Question

Argument of the complex number z = \( \frac{13-5i}{4-9i} \), i = \( \sqrt{-1} \) is

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{2} \)
• C. \( \pi \)
• D. \( \frac{3\pi}{2} \)

Hint:

To divide complex numbers, multiply by the conjugate of the denominator to make it a real number.

Answer 1

The Correct Option is A

Step 1: Simplify z
Multiply numerator and denominator by the conjugate $(4+9i)$:
$z = \frac{(13-5i)(4+9i)}{(4-9i)(4+9i)} = \frac{52 + 117i - 20i - 45i^2}{16 + 81}$
$z = \frac{52 + 97i + 45}{97} = \frac{97 + 97i}{97} = 1 + i$.
Step 2: Find Argument
For $z = x + iy$, $\arg(z) = \tan^{-1}(\frac{y}{x})$.
$\arg(z) = \tan^{-1}(\frac{1}{1}) = \tan^{-1}(1)$.
Step 3: Conclusion
$\arg(z) = \frac{\pi}{4}$.
Final Answer:(A)