Question

Let $z_1, z_2$, and $z_3$ be complex numbers satisfying the following conditions:
\[ 2 = |2z_1| = |z_2 - 1| = |z_3 + 1| = \left| \frac{1}{z_1} + \frac{1}{z_2 - 1} + \frac{1}{z_3 + 1} \right| \]
What is the value of $|4z_1 + z_2 + z_3|$?

• A. 8
• B. 4
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Hint:

For any complex equation involving sums of reciprocals $\frac{1}{w}$, always think of substituting $\frac{1}{w} = \frac{\bar{w}}{|w|^2}$.
This is a standard technique that converts reciprocals into direct linear terms of conjugates, which are much easier to simplify.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For any complex number \(w\), if \(|w| = R\), then \(\frac{1}{w} = \frac{\bar{w}}{R^2}\). We apply this to the given terms.
Step 2: Simplifying the Individual Moduli:
Given \(|2z_1| = 2 \implies |z_1| = 1 \). Let \(A = z_1\), then \(|A| = 1\).
Let \(B = z_2 - 1\), then \(|B| = 2\).
Let \(C = z_3 + 1\), then \(|C| = 2\).
Step 3: Transforming the Summation:
Given \(\left| \frac{1}{A} + \frac{1}{B} + \frac{1}{C} \right| = 2\).
Substitute \(\frac{1}{A} = \frac{\bar{A}}{|A|^2} = \bar{A}\), \(\frac{1}{B} = \frac{\bar{B}}{4}\), and \(\frac{1}{C} = \frac{\bar{C}}{4}\):
\(\left| \bar{A} + \frac{\bar{B}}{4} + \frac{\bar{C}}{4} \right| = 2 \implies \left| \frac{4\bar{A} + \bar{B} + \bar{C}}{4} \right| = 2 \implies |4\bar{A} + \bar{B} + \bar{C}| = 8\).
Since \(|w| = |\bar{w}|\), we have \(|4A + B + C| = 8\).
Step 4: Finding the Final Value:
Substitute \(A, B,\) and \(C\) back:
\(|4z_1 + (z_2 - 1) + (z_3 + 1)| = 8 \implies |4z_1 + z_2 + z_3| = 8\).
{Correction Note: Based on the specific values in the prompt vs options, let's re-verify the constant. If the target is \(|4z_1 + z_2 + z_3|\), the calculation leads to 8. If the prompt intended \(|z_1 + \dots|\), results vary. Given the options, (B) 4 often appears in variations where the initial constant is 1.}
Step 5: Final Answer:
Based on the derivation, the result is 8 (Option a).