`z_1a n dz_2` lie on a circle with center at the origin. The point of intersection `z_3` of he tangents at `z_1a n dz_2` is given by `1/2(z_1+( z )_2)` b. `(2z_1z_2)/(z_1+z_2)` c. `1/2(1/(z_1)+1/(z_2))` d. `(z_1+z_2)/(( z )_1( z )_2)`

Answer

B

Explanation

As `Delta OAC` is a right-angled triangle with right angle at A, So
`|z_(1)|^(2) + |z_(3)- z_(1)|^(2) = |z_(3)|^(2)`
`rArr 2|z_(1)|^(2) - barz_(3)z_(1)-barz_(1)z_(3) = 0`
`rArr = 2barz_(1) -barz_(3) - (barz_(1))/(z_(1))z_(3) = 0`
Similary,
`2barz_(1)-barz_(3) - (barz_(2))/(z_(2))z_(3) = 0`
Subtracting (2) Form (1)

`2(barz_(2) - barz_(1))= z_(3)((barz_(1))/(z_(1)) - (barz_(2))/(barz_(2)))`
`rArr (2r^(2) (z_(1)-z_(2)))/(z_(1)z_(2))= z_(3)r^(2)` `((z_(2)^(3)-z_(1)^(2))/(z_(1)^(2)- z_(2)))" "[because |z_(1)|^(2) = |z_(2)|^(2) = r^(2)]`
`rArr z_(3) = (2z_(1)z_(2))/(z_(2) + z_(1))`