Q108 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics
The conjugate of the multiplicative inverse of the complex number $z = \dfrac{1 + 7i}{3 + i}$ is:
- A$1 + 2i$
- Correct answerB$\dfrac{1}{5} + \dfrac{2}{5}i$
- C$\dfrac{2}{5} + \dfrac{1}{5}i$
- D$\dfrac{1}{5} - \dfrac{2}{5}i$
Solution
$z = \dfrac{1 + 7i}{3 + i}$
Multiplying the numerator and denominator by the conjugate of the denominator:
$z = \dfrac{(1 + 7i)(3 - i)}{(3 + i)(3 - i)}$
$z = \dfrac{3 - i + 21i - 7i^2}{3^2 - i^2}$
$z = \dfrac{3 + 20i + 7}{9 + 1}$
$z = \dfrac{10 + 20i}{10} = 1 + 2i$
The multiplicative inverse of $z$ is $\dfrac{1}{z}$:
$\dfrac{1}{z} = \dfrac{1}{1 + 2i}$
$\dfrac{1}{z} = \dfrac{1 - 2i}{(1 + 2i)(1 - 2i)}$
$\dfrac{1}{z} = \dfrac{1 - 2i}{1^2 + 2^2} = \dfrac{1}{5} - \dfrac{2}{5}i$
The conjugate of the multiplicative inverse is:
$\overline{\left(\dfrac{1}{z}\right)} = \dfrac{1}{5} + \dfrac{2}{5}i$
Answer: $\dfrac{1}{5} + \dfrac{2}{5}i$
Multiplying the numerator and denominator by the conjugate of the denominator:
$z = \dfrac{(1 + 7i)(3 - i)}{(3 + i)(3 - i)}$
$z = \dfrac{3 - i + 21i - 7i^2}{3^2 - i^2}$
$z = \dfrac{3 + 20i + 7}{9 + 1}$
$z = \dfrac{10 + 20i}{10} = 1 + 2i$
The multiplicative inverse of $z$ is $\dfrac{1}{z}$:
$\dfrac{1}{z} = \dfrac{1}{1 + 2i}$
$\dfrac{1}{z} = \dfrac{1 - 2i}{(1 + 2i)(1 - 2i)}$
$\dfrac{1}{z} = \dfrac{1 - 2i}{1^2 + 2^2} = \dfrac{1}{5} - \dfrac{2}{5}i$
The conjugate of the multiplicative inverse is:
$\overline{\left(\dfrac{1}{z}\right)} = \dfrac{1}{5} + \dfrac{2}{5}i$
Answer: $\dfrac{1}{5} + \dfrac{2}{5}i$
Hence, the correct answer is option: 'B'.