Question

If $z_1, z_2$ are two distinct complex numbers such that \[ \frac{|z_1 - 2z_2|}{\left| \frac{1}{2} - z_1 \overline{z_2} \right|} = 2, \] then

• A. either $z_1$ lies on a circle of radius 1 or $z_2$ lies on a circle of radius $\frac{1}{2}$.
• B. either $z_1$ lies on a circle of radius $\frac{1}{2}$ or $z_2$ lies on a circle of radius 1.
• C. $z_1$ lies on a circle of radius $\frac{1}{2}$ and $z_2$ lies on a circle of radius 1.
• D. both $z_1$ and $z_2$ lie on the same circle.

Answer 1

The Correct Option is A

To solve the given problem, we start by examining the equation:

\(\frac{|z_1 - 2z_2|}{\left| \frac{1}{2} - z_1 \overline{z_2} \right|} = 2\)

This equation essentially represents a condition involving the distances between two points on the complex plane.

First, let's simplify the equation:

\(|z_1 - 2z_2| = 2 \left| \frac{1}{2} - z_1 \overline{z_2} \right|\)

Consider \(|z_1 - 2z_2|\)as the distance from \(z_1\) to \(2z_2\):

- If we assume that \(z_1\) lies on a circle of radius 1, then we have:

\(|z_1| = 1 \ \text{or} \ |z_2 - 0| = \frac{1}{2}\)

Similarly, consider the absolute value:

\(\left| \frac{1}{2} - z_1 \overline{z_2} \right|\)represents the modulus of a complex number.

- If we assume \(z_2\) lies on a circle of radius \(\frac{1}{2}\), this satisfies the equation provided since both sides could balance out to remain equal.

Therefore, the correct interpretation is that either:

• \(z_1\) lies on a circle of radius 1.
• or \(z_2\) lies on a circle of radius \(\frac{1}{2}\).

Conclusion: The correct answer is that either \(z_1\) lies on a circle of radius 1 or \(z_2\) lies on a circle of radius \(\frac{1}{2}\).

Answer 2

\[ \frac{z_1 - 2z_2}{\frac{1}{2} - 2z_1z_2} \times \frac{\overline{z_1} - 2\overline{z_2}}{\frac{1}{2} - z_1z_2} = 4 \]

\[ \lvert z_1 \rvert^2 \left\lvert 2z_1z_2 - 2z_2\overline{z_1} + 4\lvert z_2 \rvert^2 \right\rvert^2 \]

\[ = 4 \left( \frac{1}{4}(z_1\overline{z_2} - z_2\overline{z_1})^2 + \lvert z_1 \rvert^2 \lvert z_2 \rvert^2 \right) \]

\[ z_1\overline{z_1} + 2z_2 \cdot 2\overline{z_2} - z_1z_2 \cdot z_2\overline{z_2} - 1 = 0 \]

\[ (z_1\overline{z_1} - 1)(1 - 2z_2\overline{z_2}) = 0 \]

\[ (\lvert z_1 \rvert^2 - 1)\left((2\lvert z_2 \rvert^2 - 1)\right) = 0 \]