Key Concepts and Formulas

• $|z|^2 = z\bar{z}$, where $z$ is a complex number and $\bar{z}$ is its conjugate.
• $|z_1/z_2| = |z_1|/|z_2|$ for $z_2 \neq 0$.
• $|kz| = |k||z|$ for any complex number $z$ and scalar $k$.
• $|z-c|=r$ represents a circle centered at $c$ with radius $r$ in the complex plane.

Step-by-Step Solution

• Step 1: Separate the moduli and simplify the expression. We start with the given equation: $\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2$ Using the property $|z_1/z_2| = |z_1|/|z_2|$, we separate the moduli: $\frac{|z_1-2 z_2|}{\left|\frac{1}{2}-z_1 \bar{z}_2\right|}=2$ Multiplying both sides by the denominator's modulus, we get: $|z_1-2 z_2|=2\left|\frac{1}{2}-z_1 \bar{z}_2\right|$ Moving the scalar 2 inside the modulus on the right-hand side using $|kz| = |k||z|$, we have: $|z_1-2 z_2|=\left|2\left(\frac{1}{2}-z_1 \bar{z}_2\right)\right|$ $|z_1-2 z_2|=|1-2 z_1 \bar{z}_2|$ This step simplifies the equation by isolating the moduli of expressions involving $z_1$ and $z_2$.

• Step 2: Square both sides and apply the $z\bar{z} = |z|^2$ identity. Squaring both sides of the equation $|z_1-2 z_2|=|1-2 z_1 \bar{z}_2|$, we get: $|z_1-2 z_2|^2=|1-2 z_1 \bar{z}_2|^2$ Using the identity $|z|^2 = z\bar{z}$, we have: $(z_1-2 z_2)(\overline{z_1-2 z_2})=(1-2 z_1 \bar{z}_2)(\overline{1-2 z_1 \bar{z}_2})$ Applying the conjugate properties $\overline{A \pm B} = \bar{A} \pm \bar{B}$ and $\overline{AB} = \bar{A}\bar{B}$, we get: $(z_1-2 z_2)(\bar{z}_1-2 \bar{z}_2)=(1-2 z_1 \bar{z}_2)(1-2 \bar{z}_1 z_2)$ This step eliminates the modulus signs and transforms the equation into an algebraic form using complex conjugation.

• Step 3: Expand both sides and simplify the equation. Expanding both sides of the equation: Left Hand Side (LHS): $(z_1-2 z_2)(\bar{z}_1-2 \bar{z}_2) = z_1\bar{z}_1 - 2z_1\bar{z}_2 - 2z_2\bar{z}_1 + 4z_2\bar{z}_2$ Using $z\bar{z} = |z|^2$, this becomes: $\text{LHS} = |z_1|^2 - 2z_1\bar{z}_2 - 2\bar{z}_1z_2 + 4|z_2|^2$ Right Hand Side (RHS): $(1-2 z_1 \bar{z}_2)(1-2 \bar{z}_1 z_2) = 1 - 2\bar{z}_1 z_2 - 2z_1 \bar{z}_2 + 4 z_1\bar{z}_1 z_2\bar{z}_2$ Again, using $z\bar{z} = |z|^2$: $\text{RHS} = 1 - 2\bar{z}_1 z_2 - 2z_1 \bar{z}_2 + 4|z_1|^2|z_2|^2$ Equating the simplified LHS and RHS: $|z_1|^2 - 2z_1\bar{z}_2 - 2\bar{z}_1z_2 + 4|z_2|^2 = 1 - 2\bar{z}_1 z_2 - 2z_1 \bar{z}_2 + 4|z_1|^2|z_2|^2$ Canceling the terms $- 2z_1\bar{z}_2 - 2\bar{z}_1z_2$ on both sides, we get: $|z_1|^2 + 4|z_2|^2 = 1 + 4|z_1|^2|z_2|^2$ This step involves careful algebraic expansion, leading to an expression purely in terms of the moduli.

• Step 4: Rearrange and Factorize the expression. Rearranging the terms to one side: $|z_1|^2 + 4|z_2|^2 - 1 - 4|z_1|^2|z_2|^2 = 0$ Grouping terms strategically to factorize the expression: $(|z_1|^2 - 1) + (4|z_2|^2 - 4|z_1|^2|z_2|^2) = 0$ Factoring out $4|z_2|^2$ from the second group of terms: $(|z_1|^2 - 1) + 4|z_2|^2(1 - |z_1|^2) = 0$ Rewriting $(1 - |z_1|^2)$ as $-(|z_1|^2 - 1)$: $(|z_1|^2 - 1) - 4|z_2|^2(|z_1|^2 - 1) = 0$ Factoring out the common term $(|z_1|^2 - 1)$: $(|z_1|^2 - 1)(1 - 4|z_2|^2) = 0$ This step uses factorization to simplify the equation and find possible solutions.

• Step 5: Determine the conditions for $|z_1|$ and $|z_2|$. Using the zero-product property, we have two possible cases:

  Case 1: The first factor is zero. $|z_1|^2 - 1 = 0$ $|z_1|^2 = 1$ $|z_1| = 1$

  Case 2: The second factor is zero. $1 - 4|z_2|^2 = 0$ $4|z_2|^2 = 1$ $|z_2|^2 = \frac{1}{4}$ $|z_2| = \frac{1}{2}$ Therefore, either $|z_1|=1$ or $|z_2|=\frac{1}{2}$.

Common Mistakes & Tips

• Remember the modulus properties, especially $|z|^2 = z\bar{z}$.
• Be careful with conjugate arithmetic and algebraic manipulations.
• Understand "either/or" vs. "and" conditions.

Summary

By applying the properties of complex number moduli and using careful algebraic manipulation, we deduced that either $|z_1|=1$ or $|z_2|=\frac{1}{2}$. Geometrically, this means that either $z_1$ lies on the unit circle or $z_2$ lies on a circle of radius $\frac{1}{2}$, both centered at the origin. This corresponds to option (A) in the given choices.

Final Answer

The final answer is \boxed{A}, which corresponds to option (A).