Question

Given that \( z \) is a real number and \( z = \frac{\lambda + 4i}{1+\lambda i} \) where \( \lambda \in R \), then the possible value of \( \lambda \) is:

• A. \( -2 \)
• B. \( 2i \)
• C. \( 5 \)
• D. \( \pm 2i \)

Hint:

For a complex number to be real, its imaginary part must be zeroRationalize first and then equate the imaginary part to zero.

Answer 1

The Correct Option is A

Step 1: Write the given complex number.
\[ z = \frac{\lambda + 4i}{1+\lambda i} \]
Given that \( z \) is real and \( \lambda \in R \).

Step 2: Rationalize the denominator.
Multiply numerator and denominator by the conjugate of \( 1+\lambda i \), which is \( 1-\lambda i \):
\[ z = \frac{(\lambda + 4i)(1-\lambda i)}{(1+\lambda i)(1-\lambda i)} \]

Step 3: Simplify the denominator.
\[ (1+\lambda i)(1-\lambda i) = 1 - \lambda^2 i^2 \]
Since \( i^2 = -1 \):
\[ = 1 + \lambda^2 \]

Step 4: Simplify the numerator.
\[ (\lambda + 4i)(1-\lambda i) \]
\[ = \lambda - \lambda^2 i + 4i - 4\lambda i^2 \]
Since \( i^2 = -1 \):
\[ = \lambda - \lambda^2 i + 4i + 4\lambda \]
\[ = 5\lambda + (4-\lambda^2)i \]

Step 5: Write \( z \) in real and imaginary form.
\[ z = \frac{5\lambda + (4-\lambda^2)i}{1+\lambda^2} \]
\[ z = \frac{5\lambda}{1+\lambda^2} + \frac{4-\lambda^2}{1+\lambda^2}i \]

Step 6: Use the condition that \( z \) is real.
For \( z \) to be real, imaginary part must be zero.
\[ \frac{4-\lambda^2}{1+\lambda^2} = 0 \]
\[ 4-\lambda^2 = 0 \]
\[ \lambda^2 = 4 \]
\[ \lambda = \pm 2 \]

Step 7: Match with the given options.
Among the given options, only \( -2 \) is a real possible value.
\[ \boxed{-2} \]