Question

If $z_1$, $z_2$ are the complex numbers such that $|z_1+z_2|=|z_1|+|z_2|$ then arg $z_1-$ arg $z_2$ is

• A. $-\pi$
• B. $\frac{-\pi }{2}$
• C. 0
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

0

Let
$z_1=x_1+i{y}_1$
$z_2=x_2+i{y}_2$
$z_1+z_2=(x_1+x_2)+i(y_1+y_2)$ ...(i)
Now
$|z_1+z_2|=|z_1|+|z_2|$
$\sqrt{(x_1+x_2{)}^2+(y_1+y_2{)}^2}=\sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2}$
Squaring both sides give us
$(x_1+x_2{)}^2+(y_1+y_2{)}^2=(x_1{)}^2+(x_2{)}^2+(y_1{)}^2+(y_2{)}^2+2\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}$
$(x_1{)}^2+(x_2{)}^2+(y_1{)}^2+(y_2{)}^2+2x_1{x}_2+2y_1{y}_2=(x_1{)}^2+(x_2{)}^2+(y_1{)}^2+(y_2{)}^2+2\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}$
$2x_1{x}_2+2y_1{y}_2=2\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}$
$(x_1{x}_2+y_1{y}_2{)}^2=x_1^2{x}_2^2+y_1^2{y}_2^2+x_1^2{y}_2^2+y_1^2{x}_2^2$
$x_1^2{x}_2^2+y_1^2{y}_2^2+2x_1{x}_2{y}_1{y}_2=x_1^2{x}_2^2+y_1^2{y}_2^2+x_1^2{y}_2^2+y_1^2{x}_2^2$
$2x_1{x}_2{y}_1{y}_2=x_1^2{y}_2^2+y_1^2{x}_2^2$
Or
$x_1^2{y}_2^2+y_1^2{x}_2^2-2x_1{x}_2{y}_1{y}_2=0$
Or
$(x_1{y}_2-x_2{y}_1{)}^2=0$
Or
$x_1{y}_2-x_2{y}_1=0$
$x_1{y}_2=x_2{y}_1$
Or
$\frac{x_1}{y_1}=\frac{x_2}{y_2}$
Hence $z_1$ and $z_2$ are colinear.