Question

Let $z = 1 + i$, where $i = \sqrt{-1}$. If $z - \frac{24\bar{z}}{z^2} = \lambda z$, then the value of $\lambda$ is equal to

• A. 12
• B. 13
• C. 18
• D. 23
• E. 24

Hint:

A useful property for \( z = 1+i \) is that \( z^2 = 2i \). Also, whenever you see a fraction with $i$ in the denominator, multiplying by $-i$ is often faster than multiplying by $i$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves complex number arithmetic including squares and conjugates. We need to evaluate the term \( \frac{24\bar{z}}{z^2} \) for the given value of \( z \).

Step 2: Detailed Explanation:
Given \( z = 1 + i \).
Calculate \( \bar{z} = 1 - i \).
Calculate \( z^2 = (1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i \).
Now calculate the fraction:
\[ \frac{24\bar{z}}{z^2} = \frac{24(1-i)}{2i} = \frac{12(1-i)}{i} \]
Rationalize the denominator by multiplying by \( -i/ -i \):
\[ \frac{12(1-i)(-i)}{1} = 12(-i + i^2) = 12(-i - 1) = -12(1 + i) = -12z \]
Substitute this back into the original equation:
\[ z - (-12z) = \lambda z \]
\[ z + 12z = \lambda z \]
\[ 13z = \lambda z \]
Thus, \( \lambda = 13 \).

Step 3: Final Answer:
The value of $\lambda$ is 13.