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JEE Advanced · Mathematics · 3. Complex Numbers

Let ω1 be a cube root of unity. Then the minimum of the set {a+bω+cω22:a,b,c distinct nonzero integers } equals ________

  1. A 1
  2. B 2
  3. C 3
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(C) 3

Step-by-step Solution

Detailed explanation

If z is a complex number
Then |z | 2 =z z ¯
Now,
a+bω+cω22=a+bω+cω2a+bω+cω2¯
\(\left|a+b \omega+c \omega^2\right|^2=\left(a+b \omega+c \omega^2\right)(a+b \bar{\omega}\) \(+~c \bar{\omega}^2)\)
ifaRa-=a
z1z2¯=z-1z-2
ω-=ω2,  ω-2=ω
a+bω+cω22=a+bω+cω2a+bω2+cω
=a2+b2+c2-ab-bc-ca
∵  ω3=1,  1+ω+ω2=0
\(\left|a+b \omega+c \omega^2\right|^2=\frac{1}{2}[(a-b)^2+(b-c)^2\) \(+~(c-a)^2]\)
Now for this to be minimum
Take a=1, b=2, c=3, as a, b, c are distinct and non-zero integers
Minimum of a+bω+cω22=3
From JEE Advanced
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