Question

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

• A. $\frac{\pi }{4}$
• B. $-\frac{\pi }{2}$
• C. $0$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $0$

Let $z_1=r_1(\cos {\theta }_1+i\sin {\theta }_1),z_2=r_2(\cos {\theta }_2+i\sin {\theta }_2)$
$\text{But}|z_1-z_2|=||z_1|-|z_2||,$
$\Rightarrow |(r_1\cos {\theta }_1-r_2\cos {\theta }_2)+i(r_1\sin {\theta }_1-r_2\sin {\theta }_2)|=|r_1-r_2|$
Squaring both the sides
$\Rightarrow (r_1\cos {\theta }_1-r_2\cos {\theta }_2{)}^2+(r_1\sin {\theta }_1-r_2\sin {\theta }_2{)}^2=|r_1-r_2{|}^2$
$\Rightarrow r_1^2+r_2^2-2r_1{r}_2[\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2]=r_1^2+r_2^2-2r_1{r}_2$
$\Rightarrow \cos ({\theta }_1-{\theta }_2)=1$
$\Rightarrow {\theta }_1-{\theta }_2=0\Rightarrow \arg \left(z_1\right)-\arg \left(z_2\right)=0$

$\text{Alternate Solution}$
We know that,
$|z_1-z_2{|}^2=|z_1{|}^2+|z_2{|}^2-2Re\left(z_1\overline{z_2}\right)\dots \left(1\right)$
Also $Re\left(z_1\overline{z_2}\right)=|z_1\overline{z_2}|\cos \left(\arg \right(z_1\overline{z_2}\left)\right)$
$=|z_1\overline{z_2}|\cos (\arg z_1+\arg \overline{z_2})$
$[\because |z_1{z}_2|=|z_1||z_2|,|\overline{z}|=|z|,\arg \overline{z}=-\arg z]$
$=|z_1||z_2|\cos (\arg z_1-\arg z_2)$
Now, from equation $\left(1\right)$
$|z_1-z_2{|}^2=|z_1{|}^2+|z_2{|}^2-2|z_1||z_2|\cos ({\theta }_1-{\theta }_2)\dots \left(2\right)$
where ${\theta }_1=\arg \left(z_1\right)$ and ${\theta }_2=\arg \left(z_2\right)$.
We are given that $|z_1-z_2|=||z_1|-|z_2||$
Squaring both the sides
$|z_1-z_2{|}^2=||z_1|-|z_2|{|}^2$
From equation $\left(2\right)$
$|z_1{|}^2+|z_2{|}^2-2|z_1||z_2|\cos ({\theta }_1-{\theta }_2)=|z_1{|}^2+|z_2{|}^2-2|z_1||z_2|$
$\Rightarrow \cos ({\theta }_1-{\theta }_2)=1$
$\Rightarrow {\theta }_1-{\theta }_2=0$
$\Rightarrow \arg z_1-\arg z_2=0$