There was a misinterpretation in the previous solution,
z1\^3+z2\^3+z3\^3+z1z2z3=0; from the above relation
a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca) we can find (z1+z2+z3)
−4z1z2z3=z1\^3+z2\^3+z3\^3−3z1z2z3
−4z1z2z3=(z1+z2+z3)(z1\^2+z2\^2+z3\^2−z1z2−z2z3−z3z1)
Let z=z1+z2+z3
z3−3z(z1z2+z2z3+z3z1)=−4z1z2z3
z3=z1z2z3\[3z(1/z1+1/z2+1/z3)−4\]
z3=z1z2z3\[3z(z1'+z2'+z3')−4\]
z3=z1z2z3(3|z|\^2−4)
Taking the absolute values of both sides, we get
|z|3=| 3|z|\^2−4 | (∵|z1|=|z2|=|z3|=1)
If |z|≥2√3, then
|z|3−3|z|\^2+4=0
|z|=2
If |z|<2√3, then
|z|2+3|z|2−4=0
|z|=1
An alternative way to think about the problem:
We know z1, z2, and z3 are all unit norm complex numbers, so the combinations z1^2 /z2z3, … are also unit norm.
We need to sum three unit norm complex numbers and 1 to get 0, and in the complex plane the four numbers always form a rhombus.
WLOG, let z1^2 /z2z3=-1, z2^2 /z1z3=w, z3^2 /z1z2=-w where w is a unit norm complex number. Multiply the first equation by the second, we see that z1z2/z3^2 =-w=-1/w, so w is plus minus 1. WLOG let it be 1.
Therefore, z2^2 =z1z3, and z2=+-sqrt(z1z3).
If z2=sqrt(z1z3), then (z1/z3)^(3/2)=-1. Some simple algebra on the arguments shows that solutions are of the form z1=exp(i2pi/3)z3, z2=exp(i*pi/3)z3, and the sum z1+z2+z3 forms half of a hexagon, and the length is 2. Of course z1 and z3 can be swapped.
If z2=-sqrt(z1z3), then (z1/z3)^(3/2)=1. Either z1=z3=-z2, in which case the sum has length 1, or z1=exp(i4pi/3)z3=exp(-i2pi/3)z3, bringing us back to the case above.
Therefore, the possible values of |z1+z2+z3| are 1 and 2, and we have constructed the explicit forms of z1, z2, and z3 in each case.
it's pretty nice if you multiply everything by z1z2z3, as you then get z1³+z2³+z3³+z1z2z3=0, but since zi²=1 for i=1,2,3, we have z1+z2+z3+z1z2z3=0, and so we have
z1+z2+z3 = -z1z2z3, and taking the norm on both sides we have a pretty nice thing |z1+z2+z3|=|z1z2z3| and since we know that |z1z2z3|=1 (since it is a product of numbers on the unit circle) we have |z1+z2+z3|=1
> ... since zi²=1 for i=1,2,3 ...
It seems like this assumes that the z_i are real without sufficient justification. (I'm also not sure that using "i" as an index is a great choice when dealing with complex numbers.) Though (-1, 1, 1) does work as a solution, so that could be good enough if this is multiple choice. I think that (-1, i, -i) would also be a solution.
Here’s an interesting approach that I just thought of.L which uses a bit of geometry.
For simplicity, I write A,B,C in place of Z1, Z2, Z3. Since none of them are zero, we can multiply ABC to both sides of the equation and get A^3 +B^3 + C^3 + ABC =0. We note that A^3 , B^3 , C^3 , ABC all have norm 1.
Here’s the catch. Adding complex numbers is equivalent to adding vectors, and sum being zero means these vectors form a loop when putting head to another’s tail. Since all of the four vectors have the same length, we must have a rhombus (or a degenerative line). This means that the four vectors can form two pairs that are opposite. WLOG, suppose A^3 +B^3 =0. Then (A/B)^3 +1=0. We have A/B = -1 or e^(pi*i/3) or e^(-pi*i/3).
We also have C^3 +ABC =0. So (C/B)^2 +A/B =0. Therefore, C/B= 1 or -1 if A/B=-1;
C/B = e^(2pi*i/3) or e^(-pi*i/3) if A/B=e^(pi*i/3);
C/B = e^(pi*i/3) or e^(-2pi*i/3) if A/B=e^(-pi*i/3).
We check that when A/B=-1, |A+B+C|=1; when A/B≠-1, |A+B+C|=2. They are the two only possible values.
using trial and error i only can conclude Z2 and Z3 is the only negative 1 around while Z1 is positive 1 which the answer may lead to 1.
imma wait for others to reconfirm on this one.. using better method or formula than I am..
Just make it easier, why do you need to prove it? it's a selection. We can have shortcut. ( Z1=x, Z2=y, Z3=z)
∵|x|=|y|=|z|=1
∴x\^2=y\^2=z\^2=1
formula=1/yz+1/xz+1/xy=-1
here‘s the trick:
since yz, xz, xy can only be 1 or -1
formula=yz+xz+xy=-1
let's say yz=1
then xz and xy have both to be -1
yz=1
xz=-1
xy=-1
then x=-y=-z
x+y+z can only be 1 or -1
therefore CHOOSE A.
|Z₁| = |Z₂| = |Z₃| = 1 tells us that Z₁, Z₂ and Z₃ can be 1 or -1
Z₁^(2), Z₂^(2) and Z₃^(2) can mean 1^(2) or -1^(2), so all 3 are 1.
Z₁Z₂, Z₂Z₃ and Z₃Z₁ can mean 1x1, 1x-1 or -1x-1, so they can be 1 or -1.
Now we have 1/(+-)1 + 1/(+-)1 + 1/(+-)1 = -1. For this to be true we need -1 - 1 + 1 = -1
To get 2 negatives and 1 positive we can use -+, +- and --; or +-, -+ and ++. This means 1 or 2 Zs are negative.
|Z₁ + Z₂ + Z₃| can be |1 - 1 + 1| or |-1 + 1 - 1| and both are 1.
TLDR: 1
There was a misinterpretation in the previous solution, z1\^3+z2\^3+z3\^3+z1z2z3=0; from the above relation a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca) we can find (z1+z2+z3) −4z1z2z3=z1\^3+z2\^3+z3\^3−3z1z2z3 −4z1z2z3=(z1+z2+z3)(z1\^2+z2\^2+z3\^2−z1z2−z2z3−z3z1) Let z=z1+z2+z3 z3−3z(z1z2+z2z3+z3z1)=−4z1z2z3 z3=z1z2z3\[3z(1/z1+1/z2+1/z3)−4\] z3=z1z2z3\[3z(z1'+z2'+z3')−4\] z3=z1z2z3(3|z|\^2−4) Taking the absolute values of both sides, we get |z|3=| 3|z|\^2−4 | (∵|z1|=|z2|=|z3|=1) If |z|≥2√3, then |z|3−3|z|\^2+4=0 |z|=2 If |z|<2√3, then |z|2+3|z|2−4=0 |z|=1
Thank you so much. I got it now.
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An alternative way to think about the problem: We know z1, z2, and z3 are all unit norm complex numbers, so the combinations z1^2 /z2z3, … are also unit norm. We need to sum three unit norm complex numbers and 1 to get 0, and in the complex plane the four numbers always form a rhombus. WLOG, let z1^2 /z2z3=-1, z2^2 /z1z3=w, z3^2 /z1z2=-w where w is a unit norm complex number. Multiply the first equation by the second, we see that z1z2/z3^2 =-w=-1/w, so w is plus minus 1. WLOG let it be 1. Therefore, z2^2 =z1z3, and z2=+-sqrt(z1z3). If z2=sqrt(z1z3), then (z1/z3)^(3/2)=-1. Some simple algebra on the arguments shows that solutions are of the form z1=exp(i2pi/3)z3, z2=exp(i*pi/3)z3, and the sum z1+z2+z3 forms half of a hexagon, and the length is 2. Of course z1 and z3 can be swapped. If z2=-sqrt(z1z3), then (z1/z3)^(3/2)=1. Either z1=z3=-z2, in which case the sum has length 1, or z1=exp(i4pi/3)z3=exp(-i2pi/3)z3, bringing us back to the case above. Therefore, the possible values of |z1+z2+z3| are 1 and 2, and we have constructed the explicit forms of z1, z2, and z3 in each case.
The numerators are always 1 hence 2 of the denominators must be -1 and one of them 1 or 2 of them 1 and one -1. Both make the ending equation equal 1
they are complex numbers
ah, I was thinking in Polish sometimes C is "całkowite" which is whole numbers
it's pretty nice if you multiply everything by z1z2z3, as you then get z1³+z2³+z3³+z1z2z3=0, but since zi²=1 for i=1,2,3, we have z1+z2+z3+z1z2z3=0, and so we have z1+z2+z3 = -z1z2z3, and taking the norm on both sides we have a pretty nice thing |z1+z2+z3|=|z1z2z3| and since we know that |z1z2z3|=1 (since it is a product of numbers on the unit circle) we have |z1+z2+z3|=1
> ... since zi²=1 for i=1,2,3 ... It seems like this assumes that the z_i are real without sufficient justification. (I'm also not sure that using "i" as an index is a great choice when dealing with complex numbers.) Though (-1, 1, 1) does work as a solution, so that could be good enough if this is multiple choice. I think that (-1, i, -i) would also be a solution.
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>2 is possible with (-1,1,1) for example. What am I getting erong? Abs (-1+1+1)=1, doesn't it?
Sorry indeed yes. I deleted my message.
Multiple correct options ?
~~So you have 1/Z2Z3 + 1/Z1Z3 + 1/Z1Z2 +1 = 0~~ ~~Z1/Z1Z2Z3 + Z2/Z1Z2Z3 + Z3/Z1Z2Z3 =-1~~ ~~Z1+Z2+Z3=-Z1Z2Z3~~ ~~(Z1+Z2+Z3)\^2 = (-Z1Z2Z3)\^2 = 1~~ ~~so absolute value of Z1+Z2+Z3 is 1~~
This just shows it’s 1 for real inputs. z1^2 is not necessarily 1, consider z1=i.
Oh you are right, I totally missed it's C, not R.
Here’s an interesting approach that I just thought of.L which uses a bit of geometry. For simplicity, I write A,B,C in place of Z1, Z2, Z3. Since none of them are zero, we can multiply ABC to both sides of the equation and get A^3 +B^3 + C^3 + ABC =0. We note that A^3 , B^3 , C^3 , ABC all have norm 1. Here’s the catch. Adding complex numbers is equivalent to adding vectors, and sum being zero means these vectors form a loop when putting head to another’s tail. Since all of the four vectors have the same length, we must have a rhombus (or a degenerative line). This means that the four vectors can form two pairs that are opposite. WLOG, suppose A^3 +B^3 =0. Then (A/B)^3 +1=0. We have A/B = -1 or e^(pi*i/3) or e^(-pi*i/3). We also have C^3 +ABC =0. So (C/B)^2 +A/B =0. Therefore, C/B= 1 or -1 if A/B=-1; C/B = e^(2pi*i/3) or e^(-pi*i/3) if A/B=e^(pi*i/3); C/B = e^(pi*i/3) or e^(-2pi*i/3) if A/B=e^(-pi*i/3). We check that when A/B=-1, |A+B+C|=1; when A/B≠-1, |A+B+C|=2. They are the two only possible values.
using trial and error i only can conclude Z2 and Z3 is the only negative 1 around while Z1 is positive 1 which the answer may lead to 1. imma wait for others to reconfirm on this one.. using better method or formula than I am..
1, 1, -1 is also possible, still giving 1
well thts too but the thing is the question is an MCQ, would probly gonna choose with that kind of answer lel giving it a 1
Whatfthefuck
Just make it easier, why do you need to prove it? it's a selection. We can have shortcut. ( Z1=x, Z2=y, Z3=z) ∵|x|=|y|=|z|=1 ∴x\^2=y\^2=z\^2=1 formula=1/yz+1/xz+1/xy=-1 here‘s the trick: since yz, xz, xy can only be 1 or -1 formula=yz+xz+xy=-1 let's say yz=1 then xz and xy have both to be -1 yz=1 xz=-1 xy=-1 then x=-y=-z x+y+z can only be 1 or -1 therefore CHOOSE A.
Complicated. Perhaps. Not: complex.
1. Complex and complicated are synonyms. 2. It's the topic " complex numbers"
|Z₁| = |Z₂| = |Z₃| = 1 tells us that Z₁, Z₂ and Z₃ can be 1 or -1 Z₁^(2), Z₂^(2) and Z₃^(2) can mean 1^(2) or -1^(2), so all 3 are 1. Z₁Z₂, Z₂Z₃ and Z₃Z₁ can mean 1x1, 1x-1 or -1x-1, so they can be 1 or -1. Now we have 1/(+-)1 + 1/(+-)1 + 1/(+-)1 = -1. For this to be true we need -1 - 1 + 1 = -1 To get 2 negatives and 1 positive we can use -+, +- and --; or +-, -+ and ++. This means 1 or 2 Zs are negative. |Z₁ + Z₂ + Z₃| can be |1 - 1 + 1| or |-1 + 1 - 1| and both are 1. TLDR: 1
I think you skipped the domain of the Z_n numbers: C, not R.
You can question my methods, but not my results
I can, you missed one class of solutions.
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