Question

If $z_1,z_2$ are two complex numbers such that $arg(z_1+z_2)=0$ and $Im\left(z_1{z}_2\right)=0$, then

• A. $z_1=-z_2$
• B. $z_1=z_2$
• C. $z_1=\overline{z_2}$
• D. $noneofthese$

Answer 1

The correct option is C $z_1=\overline{z_2}$

Let the complex numbers be,

$z_1=a+ib,z_2=x+iy$

$arg(z_1+z_2)=arg(a+x+i(b+y\left)\right)=0$

${\tan }^{-1}\left(\frac{b+y}{a+x}\right)=0$

$\frac{b+y}{a+x}=0$

$b+y=0$.....(1)

$Im\left(z_1{z}_2\right)=Im(ax-by+i(ay+bx\left)\right)=0$

$(ay+bx)=0$

from (1),

$ay=xy\Rightarrow a=x$

$\Rightarrow b=-y,a=x$

$\therefore z_1=x-iy,z_2=x+iy$

Hence proved, $z_1={\overline{z}}_2$