`|z_(1)+z_(2)|=|z_(1)|+|z_(2)|` is possible, ifA. `z_(2) = barz_(1)`B. `z_(2)=(1)/(z_(1))`C. arg`(z_(1))=arg(z_(2))`D. `|z_(1)|+|z_(2)|`

`|z_(1)+z_(2)|=|z_(1)|+|z_(2)|` is possible, ifA. `z_(2) = barz_(1)`B.

1.	`\|z_(1)+z_(2)\|=\|z_(1)\|+\|z_(2)\|` is possible, ifA. `z_(2) = barz_(1)`B. `z_(2)=(1)/(z_(1))`C. arg`(z_(1))=arg(z_(2))`D. `\|z_(1)\|+\|z_(2)\|`
Answer» Correct Answer - C Given that, `rArr \|r_(1)(costheta_(1)+isintheta_(1))+r_(2)(costheta_(2)+isintheta_(2))\|=r_(1)(costheta_(1)+isintheta_(2))\|+r_(2)(costheta_(2) +isintheta_(2))\|` `rArr \|r_(1)(costheta_(1)+r_(2)costheta)+i(r_(1)sintheta_(1)+r_(2)sintheta_(2)\|=r_(1)+r_(2)` `rArr sqrt(r_(1)^(2)cos^(2)theta_(1)+r_(2)^(2)costheta+2r_(1)r_(2)costheta_(1)costheta_(2) + r_(1)^(2)sin^(2)theta_(1)+r_(2)^(2)sin^(2)theta_(2))` `rArr sqrt(+2r_(1)r_(2)sintheta_(1)+sintheta_(2))=r_(1)+r_(2)` `rArr sqrt(r_(1)^(2)+r_(2)^(2)+ 2r_(1)r_(2)[ cos(theta_(1)-theta_(2))]=r_(1)+r_(2)` On squaring both sides, we get ` r_(1)^(2)+r_(2)^(2)+ 2r_(1)r_(2) cos(theta_(1)-theta_(2))=r_(1)+r_(2)+2r_(1)r_(2)` `rArr 2r_(1)r_(2)[1 -cos(theta_(1)-theta_(2))]=0` `rArr 1-cos(theta_(1)-theta_(2))=1` `rArr cos(theta _(1)-theta_(2)) = cos 0^(@)` `rArr theta_(1)= theta_(2)` arg`(z_(1)) = arg (z_(2))