A complex number $$z$$ is said to be unimodular if $$|z| = 1$$. Let $$z_1$$ and $$z_2$$ are complex numbers such that $$\frac{z_1 - 2z_2}{2 - z_1\bar{z_2}}$$ is unimodular and $$z_2$$ is not unimodular, then the point $$z_1$$ lies on a

First, recall the definition: a complex number $$w$$ is called unimodular if its modulus equals 1, that is, $$|w| = 1$$.

According to the statement, the complex fraction

$$\frac{z_1 - 2z_2}{\,2 - z_1\bar z_2\,}$$

is unimodular. Therefore its modulus is 1, and we can write

$$\left|\frac{z_1 - 2z_2}{\,2 - z_1\bar z_2\,}\right| = 1.$$

Using the basic property of moduli $$\left|\dfrac{a}{b}\right| = \dfrac{|a|}{|b|}$$, we obtain

$$|z_1 - 2z_2| \;=\; |\,2 - z_1\bar z_2\,|.$$

To remove the square roots, we square both sides. (Since moduli are non-negative, squaring keeps the equality valid.) We get

$$(z_1 - 2z_2)\;(\bar z_1 - 2\bar z_2) \;=\; (\,2 - z_1\bar z_2)(\,2 - \bar z_1 z_2).$$

Now we expand each side term by term.

Left side:

$$$ \begin{aligned} (z_1 - 2z_2)(\bar z_1 - 2\bar z_2) &= z_1\bar z_1 - 2z_1\bar z_2 - 2z_2\bar z_1 + 4 z_2\bar z_2 \\ &= |z_1|^{2} \;-\; 2z_1\bar z_2 \;-\; 2\bar z_1 z_2 \;+\; 4|z_2|^{2}. \end{aligned} $$$

Right side:

$$$ \begin{aligned} (2 - z_1\bar z_2)(2 - \bar z_1 z_2) &= 4 \;-\; 2z_1\bar z_2 \;-\; 2\bar z_1 z_2 \;+\; |z_1|^{2}|z_2|^{2}. \end{aligned} $$$

Setting the two expansions equal gives

$$|z_1|^{2} - 2z_1\bar z_2 - 2\bar z_1 z_2 + 4|z_2|^{2} \;=\; 4 - 2z_1\bar z_2 - 2\bar z_1 z_2 + |z_1|^{2}|z_2|^{2}.$$

Notice that the mixed terms $$-2z_1\bar z_2 - 2\bar z_1 z_2$$ appear on both sides; they cancel out immediately. After cancelling, we have

$$|z_1|^{2} \;+\; 4|z_2|^{2} \;=\; 4 \;+\; |z_1|^{2}|z_2|^{2}.$$

We rearrange all terms to one side:

$$$ \begin{aligned} |z_1|^{2} + 4|z_2|^{2} - 4 - |z_1|^{2}|z_2|^{2} &= 0,\\ |z_1|^{2} - |z_1|^{2}|z_2|^{2} + 4|z_2|^{2} - 4 &= 0. \end{aligned} $$$

Factorising, first pull $$|z_1|^{2}$$ from the first two terms and $$4$$ from the last two terms:

$$|z_1|^{2}\bigl(1 - |z_2|^{2}\bigr) \;+\; 4\bigl(|z_2|^{2} - 1\bigr) = 0.$$

Next, observe that the bracket $$(|z_2|^{2} - 1)$$ is common (but with opposite sign) in both parts. We take it out to get

$$(|z_2|^{2} - 1)\,\bigl(\!-|z_1|^{2} + 4\bigr) = 0.$$

At this stage, the product of two factors equals zero, so at least one factor must be zero.

• If $$|z_2|^{2} - 1 = 0$$, then $$|z_2| = 1$$, i.e. $$z_2$$ would be unimodular. But the problem explicitly states that $$z_2$$ is not unimodular. Hence this factor cannot vanish.
• Therefore the only possibility is $$-\,|z_1|^{2} + 4 = 0.$$

Solving for $$|z_1|$$ gives

$$|z_1|^{2} = 4 \quad\Longrightarrow\quad |z_1| = 2.$$

The condition $$|z_1| = 2$$ describes all complex numbers $$z_1$$ whose distance from the origin is 2. Geometrically, this is a circle with centre at the origin and radius 2 in the complex plane (which corresponds to the Cartesian plane with $$x$$-axis as real part and $$y$$-axis as imaginary part).

Hence, the correct answer is Option D.