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

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

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

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

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

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

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