Question

Let \(Z_1\) and \(Z_2\) be two complex numbers, then:

• A. \(\overline{Z_1 + Z_2} \le \overline{Z_1} + \overline{Z_2}\)
• B. \(\overline{Z_1 + Z_2} \ge \overline{Z_1} + \overline{Z_2}\)
• C. \(\overline{Z_1 Z_2} = \overline{Z_1} \cdot \overline{Z_2}\)
• D. \(\overline{Z_1 Z_2} < \overline{Z_1} \cdot \overline{Z_2}\)

Hint:

Conjugation is a distributive operation over standard arithmetic.
This means that for any complex numbers \(Z_1\) and \(Z_2\):
- \(\overline{Z_1 \pm Z_2} = \overline{Z_1} \pm \overline{Z_2}\)
- \(\overline{Z_1 \cdot Z_2} = \overline{Z_1} \cdot \overline{Z_2}\)
- \(\overline{(Z_1 / Z_2)} = \overline{Z_1} / \overline{Z_2}\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem relates to the fundamental algebraic properties of complex conjugates.
A complex conjugate is formed by changing the sign of the imaginary part of a complex number.

Step 3: Detailed Explanation:
Let us define two arbitrary complex numbers:
\[ Z_1 = a + ib \]
\[ Z_2 = c + id \]
where \(a, b, c, d \in \mathbb{R}\) and \(i = \sqrt{-1}\).
The conjugates of these numbers are:
\[ \overline{Z_1} = a - ib \]
\[ \overline{Z_2} = c - id \]
Let us evaluate the product \(Z_1 Z_2\):
\[ Z_1 Z_2 = (a + ib)(c + id) = ac + iad + ibc + i^2 bd \]
Since \(i^2 = -1\):
\[ Z_1 Z_2 = (ac - bd) + i(ad + bc) \]
The conjugate of this product is:
\[ \overline{Z_1 Z_2} = (ac - bd) - i(ad + bc) \quad \text{--- (Equation 1)} \]
Now, let us calculate the product of the individual conjugates:
\[ \overline{Z_1} \cdot \overline{Z_2} = (a - ib)(c - id) \]
\[ \overline{Z_1} \cdot \overline{Z_2} = ac - iad - ibc + i^2 bd \]
Using \(i^2 = -1\):
\[ \overline{Z_1} \cdot \overline{Z_2} = (ac - bd) - i(ad + bc) \quad \text{--- (Equation 2)} \]
Comparing Equation 1 and Equation 2, we see that they are identical:
\[ \overline{Z_1 Z_2} = \overline{Z_1} \cdot \overline{Z_2} \]
This identity is a standard property of complex numbers, corresponding to Option (C).
The expressions in Options (A) and (B) are incorrect because the conjugate of a sum is strictly equal to the sum of the conjugates, with no inequalities.

Step 4: Final Answer:
Therefore, the correct option is (C).