For two non-zero complex numbers z1 and z2, if \[\operatorname{Re}\left(z_{1} z_{2}\right)=0\] and \[\operatorname{Re}\left(z_{1}+z_{2}\right)=0\], then which of the following are possible?
A. \[\operatorname{Im}\left(z_{1}\right)>0\] and \[\operatorname{Im}\left(z_{2}\right) > 0\]
B. \[\operatorname{Im}\left(z_{1}\right) 0\]
C. \[\operatorname{Im}\left(z_{1}\right) > 0\]
Solutions · 1
Official solution
Let, z1=x1+iy1
and z2=x2+iy2
\[\therefore\] \[{z_1}{z_2} = {x_1}{x_2} - {y_1}{y_2} + i({x_1}{y_2} + {x_2}{y_1})\]
Given, \[{\mathop{\rm Re}\nolimits} ({z_1} + {z_2}) = 0\]
\[ \Rightarrow {x_1} + {x_2} = 0\] ...... (1)
also given, \[{\mathop{\rm Re}\nolimits} ({z_1}{z_2}) = 0\]
\[ \Rightarrow {x_1}{x_2} - {y_1}{y_2} = 0\]
\[ \Rightarrow {x_1}{x_2} = {y_1}{y_2}\]
\[ \Rightarrow {y_1}{y_2} = - x_1^2\] [\[\because\] x2=−x1]
So, multiplication of imaginary part's of z1 and z2 is negative. It means sign of y1 and y2 are opposite of each other.
\[ \therefore \] B and C are correct.