z3*z1), and everything cancels out.
A nice and elegant problem, thank you! :)
Yes! And that makes it easier to get λ1λ2+λ2λ3+λ3λ1 from knowing 1/λ1+1/λ2+1/λ3. Just substitute all three ones in the numerators with λ1λ2λ3
While the z's play symmetric roles in the setup of the problem, they do *not* do so in the quantity we want to calculate. If you set lambda_2 = 1 instead, you get z_2 = z_3 = a and z_1 = +/- ia, so our desired quantity is |a(5 +/- i)| = sqrt(26). And if you set lambda_3 = 1, you get z_3 = z_1 = a and z_2 = +/- ia, so our desired quantity is |a(4 +/- 2i)| = 2sqrt(5)
They don't actually play a symmetric role even in the given conditions. The first one is cyclically symmetric, but e.g. (z_1, z_2, z_3) |--> (z_3, z_2, z_1) does not preserve the equation.
@@zadsar3406 Sorry, yeah, cyclic, not symmetric
6:16 seriously?!
Just multiply the lambdas and you get 1.
Plus your answer is not complete.
The fact that the Zs are symmetric (and thus the lambdas also) doesn't mean that the target function is also.
In fact it is not.
For the othr cases you get
3|1+i| (edited because i forgot the 3)
|1+5i|
|1-5i|
|4+2i|
|4-2i|
Are the modulus of all those 1?
@@evankalis they are, in order, 2, 26, 26 20, 20 (all square rooted)
You forgot that Michael Penn took out a factor of 3 (for some reason). The six cases give |3 ± 3i| = √18; |4 ± 2i| = √20; |5 ± i| = √26.
an alternate approach for finding the "lambdas" that I haven't seen mentioned: all the z's having modulus 1 means all the lambdas have modulus 1. From a graphical perspective, adding up 3 complex numbers of length 1 to make the number 1 forms a rhombus - 3 sides of length 1 plus a line from 0 to 1. Rhombus' sides come in parallel pairs, so one of the lambdas must be 1 and the others must be a positive/negative pair, from which it's easy to find the rest.
that's a very clever idea
Yeah this way it solves in 30 secs
1. Find all complex triplets (z1, z2, z3) where z1/z2+z2/z3+z3/z1=1 and |z1|=|z2|=|z3|=1
Write z1=e^a1•i etc, so that the 2nd condition is fulfilled automatically. Note that 1/z1=e^-a1i etc. So the 1st condition becomes
e^(a1-a2)i + e^(a2-a3)i + e^(a3-a1)i = 1
Two equations and 3 variables leave one degree of freedom. Since only angle *differences* occur in the exponents, we could add the same angle to each a without changing the result. Which gives us the freedom to rotate solutions.
A particular solution is a1=a2=0, a3=π/2, giving (z1, z2, z3) = (1,1,i).
Then also its conjugate (1,1,-i) is a solution, and their permutational symmetries.
General solutions then are (α,α,+/-iα), (α,+/-iα,α) and (+/-iα,α,α) for any complex α of norm 1.
2. |z1+2z2+3z3| = sqrt 18 = 3 sqrt(2)
Multiplying the definitions of lambas would have yielded the triple product equals one also.
I'm surprised at how cool it was to see an algebra problem on here, given how the channel is usually only more "advanced" topics-- but this feels like your number theory videos, except with one of my favorite fields (heh) of math!! What a great puzzle!
Am I missing something? Why conjugate @3:40 only to "multiply by 1" @4:50, just do your "multiply by 1" trick at @3:38? You still arrive at the linear coefficient's, with less work.
Yep, I also got a little confused. But your way definitely works and it's easier
Yep, you're missing the fact that rock climbers love to do things in a complicated way.
@@allykid4720 lol " I will because I can"
Why use Vieta at all would anyone even think of that??
10:37
NOT a good place to stop in this case, given that Michael missed out 2 other solutions by incorrectly assuming the final problem was symmetric 😂
There is issue with this ques
let z1=1 z2=1 z3=i it will satisfy all the given conditions.
we have |z1+2z2+3z3|= 18^(1/2)
but z1=i z2=1 z3=1 will also satisfy all the conditions but we will get different ans
Those choices don't satisfy the reciprocal sum condition.
@@MushookieMan check again bro
@@TechyMage I did
@@MushookieMan I also checked and you're wrong. The first condition is z1/z2 + z2/z3 + z3/z1 = 1 and we can see that 1/1 + 1/i + i/1 = 1 - i + i = 1. So first condition satisfied. It's also obvious that |z1| = |z2| = |z3| = 1 because |i| = 1. So second condition is satisfied. Check that both { 1, 1, i } and {1, 1, -i } work for those conditions.
Note that 18^(1/2) = √18 = √9√3 = 3√2 as found in the video when z3 = ±i.
Cyclically permuting { 1, 1, i } for { z1, z2, z3 } makes |z1+2*z2+3*z3| equal to |4+2i| = √20 = 2√5 when z2 = i; and equal to |5+i| = √26 when z1= i.
6:22 Wouldn't it be much easier to calculate the triple product by noticing that all three z terms are both at top and at bottom so it will be also 1, instead of multiply it through the things out as you did?
If by "easier", you mean fewer steps, then yes.
What you noticed, follows from the cyclic nature of the expressions, right?
without algebra there is also a trignometric solution: let Z1/Z2 = e^iA, Z2/Z3=e^iB, Z3=e^{-i(A+B)) now according to given conditions we get sinA+sinB=sin(A+B) and cosA+cosB=1-cos(A+B) now squaring and adding and then simplifying both equations we get cosAcosB=0 now from this we get different cases (5 in total) and we can easily eliminate 2 of them
Nice to see a "normal" algebra problem after a while.
Always interesting to follow through your solution - figure out what I find surprising in it - and then see if the comments section agrees!
A bit too harsh on the method of multplication, just multiply them and it becomes one.
The meaning of "interesting" mathematically is "something I don't quite understand yet".
It seems simpler to calculate z1/z2 * z2/z3 = z1/z3 = 1/lambda3 = conjugate of lambda3. Similar results for the other terms then the 2nd parenthesis = conj. of 1=1.
You can solve the problem by applying the algebra of complex numbers.
Zk=exp (i*φk), k=1,2,3.
exp(i*α)+ exp(i*β)+ exp[-i*(α+β)]=1, (1)
α=φ1-φ2, β= φ2-φ3, -(α+β) = φ3-φ1.
|Z1 +2*Z2+3*Z3|= |exp(i*φ1)+2*exp(i*φ2 )+3*exp(i*φ3)]|= |exp(i*φ2)*[exp(i*α)+2+
+3*exp(-i*β )|=|exp(i*α)+2+3*exp(-i*β )|. (2)
From (1) follow the conditions for the angles α and β:
1) sin[(α+β)/2] =0 & cos[(α-β)/2]=0.
2) sin( α/2)=0 & tg(β/2)= ±1.
3) sin( β/2)=0 & tg(α/2)= ±1.
Finding the corresponding pairs of values α and β from these conditions, we get the answers:
2√5 in case 1) ; 3√2 in case 2); √26 in case 3).
This solution method turns out to be more cumbersome. But don't come up with a trick.
I truly believe you should be one of the biggest math channels on RUclips. i hope yiu never stop 🙏🏻
Lambda1 * lambda2 * lambda3 = 1 just by looking at the first step, the z’s cancel out
That was so interesting!
What about solving by using Euler's transformation?
6:18 Or you can just go back to the definition of the lambdas and multiply them trivially to get 1...
Ah, that pesky sqrt(2) is as omnipresent as pi/2.
Thank you, professor.
Hi Dr. Penn!
If there is a unique answer, then it must be independent of the choice of z1, z2, z3 that satisfy the conditions.
One such choice is z1 = 1, z2 = 1, z3 = -i. That choice will make |z1 + 2 z2 + 3 z3| = |3 - 3i| = 3√2.
Spoiler alert, there ISN'T a unique answer. Other commenters have skyway mentioned there are 3 different answers because the final question isn't symmetric.
But unfortunately there isn't a unique answer. (Michael screwed up here! Perhaps deliberately, I don't know...) The answer *does* depend on the choice of z1, z2, and z3. It can be sqrt(18), sqrt(20), or sqrt(26). Michael only found the first.
@@JosBergervoetbut if the initial conditions are met, then the answer has to be the same, right?
@@zunaidparker it doesn't matter there isn't a unique answer, as long as the initial conditions are met, the final result has to be the same, as the final result does not depend on the choice of z1, z2, z3.
@@11pupona read some of the other comments. The final question is not symmetric.
great video! but i have a question…what if z1=i, and z2=z3=1; then z1/z2+z2/z3+z3/z1= i/1+1/1+1/i=i+1-i=1, so the first condition is satisfied. as well, |z1|=|i|=1 and |z2|=|z3|=1, so the second condition is satisfied as well.
so the modulus we want will be |i+2+3|=|i+5| which is 26^1/2.
100% right. Other commenters have pointed out this mistake by Michael as well. It's blatantly obvious looking at the asymmetry of the final question.
No, your last line should have the expression 26^(1/2), due to the Order of Operations.
@@forcelifeforce ah, ok, thanks!
bruh
First comment after good place to stop
Second comment after good place to stop.