InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
If the complex number `Z_(1)` and `Z_(2), arg (Z_(1))- arg(Z_(2)) =0`. then show that `|z_(1)-z_(2)| = |z_(1)|-|z_(2)|`. |
|
Answer» Let `z_(1) = r_(1) (costheta_(1) + isin theta_(1))` and `z_(2) = r_(2) (costheta_(2) + isin theta_(2))` `rArr arg (z_(1)) = theta_(1)arg(z_(2)) = theta_(2)` Given that, arg `(z_(1)) - arg (z_(2)) = 0` `theta _(1)-theta _(2) -0 rArr theta _(1) = theta _(2)` ` z_(2) = r_(2) (costheta_(2) + isintheta_(2)) [:.theta_(1) = theta_(2)]` ` z_(1)-z_(2) = (r_(1) costheta_(1)-r_(2) costheta_(1)) + (r_(1) isintheta_(1)- r_(2)isintheta_(1)) ` ` |z_(1)-z_(2)| = sqrt((r_(1) costheta_(1)-r_(2) costheta_(1))^(2) + (r_(1) isintheta_(1)- r_(2)isintheta_(1))^(2)) ` `=sqrt(r_(1)^(2) +r_(1)^(2)-2 r_(1)r_(2)cos^(2)theta_(1) -2 r_(1)r_(2)sin^(2)theta_(1))` `=sqrt(r_(1)^(2) +r_(1)^(2)-2 r_(1)r_(2)(sin^(2)theta_(1) + cos^(2)theta_(1)))` `=sqrt(r_(1)^(2) +r_(1)^(2)-2 r_(1)r_(2))=sqrt((r_(1)-r_(2))^(2))` `rArr |z_(1) - z_(2)| =r_(1)-r_(2) [:. r = |z|]` `|z_(1) - z_(2)|` " "Hence proved |
|