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)|`.

Explanation

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