Question

If $z_1$ and $z_2$ are two non-zero complex numbers such that $\left|z_1-z_2\right|=\left|z_1\right|-\left|z_2\right|$ then $\arg z_1-\arg z_2$ is equal to

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

Answer 1

Answer: (A)

$0$

Let $z_1=x_1+y_{1}i$

$z_2=x_2+y_{2}i$

$|z_1-z_2|=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$

$|z_1|=\sqrt{x_1^2+y_1^2}$

$|z_2|=\sqrt{x_2^2+y_2^2}$

$|z_1-z_2|=|z_1|-|z_2|$

$\Rightarrow (x_1-x_2{)}^2+(y_1-y_2{)}^2=x_1^2+y_1^2+x_2^2+y_2^2-2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}$

$\Rightarrow x_1^2+x_2^2-2x_1{x}_2+y_1^2+y_2^2-2y_1{y}_2$$=x_1^2+y_1^2+x_2^2+y_2^2+2\sqrt{(x_1^2+y_1{)}^2(x_2^2+y_2^2)}$

$\Rightarrow -2(x_1{x}_2+y_1{y}_2)=2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}$

Square both sides

$\Rightarrow x_1^2{x}_2^2+y_1^2{y}_2^2+2x_1{x}_2{y}_1{y}_2=x_1^2{x}_2^2+y_1^2{x}_2^2+x_1^2{y}_2^2+y_1^2{y}_2^2$

$\Rightarrow 2x_1{x}_2{y}_1{y}_2=y_1^2{x}_2^2+x_1^2{y}_2^2$ ___- (1)

$arg{z}_1-arg{z}_2=\frac{y_1}{x_1}-\frac{y_2}{x_2}$

$A=\frac{y_1{x}_2-x_1{y}_2}{x_1{x}_2}$

$A^2=\frac{y_1^2{x}_2^2+x_1^2{y}_2^2-2x_1{x}_2{y}_1{y}_2}{x_1{x}_2}$

$A^2=\frac{0}{x_1{x}_2}$ (from 1)

$A=0$

A is correct