JEE Advanced 2019 · Paper 1 · Mathematics Q13

MathematicsNumerical+3Question 49 of 54

Let $\omega \ne 1$ be a cube root of unity. Then the minimum of the set

$\left\{|a + b\omega + c\omega^2|^2 : a, b, c \text{ distinct non-zero integers}\right\}$

equals ______

Your answer

Answer and worked solution

Official answer: 3

Using $1+\omega+\omega^2=0$ :
$|a+b\omega+c\omega^2|^2=a^2+b^2+c^2-ab-bc-ca=\tfrac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right].$

For distinct integers the differences are at least $1,1,2$ , which gives $\frac12(1+1+4)=3$ . This is attained by $(a,b,c)=(1,2,3)$ , all non-zero.

Answer: 3

Topics:Roots of unity

JEE Advanced 2019 · Paper 1, Mathematics question 13