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Complex Numbers question

2019 · Shift 1 · Q33
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  5. /2019 · Shift 1 · Q33

Complex Numbers question

2019 · Shift 1 · Q33

JEE AdvancedMathematicsComplex NumbersNumerical+3 / −1
Let ωe1\omega e 1ωe1 be a cube root of unity. Then the minimum of the set {∣a+bω+cω2∣2:a,b,c\{ {\left| {a + b\omega + c{\omega ^2}} \right|^2}:a,b,c{​a+bω+cω2​2:a,b,c distinct non-zero integers} equals ..................
Numerical answer
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Correct answer: 3

  1. Use the cube root of unity identities

Let ω≠1\omega \ne 1ω=1 be a cube root of unity. Then

ω3=1,1+ω+ω2=0,ω‾=ω2.\omega^3=1, \qquad 1+\omega+\omega^2=0, \qquad \overline{\omega}=\omega^2.ω3=1,1+ω+ω2=0,ω=ω2.

We need the minimum of

∣a+bω+cω2∣2\left|a+b\omega+c\omega^2\right|^2​a+bω+cω2​2

where a,b,ca,b,ca,b,c are distinct non-zero integers.

  1. Expand the modulus square

Let

z=a+bω+cω2.z=a+b\omega+c\omega^2.z=a+bω+cω2.

Then

∣z∣2=zzˉ=(a+bω+cω2)(a+bω2+cω).|z|^2=z\bar z=(a+b\omega+c\omega^2)(a+b\omega^2+c\omega).∣z∣2=zzˉ=(a+bω+cω2)(a+bω2+cω).

Expanding,

∣z∣2=a2+b2+c2+ab(ω+ω2)+ac(ω+ω2)+bc(ω+ω2).|z|^2=a^2+b^2+c^2+ab(\omega+\omega^2)+ac(\omega+\omega^2)+bc(\omega+\omega^2).∣z∣2=a2+b2+c2+ab(ω+ω2)+ac(ω+ω2)+bc(ω+ω2).

Since

ω+ω2=−1,\omega+\omega^2=-1,ω+ω2=−1,

we get

∣a+bω+cω2∣2=a2+b2+c2−ab−bc−ca.|a+b\omega+c\omega^2|^2=a^2+b^2+c^2-ab-bc-ca.∣a+bω+cω2∣2=a2+b2+c2−ab−bc−ca.

So we must minimize

S=a2+b2+c2−ab−bc−ca.S=a^2+b^2+c^2-ab-bc-ca.S=a2+b2+c2−ab−bc−ca.
  1. Rewrite in a useful form

A standard identity is

(a−b)2+(b−c)2+(c−a)2=2(a2+b2+c2−ab−bc−ca).(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2-ab-bc-ca).(a−b)2+(b−c)2+(c−a)2=2(a2+b2+c2−ab−bc−ca).

Hence

S=(a−b)2+(b−c)2+(c−a)22.S=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}.S=2(a−b)2+(b−c)2+(c−a)2​.

Now a,b,ca,b,ca,b,c are distinct integers, so each difference a−ba-ba−b, b−cb-cb−c, c−ac-ac−a is a non-zero integer.

We want the minimum possible value of

(a−b)2+(b−c)2+(c−a)22.\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}.2(a−b)2+(b−c)2+(c−a)2​.
  1. Find the minimum possible sum of squared differences

For three distinct integers, the smallest spacing occurs when they are consecutive, say

a,b,c=n−1,n,n+1a,b,c = n-1,n,n+1a,b,c=n−1,n,n+1

(in some order). Then the pairwise differences are 1,1,21,1,21,1,2 in magnitude, so

(a−b)2+(b−c)2+(c−a)2=12+12+22=6.(a-b)^2+(b-c)^2+(c-a)^2 = 1^2+1^2+2^2=6.(a−b)2+(b−c)2+(c−a)2=12+12+22=6.

Thus

Smin⁡=62=3.S_{\min}=\frac{6}{2}=3.Smin​=26​=3.
  1. Check that this is attainable with non-zero distinct integers

Take for example

(a,b,c)=(1,2,3).(a,b,c)=(1,2,3).(a,b,c)=(1,2,3).

These are distinct non-zero integers. Then

∣1+2ω+3ω2∣2=12+22+32−(1⋅2+2⋅3+3⋅1)|1+2\omega+3\omega^2|^2=1^2+2^2+3^2-(1\cdot 2+2\cdot 3+3\cdot 1)∣1+2ω+3ω2∣2=12+22+32−(1⋅2+2⋅3+3⋅1) =1+4+9−(2+6+3)=14−11=3.=1+4+9-(2+6+3)=14-11=3.=1+4+9−(2+6+3)=14−11=3.

So the minimum value is indeed achieved.

  1. Conclusion

Therefore,

min⁡{∣a+bω+cω2∣2:a,b,c distinct non-zero integers}=3.\min \left\{\left|a+b\omega+c\omega^2\right|^2 : a,b,c \text{ distinct non-zero integers}\right\}=3.min{​a+bω+cω2​2:a,b,c distinct non-zero integers}=3.
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