Question

Locus of z if $|z-z_1|=|z-z_2|$, where $z_1$ and $z_2$ are complex numbers with the greatest and the least moduli, is

• A. Line parallel to the real axis
• B. Line parallel to the imaginary axis
• C. Line having a positive slope
• D. Line having a negative slope

Answer 1

The correct option is B Line parallel to the imaginary axis

We have
$||z|-\left|\frac{4}{z}\right||\le |z-\frac{4}{z}|=2$

$\Rightarrow -2\le |z|-\frac{4}{|z|}\le 2$

$\Rightarrow |z{|}^2+2|z|-4\ge 0$ and $|z{|}^2-2|z|-4\le 0$

$\Rightarrow (|z|+1{)}^2-5\ge 0$ and $(|z|-1{)}^2\le 5$

$\Rightarrow (|z|+1+\sqrt{5})(|z|+1-\sqrt{5})\ge 0$

and $(|z|-1+\sqrt{5})\times (|z|-1-\sqrt{5})\le 0$

$\Rightarrow |z|\le -\sqrt{5}-1$ or $|z|\ge \sqrt{5}-1$ and $\sqrt{5}-1\le |z|\le \sqrt{5}+1$

$\Rightarrow \sqrt{5}-1\le |z|\le \sqrt{5}+1$

Hence, the least modulus is $\sqrt{5}-1$ and the greatest modulus is $\sqrt{5}+1$, Also,

$|z|=\sqrt{5}+1\Rightarrow \frac{4}{|z|}=\sqrt{5}-1$

Now, $\frac{4}{z}=\frac{4\overline{z}}{|z{|}^2}$

Hence, 4/z lies in the direction of $\overline{z}$.

$|z-\frac{4}{z}|=PR=2$ (given)

We have
$OP=\sqrt{5}+1$ and $OR=\sqrt{5}-1$

$\Rightarrow cos2\theta =\frac{O{P}^2+O{R}^2-P{R}^2}{2OP\cdot OR}$

$=\frac{(\sqrt{5}+1{)}^2(\sqrt{5}-1{)}^2-4}{2(5-1)}=1$

$\Rightarrow 2\theta =0,2\pi$

$\Rightarrow \theta =0,\pi$

$\Rightarrow$ z is purely real

$\Rightarrow z=\pm (\sqrt{5}+1)$

Similarly for $|z|=\sqrt{5}-1$, we have $z=\pm (\sqrt{5}-1)$.