Let `z_1a n dz_2` be complex numbers such that `z_1!=z_2` and `|z_1|=|z_2|dot` If `z_1` has positive real part and `z_2` has negative imaginary part, then `(z_1+z_2)/(z_1-z_2)` may be zero (b) real and positive real and negative (d) purely imaginary

Options

A. purely imaginary
B. real and positive
C. real and negative
D. none of these

Answer

A

Explanation

Let `z_(1)=a+ibandz_(2)=c-"id", "where "agt0 and dgt0`. Then,
`|z_(1)|=|z_(2)|rArra^(2)+b^(2)=c^(2)+d^(2)" "(1)`
Now, `(z_(1)+z_(2))/(z_(1)-z_(2))=((a+ib)+(c-"id"))/((a+ib)-(c-"id"))`
`=([(a+c)+i(b-d)][(a-c)-i(b+d)])/([(a-c)+i(b+d)][(a-c)-i(b+d)])`
`=((a^(2)+b^(2))-(c^(2)+d^(2))-2(ad+bc)i)/(a^(2)+c^(2)-2ac+b^(2)+d^(2)+2bd)`
`=(-(ad+bc)i)/(a^(2)+b^(2)-ac+bd)" "["Using" (1)]`
Hence, `(z_(1)+z_(2))//(z_(1)-z_(2))` is purely imaginary. However, if `ad+bc=0`, then `(z_(1)+z_(2))//(z_(1)-z_(2))` will be equal to zero. According to the conditions of the equation, we can have `ad+bc=0`