Q98 / 180 · COMEDK 2021 · Mathematics
What is the argument of the complex number $\dfrac{(1+i)(2+i)}{3-i}$, where $i=\sqrt{-1}$ ?
- A0
- B$\dfrac{\pi}{4}$
- C$-\dfrac{\pi}{4}$
- Correct answerD$\dfrac{\pi}{2}$
Solution
Let $z = \dfrac{(1+i)(2+i)}{3-i}$.
First, simplify the numerator:
$(1+i)(2+i) = 2 + i + 2i + i^{2} = 2 + 3i - 1 = 1 + 3i$.
Now, multiply the numerator and denominator by the conjugate of the denominator $(3+i)$:
$z = \dfrac{(1+3i)(3+i)}{(3-i)(3+i)} = \dfrac{3 + i + 9i + 3i^{2}}{3^{2} + 1^{2}} = \dfrac{3 + 10i - 3}{9 + 1} = \dfrac{10i}{10} = i$.
The complex number $z = i$ can be written in polar form as $0 + 1i$.
The argument of $z = i$ is $\text{arg}(i) = \dfrac{\pi}{2}$.
Answer: $\dfrac{\pi}{2}$
First, simplify the numerator:
$(1+i)(2+i) = 2 + i + 2i + i^{2} = 2 + 3i - 1 = 1 + 3i$.
Now, multiply the numerator and denominator by the conjugate of the denominator $(3+i)$:
$z = \dfrac{(1+3i)(3+i)}{(3-i)(3+i)} = \dfrac{3 + i + 9i + 3i^{2}}{3^{2} + 1^{2}} = \dfrac{3 + 10i - 3}{9 + 1} = \dfrac{10i}{10} = i$.
The complex number $z = i$ can be written in polar form as $0 + 1i$.
The argument of $z = i$ is $\text{arg}(i) = \dfrac{\pi}{2}$.
Answer: $\dfrac{\pi}{2}$
Hence, the correct answer is option: 'D'.