Grade 12Algebra

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 /.

Andy !@

15 Years agoGrade 12

3 Answers

pratish shrivastava

15 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

15 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}$$