nth-root of unity question (1 Viewer)

[OLDMAN]

This question is from Moriah College 2003 Trial Q8.

P_1,P_2,P_3,...,P_n represent the complex numbers z1,z2,z3,...,zn (zn=1) and are the vertices of a regular polygon on a unit circle. Prove that
(z1-z2)^2+(z2-z3)^2+(z3-z4)^2+...+(zn-z1)^2=0

 

[ezzy85]

z1 = cis@
z2 = cis@ cis(2pi/n)

(z1 - z2)² = (z1-z2)(z1 - z2)^---

i think im abit off....

 

[ND]

(z1-z2)^2+(z2-z3)^2+(z3-z4)^2+...+(zn-z1)^2
= 2(z1^2+z2^2+...+zn^2) - 2(z1z2+z2z3+...+znz1)
= 2[(z1+z2+...+zn)^2-2(z1z2+z2z3+...+znz1)] - 2(z1z2+z2z3+...+znz1)
= 2(z1+z2+...+zn)^2 - 6(z1z2+z2z3+...+znz1)

Now z^n-1=0
so z1+z2+...+zn=0 and z1z2+z2z3+...+znz1=0

.'. (z1-z2)^2+(z2-z3)^2+(z3-z4)^2+...+(zn-z1)^2 = 2(0)^2 - 6(0)
= 0

 

[underthesun]

When you expand (z1 - z2)², you'll get
(z1-z2)² = z1² + z2² + 2z1z2.

Now, that means the original equation equals

(z1-z2)^2+(z2-z3)^2+(z3-z4)^2+...+(zn-z1)^2
= 2(z1² + z2² + ... + zn²) - 2(z1z2 + z2z3 + ... + znz1) .........(bleh)

Given that z1 to zn are vertices of a regular polygon on a unit circle, then it can be assumed (?) that z1 to zn are roots of unity.

Hence, z1 to zn are roots of the following equation.

zⁿ = 1
zⁿ - 1 = 0

By polynomials, the sum of roots equal to zero. Hence z1 + all the way to + zn = 0. Also The sum of products of two roots are also zero

(z1 + z2 + ... + zn) = 0
(z1z2 + z1z3 + z1z4 + ... + z2z3 + z2z4 + ... + z(n-1)zn) = 0

in (bleh), the sum of z1² equals to zero, because:
z1² + z2² + ... + zn²
= (z1 + z2 + ... + zn)² - 2(z1z2 + z1z3 + z1z4 + .... + z2z3 + z2z4 + .... + z(n-1)zn)

now the above equals zero. But 2(z1z2 + z2z3 + ... + znz1) from (bleh).. how do I prove that it equals zero? Or did I go the wrong way?

edit : looking at andy's post, did I get something wrong?

 

[ND]

Originally posted by underthesun

now the above equals zero. But 2(z1z2 + z2z3 + ... + znz1) from (bleh).. how do I prove that it equals zero? Or did I go the wrong way?

It equals zero because z1z2+z2z3+...+znz1 is the sum of roots taken two at a time (i.e. the coefficient of z^(n-2)).

 

[underthesun]

Oh sorry it's ND. Unless your real name is andy

 

[underthesun]

Originally posted by ND
It equals zero because z1z2+z2z3+...+znz1 is the sum of roots taken two at a time (i.e. the coefficient of z^(n-2)).

Yeah it is, but i thought:

z1z2 + z1z3 + z1z4 + z1z5 + ... + z1zn +
z2z3 + z2z4 + z2z5 + ... + z2zn +
all the way
+ z(n-1)zn = 0

but then, your equation only involves with z1z2, z2z3, z3z4, z4z5.. and does not involve z1z3 (number of the root that is not next to each other)..

 

[ND]

Originally posted by underthesun
Yeah it is, but i thought:

z1z2 + z1z3 + z1z4 + z1z5 + ... + z1zn +
z2z3 + z2z4 + z2z5 + ... + z2zn +
all the way
+ z(n-1)zn = 0

but then, your equation only involves with z1z2, z2z3, z3z4, z4z5.. and does not involve z1z3 (number of the root that is not next to each other)..

Yeh you're right. Silly me.

 

[Constip8edSkunk]

what about this way?

z1, z2, ... zn represent the solution to the equation z^n-1 =0
z^n=1
=cis(2kpi)
z_k=cis(2kpi/n), as z_n=cis(2pi)

consider points z_m and z_m+1

[z_m-z_m+1]²
= z_m²+z_m+1²-2z_mz_m+1
=z_m²+z_m+1² - 2cis (2pim)/n . cis 2pi(m+1)/n
=z_m²+z_m+1² - 2cis [(2pim)+(2pi(m+1)]/n
=z_m²+z_m+1² - 2cis (4pim/n +2pi/n)
=z_m²+z_m+1²-2z_m²z₁

.'. (z1-z2)^2+(z2-z3)^2+(z3-z4)^2+...+(zn-z1)^2
= 2[z₁²+z₂²+...+z_n²-z₁²z₁+z₂²z₁+...+z_n²z₁]
=2[(z₁²+z₂²+...+z_n²)-z₁(z₁²+z₂²z+...+z_n²)]
=2(z₁²+z₂²+...+z_n²)(1-z₁)
=2(1-z₁)[(z₁+z₂+...+z_n)² - 2(z₁z₂+z₁z₃+...+z_n-1z_n)]
=2(1-z₁)[(0)-2(0)]
=0

 

[freaking_out]

Originally posted by underthesun
Oh sorry it's ND. Unless your real name is andy

lol, thats what i thought.

 

[Archman]

z1-z2, z2-z3.. etc are really the edges of a regular n-gon.
if we just rename them:w1,w2,w3 etc
if n is even, the w1^2,w3^2 (odd ones) will for a regular n/2-gon and so will the w2^2,w4^2 (eveno ones) so they must sum to 0.
if n is odd, w1^2, w2^2 ... wn^2 will form another regular n-gon. so the sum must be 0 as well.

 

[ND]

Originally posted by underthesun
Yeah it is, but i thought:

z1z2 + z1z3 + z1z4 + z1z5 + ... + z1zn +
z2z3 + z2z4 + z2z5 + ... + z2zn +
all the way
+ z(n-1)zn = 0

but then, your equation only involves with z1z2, z2z3, z3z4, z4z5.. and does not involve z1z3 (number of the root that is not next to each other)..

Ok well i've got a way of explaining it, but i'm not really sure how to put it into words, i'll have a go though:

z1z2 = z3 since |z1|=|z2|=|zn|, and arg(z2/z1)=arg(z3/z2)=arg(zn/zn-1), so z1z2+z2z3+...+znz1 = z3+z5+...+z(n+1). Now if n is even, after z(n/2), they will start to equal -z3, -z5 etc. and so will cancel. If z is odd, they will equal z2, z4 etc. and so will be the sum z1+z2+z3+...+zn = 0.

 

[underthesun]

Skunk's solution looks good, it seems to be right so far..

 

[underthesun]

Originally posted by ND
Ok well i've got a way of explaining it, but i'm not really sure how to put it into words, i'll have a go though:

z1z2 = z3 since |z1|=|z2|=|zn|, and arg(z2/z1)=arg(z3/z2)=arg(zn/zn-1), so z1z2+z2z3+...+znz1 = z3+z5+...+z(n+1). Now if n is even, after z(n/2), they will start to equal -z3, -z5 etc. and so will cancel. If z is odd, they will equal z2, z4 etc. and so will be the sum z1+z2+z3+...+zn = 0.

It should be right, because skunk proved that it's equal to zero. Maybe i'll have to find a quick way of explaining that, for exam conditions...

If only everyone's active in the morning instead of late night, my eyes are so sleepy

 

[MyLuv]

In ND solution:
z1^2 + z2^2 +...+zn^2 = [Sum(zi)]^2 - 2Sum(zi*zj)
eg: a2+b2+c2= (a+b+c)2 - 2(ab+bc+ac)
---> so U dont have to worry 'bout missing z1z3 or z1z4
Btw do we need to prove this or just assume

 

[ND]

Yeh that gets rid of teh z1^2+z2^2+...+zn^2, but underthesun was referring to z1z2+z2z3+...+znz1 which does not equal sum of the roots taken 2 at a time.

 

[maniacguy]

Observe that the complex numbers form the vertices of a regular n-gon.

Hence z_(k+1) = cis(2*pi/n)*z_k

Also, z_1 = cis(2*pi/n)*z_n

Hence:
(z_1 - z_2)^2 + ... + (z_n - z_1)^2
= (z_1 - cis(2*pi/n)*z_1)^2 + ... + (z_n - cis(2*pi/n)*z_n)^2
= (1-cis(2*pi/n))^2*z_1^2 + ... + (1-cis(2*pi/n))^2*z_n^2
= (1-cis(2*pi/n))^2*(z_1^2 + z_2^2 + ... + z_n^2)
= (1-cis(2*pi/n))^2*0 [proven in several posts...]
= 0 as desired

Constip8edSkunk - your solution is right, but it's a little longer than necessary...

 

[Constip8edSkunk]

hmm yeah i kinda went around it abit hmmm....

 

[OLDMAN]

excellent solution from maniacguy, stuck close to the plot of the nth-root unity story. Nice effort from skunk as well.

Would be good exercise to prove the same for a regular n-gon inscribed in a "floating" unit circle with center zc, where zn-zc not necessarily =1.