`|z_1|=|z_2|` and `arg((z_1)/(z_2))=pi`, then `z_1+z_2` is equal to

Options

A. 0
B. purely imaginary
C. purely real
D. none of these

Answer

A

Explanation

We have
`arg((z_(1))/(z_(2)))=pi`
or `arg(z_(1))-arg(z_(2))=pi`
or `arg(z_(1))=arg(z_(2))+pi`
Let `arg(z_(2))=theta."Then "arg(z_(1))=pi+theta`.
`therefore z_(1)=|z_(1)|[cos(pi+theta)+i sin(pi+theta)]`
`=|z_(1)|(-costheta-isintheta)` and `z_(2)=|z^(2)|(costheta+i sinetheta)`
`=|z_(1)|(costheta+i sintheta)" " (therefore|z_(1)|=|z_(2)|)`
`=-z_(1)`
`rArr z_(1)+z_(2)=0`