If `|z_1/z_2|=1` and `arg (z_1z_2)=0` , then

Options

A. `z_(1) = z_(2)`
B. `|z_(2)|^(2) = z_(1)z_(2)`
C. `z_(1)z_(2)= 1`
D. more than 8

Answer

B

Explanation

Let `z_(1)=|z_(1)|(costheta_(1)+I sintheta_(1))`.
Now, `|(z_(1))/(z_(2))|=1rArr|z_(1)|=|z_(2)|`
Also, `arg(z_(1)z_(2))=0orarg(z_(1)+arg(z_(2))=0`
`rArr arg(z_(2))=-theta_(1)`
`rArr z_(2)=|z_(2)|(cos(-theta_(1))+i sin(-theta_(1)))`
`=|z_(1)|(costheta_(1)-isintheta_(1))=barz_(1)`
`rArr |z_(2)|=bar((barz_(1)))=z_(1)`
`rArr|z_(2)|^(2)=z_(1)z_(2)`