Question

Complex number $z$ satisfy the equation $|z-(4/z\left)\right|=2$. 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

Answer: (B)

line parallel to the imaginary axis

Squaring the given equation we have,

$(z-\frac{4}{z})(\overline{z}-\frac{4}{\overline{z}})=4$

$\therefore |z{|}^2-4(\frac{z}{\overline{z}}+\frac{\overline{z}}{z})+\frac{16}{|z{|}^2}=4$

Let,

$z=r{e}^{i\theta }$

$\therefore r^2-4(e^{i2\theta }+e^{-i2\theta })+\frac{16}{r^2}=4$

$\therefore r^2-8\left(\cos 2\theta \right)+\frac{16}{r^2}=4$

$\therefore r^2+\frac{16}{r^2}=4(1+2\left(\cos 2\theta \right))\dots \dots$(1)

For greatest or least modulus as $r=f\left(\theta \right)$, $\frac{dr}{d\theta }=0$.

This gives $\sin 2\theta =0$

$\therefore \cos 2\theta =1\dots \dots$($\cos 2\theta$ cannot be $-1$ as it will not satisfy equation (1))

$\therefore r^2+\frac{16}{r^2}=12$

Solving for $r$, we get $r=\sqrt{5}\pm 1$

$r_{max}=\sqrt{5}+1$ and $r_{min}=\sqrt{5}-1$

These numbers lie on real axis as $2\theta =2n\pi$, $\theta =n\pi$

$|z-z_1|=|z-z_2|$ is locus of line equidistant from $z_1,z_2$ which is perpendicular bisector. Hence, it is parallel to imaginary axis.