Question

If $z_1$ and $z_2$ are non zero complex numbers such that $|z_1-z_2|=|z_1|+|z_2|$ then

• A. $|arg{z}_1-arg{z}_2|=\pi$
• B. $arg{z}_1=arg{z}_2$
• C. $z_1+k{z}_2$ for some positive number $k$
• D. $z_1\overline{z_1}+\overline{z_2}<0$

Answer 1

Answer: (A), (C), (D)

$|arg{z}_1-arg{z}_2|=\pi$
$z_1+k{z}_2$ for some positive number $k$
$z_1\overline{z_1}+\overline{z_2}<0$

$A.$

We have,

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

$\Rightarrow$ $|z_1-z_2{|}^2={(|z_1|+|z_2|)}^2$

$\Rightarrow$ $|z_1{|}^2+|z_2{|}^2-2|z_1||z_2|\cos \theta =|z_1{|}^2+|z_2{|}^2+2|z_1||z_2|$

$\Rightarrow$ $\cos \theta =-1$ where, where $\theta =arg{z}_1-\arg z_2$

$\Rightarrow$ $\theta =\pi$

$\therefore$ $|arg{z}_1-\arg z_2|=\pi$

$C.$

$R\ni k>0$

and

$z_1=k{z}_2$

then

$z_1\overline{z_2}=k{z}_2\overline{z_2}>0$

since

$z_2\overline{z_2}>0$

going the other way if,

$z_2\overline{z_2}=l>0$

then

$z_1{z}_2\overline{z_2}=l{z}_2$

from which

$z_1=\frac{l}{z_2\overline{z_2}}{z}_2$ ---- ( 1 )

we take

$k=\frac{1}{z_2\overline{z_2}}{z}_2$

Then ( 1 ) becomes,

$z_1=k{z}_2,$ $k>0$

$D.$

$Re\left(\frac{z_1}{z_2}\right)\le 0$

$\Rightarrow$ $\frac{z_1}{z_2}+\frac{\overline{z_1}}{\overline{z_2}}\le 0$ or $z_1\overline{z_2}+z_2\overline{z_1}\le 0$