If a , b , c are integers, not all simultaneously equal and is a cube root of unity ( 1 ) , then minimum value of | a + b + c 2 | is [2005]

Question

If $a, b, c$ are integers, not all simultaneously equal and $ω$ is a cube root of unity $(ω \neq 1)$ , then minimum value of $| a + b ω + c ω^{2} |$ is [2005]

Smart Achievers · Accepted Answer

(2)

$Given that a, b, c are integers not all equal and ω is cube root of unity \neq 1,$

$then$ $| a + b ω + c ω^{2} |$

$=$ $| a + b (\frac{- 1 + i \sqrt{3}}{2}) + c (\frac{- 1 - i \sqrt{3}}{2}) |$

$=$ $| (\frac{2 a - b - c}{2}) + i (\frac{b \sqrt{3} - c \sqrt{3}}{2}) |$

$= \frac{1}{2} \sqrt{{(2 a - b - c)}^{2} + 3 {(b - c)}^{2}}$

$= \sqrt{\frac{1}{2} [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}]}$

$R.H.S. will be minimum when a = b = c, but according to the question, we cannot take a = b = c .$

$∴$ $The minimum value is obtained when any two are zero and third is a minimum magnitude integer i.e. 1.$

$∴$ $b = c = 0, a = 1 \Rightarrow gives us the minimum value = 1 .$

Solution

Q.
If $a, b, c$ are integers, not all simultaneously equal and $ω$ is a cube root of unity $(ω \neq 1)$ , then minimum value of $| a + b ω + c ω^{2} |$ is [2005]

1 $0$

2 $1$

3 $\frac{\sqrt{3}}{2}$

4 $\frac{1}{2}$

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