Question

If \[ \left|\frac{z_1-7z_2}{\,7-z_1\overline{z_2}\,}\right|=1 \quad \text{and} \quad |z_2|\neq 1, \text{ then } |z_1|\neq \]

• A. \(0\)
• B. \(1\)
• C. \(7\)
• D. \(\dfrac{1}{7}\)

Hint:

If \[ \left|\frac{z-a}{a-\bar z\,w}\right|=1 \] and \(|w|\neq 1\), then typically it leads to: \[ |z|=|a| \] Always square moduli and cancel common real parts.

Answer 1

The Correct Option is C

Step 1: Given \[ \left|\frac{z_1-7z_2}{7-z_1\overline{z_2}}\right|=1 \] which implies \[ |z_1-7z_2|=|7-z_1\overline{z_2}| \]
Step 2: Square both sides: \[ |z_1-7z_2|^2=|7-z_1\overline{z_2}|^2 \] \[ |z_1|^2-14\Re(z_1\overline{z_2})+49|z_2|^2 =49-14\Re(z_1\overline{z_2})+|z_1|^2|z_2|^2 \]
Step 3: Cancel the common term \( -14\Re(z_1\overline{z_2}) \) from both sides: \[ |z_1|^2+49|z_2|^2=49+|z_1|^2|z_2|^2 \]
Step 4: Rearranging, \[ |z_1|^2(1-|z_2|^2)=49(1-|z_2|^2) \]
Step 5: Since \( |z_2|\neq 1 \), we divide both sides by \( (1-|z_2|^2) \): \[ |z_1|^2=49 \] \[ \Rightarrow |z_1|=7 \]
Step 6: Hence, \( |z_1| \neq 7 \) is false, so the required answer is: \[ \boxed{7} \]