Question

$\left|z_1\right|$ and $\left|z_2\right|$ are two distinct points in an Argand plane. If $a\left|z_1\right|=b\left|z_2\right|$ (where $a,b\in R$), then the point $\left(a{z}_1/b{z}_2\right)+\left(b{z}_2/a{z}_1\right)$ is a point on the

• A. line segment [-2, 2] of the real axis
• B. line segment [-2, 2] of the imaginary axis
• C. unit circle $\left|z\right|$ = 1
• D. the line with arg z = tan ${}^{-1}$2

Answer 1

The correct option is A line segment [-2, 2] of the real axis

$a\left|z_1\right|=b\left|z_2\right|(wherea,b\in R)$ ...(1)
$z_1=r_1(\cos A+i\sin A)\&z_2=r_2(\cos B+i\sin B)$
Since, $a\left|z_1\right|=b\left|z_2\right|$
$\Rightarrow a{r}_1=b{r}_2$
let $z_1=r_1(\cos A+i\sin A)=r_{1}cisA\&z_2=r_2(\cos B+i\sin B)=r_{2}cisB$
$w=\left(\frac{a{z}_1}{b{z}_2}\right)+\left(\frac{b{z}_2}{a{z}_1}\right)=\left(\frac{a{r}_{1}cisA}{b{r}_{2}cisB}\right)+\left(\frac{b{r}_{2}cisB}{a{r}_{1}cisA}\right)$
$\Rightarrow w=cis(A-B)+cis(-A+B)=2\cos (A-B)$ ...{$\because a{r}_1=b{r}_2$}

Since $w$ is real.
Therefore $w$ lies on a real axis.
Hence, option 'A' is correct.