Complex Numbers Ques 106
The minimum value of $\left|a+b \omega+c \omega^{2}\right|$, where $a, b$ and $c$ are all not equal integers and $\omega(\neq 1)$ is a cube root of unity, is
(a) $\sqrt{3}$
(b) $\frac{1}{2}$
(c) $1$
(d) $0$
$(2005,1 M)$
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Answer:
Correct Answer: 106.(c)
Solution:
Formula:
- Let $z=\left|a+b \omega+c \omega^{2}\right|$
$\Rightarrow z^{2}=\left|a+b \omega+c \omega^{2}\right|^{2}=\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
or $z^{2}=\frac{1}{2}\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \}$ $\quad$ …….(i)
Since, $a, b, c$ are all integers but not all simultaneously equal.
$\Rightarrow$ If $a=b$ then $a \neq c$ and $b \neq c$
Because difference of integers $=$ integer
$\Rightarrow(b-c)^{2} \geq 1$ (as minimum difference of two consecutive integers is $( \pm 1))$ also $(c-a)^{2} \geq 1$
and we have taken $a=b \Rightarrow(a-b)^{2}=0$
From Eq. (i), $\quad z^{2} \geq \frac{1}{2}(0+1+1)$
$\Rightarrow \quad z^{2} \geq 1$
Hence, minimum value of $|z|$ is 1 .
BETA
