# let a, b , c are distinct integers and w (omega) and w^2 are imaginary cube root of unity . then    minimum value of  / a+bw +c(w)^2 / + /  a+b (w)^2+cw /.

pratish shrivastava
31 Points
11 years ago
/a+bw+cw^2/+a+b(w)^2+cw=/2a+b(w+(w)^2)+c(w+(w)^2)=/2a-b-c/
jagdish singh singh
173 Points
11 years ago

$Let z_{1}=\left|a+b\omega+c\omega^2\right|\Leftrightarrow |z_{1}|=\left|a+b\omega+c\omega^2\right|\Leftrightarrow |z_{1}|^2=\left|a+b\omega+c\omega^2\right|^2,\\\\ Then z_{1}\bar{z_{1}}=(a+b\omega+c\omega^2).(a+b\omega^2+c\omega)=a^2+b^2+c^2-ab-bc-ca\\\\ So |z_{1}|=\sqrt{a^2+b^2+c^2-ab-bc-ca}=\sqrt{\frac{1}{2}\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)}\\\\ So |z_{1}|=\sqrt{\frac{1}{2}.\left\{(a-b)^2+(b-c)^2+(c-a)^2\right\}}................................................(1)\\\\ Similarly \\\\ z_{2}=\left|a+b\omega^2+c\omega\right|\Leftrightarrow |z_{2}|=\left|a+b\omega^2+c\omega\right|\Leftrightarrow |z_{2}|^2=\left|a+b\omega^2+c\omega\right|^2,\\\\ Then z_{2}\bar{z_{2}}=(a+b\omega^2+c\omega).(a+b\omega+c\omega^2)=a^2+b^2+c^2-ab-bc-ca\\\\ So |z_{2}|=\sqrt{a^2+b^2+c^2-ab-bc-ca}=\sqrt{\frac{1}{2}\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)}\\\\ So |z_{2}|=\sqrt{\frac{1}{2}.\left\{(a-b)^2+(b-c)^2+(c-a)^2\right\}}................................................(2)\\\\$

$|z_{1}|+|z_{2}|=\sqrt{\frac{1}{2}.\left\{(a-b)^2+(b-c)^2+(c-a)^2\right\}}+\sqrt{\frac{1}{2}.\left\{(a-b)^2+(b-c)^2+(c-a)^2\right\}}\\\\ |z_{1}|+|z_{2}|=\sqrt{2}.\sqrt{\left\{(a-b)^2+(b-c)^2+(c-a)^2\right\}}\\\\ Now Here a,b,c are Distinct Integer. So Let a=k-1,b=k,c=k+1, where k\in Z\\\\ Then a-b=-1,b-c=-1,c-a=2\\\\ Put These value in |z_{1}|+|z_{2}|=\sqrt{2}.\sqrt{(-1)^2+(-1)^2+(2)^2}=\sqrt{2}.\sqrt{6}=4\sqrt{3}$

Nikhil verma
13 Points
2 years ago
We habe to find the value of k=|a+bw+cw2|
Here a,b,c are consecutive integer

So lets take a=n-1 ,b=n ,c=n+1
K=|n-1+n(w)+(n+1)(w2))|
=|n-1+nw+nw2+w2|
=|n(1+w+w2)+w2-1|
[1+w+w2=0]
=|w2-1|
=|(-1-√3i)/2-1|
=|(-3-√3i)/2|
=√{(3/2)2+(√3/2)2}
=√(9/4)+(3/4)
=√12/4
=√3