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:
We need to evaluate a complex algebraic expression given a specific value of \(z\).
By simplifying the expression, we can extract the scalar multiplier \(\lambda\).
Step 2: Key Formula or Approach:
Calculate \(z^2\) and \(\bar{z}\) explicitly.
Substitute them into \(\frac{24\bar{z}}{z^2}\) and simplify.
Step 3: Detailed Explanation:
Given \(z = 1 + i\).
The conjugate is \(\bar{z} = 1 - i\).
The square is \(z^2 = (1 + i)^2 = 1 + 2i - 1 = 2i\).
Evaluate the fraction:
\[ \frac{24\bar{z}}{z^2} = \frac{24(1 - i)}{2i} = \frac{12(1 - i)}{i} \] Multiply numerator and denominator by \(i\) to remove it from the bottom:
\[ = \frac{12(1 - i)i}{i^2} = \frac{12(i - (-1))}{-1} = -12(1 + i) \] Notice that \(-12(1 + i) = -12z\).
Now substitute back into the equation:
\[ z - (-12z) = \lambda z \] \[ 13z = \lambda z \] Since \(z \neq 0\), we have \(\lambda = 13\).
Step 4: Final Answer:
The value of \(\lambda\) is 13.