Question

If z₁ and z₂ are two different complex numbers with |z₂|=1, then \(|\frac{1-\bar{z_1}z_2}{z_1-z_2}|\)is equal to

• A. 0
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{3}\)
• D. \(\frac{1}{4}\)
• E. 1

Answer 1

Answer: (E)

Given that \( |z_2| = 1 \), we know that \( z_2 \) lies on the unit circle in the complex plane. Also, \( z_1 \) and \( z_2 \) are different complex numbers. We need to evaluate the expression \( \left| \frac{1 - \overline{z_1}z_2}{z_1 - z_2} \right| \).
Let’s consider the geometric interpretation of this expression. The numerator involves \( \overline{z_1}z_2 \), which is a product of the conjugate of \( z_1 \) and \( z_2 \).
The denominator \( z_1 - z_2 \) represents the difference between \( z_1 \) and \( z_2 \).
Using properties of complex numbers and their magnitudes, this expression simplifies to 1.
The intuition behind this is that when \( |z_2| = 1 \), the distance between the points corresponding to \( z_1 \) and \( z_2 \) on the complex plane is normalized such that the magnitude of the given expression equals 1.

The correct option is (E) : \(1\)

Answer 2

The given expression is:
\[ \left|\frac{1 - \bar{z_1}z_2}{z_1 - z_2}\right| \]
Since \(|z_2| = 1\), we have:
\[ z_2^{-1} = \bar{z_2} \]
Thus, the expression simplifies to:
\[ \frac{|1 - \bar{z_1}z_2|}{|z_1 - z_2|} = \frac{|z_2^{-1}-\bar{z_1}|}{|z_1 - z_2|} = \frac{|\bar{z_2}-\bar{z_1}|}{|z_1 - z_2|} \]
Taking the modulus property:
\[ |\bar{z_2}-\bar{z_1}| = |\overline{z_2 - z_1}| = |z_2 - z_1| \]
Hence, we have:
\[ \frac{|z_2 - z_1|}{|z_1 - z_2|} = 1 \]
Thus, the value of the given expression is:
1