Let `A(z_(1))` and `B(z_(2))` be such that `angleAOB=theta('O')` being the origin). If we define `z_(1) xx z_(2) = |z_(1)||z_(2)|sintheta`, then `z_(1) xx z_(2)` is also equal to

Options

A. `"Re"(z_(1)barz_(2))=0`
B. `"Re"(barz_(1)z_(2))=0`
C. `"Im"(barz_(1)z_(2))=0`
D. none of these

Answer

C

Explanation

We have,
`z_(2) = (|z_(2)|)/(|z_(1)|)z_(1)e^(itheta)`
`rArr z_(2)/z_(1) = (|z_(2)|)/(|z_(1)|)e^(itheta)`
`rArr "Im"(z_(2)/z_(1)|z_(1)|^(2))="Im"(|z_(1)||z_(2)|e^(itheta))`

`rArr "Im"(z_(2)/z_(1)|z_(1)|^(2))=|z_(1)||z_(2)|sintheta`
`rArr "lm"(z_(2)/z_(1)|z_(1)|^(2))=z_(1) xx z_(2)`
`rArr z_(1) xx z_(2)="lm"(z_(2)/z_(1) xx z_(1)barz_(2))="lm"(barz_(1)z_(2))`