Q104 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\text { The modulus of the following complex number } \dfrac{1+i}{1-i}-\dfrac{1-i}{1+i} \text { is }$
- Correct answerA2
- B1
- C16
- D0
Solution
Let $z = \dfrac{1+i}{1-i} - \dfrac{1-i}{1+i}$.
Simplify the first term by multiplying the numerator and denominator by the conjugate of the denominator $(1+i)$:
$\dfrac{1+i}{1-i} = \dfrac{(1+i)(1+i)}{(1-i)(1+i)} = \dfrac{1 + 2i + i^2}{1^2 - i^2} = \dfrac{1 + 2i - 1}{1 + 1} = \dfrac{2i}{2} = i$.
Simplify the second term by multiplying the numerator and denominator by the conjugate of the denominator $(1-i)$:
$\dfrac{1-i}{1+i} = \dfrac{(1-i)(1-i)}{(1+i)(1-i)} = \dfrac{1 - 2i + i^2}{1^2 - i^2} = \dfrac{1 - 2i - 1}{1 + 1} = \dfrac{-2i}{2} = -i$.
Substitute these values back into the expression for $z$:
$z = i - (-i) = i + i = 2i$.
The modulus of $z$ is $|z| = |2i| = \sqrt{0^2 + 2^2} = 2$.
Answer: 2
Simplify the first term by multiplying the numerator and denominator by the conjugate of the denominator $(1+i)$:
$\dfrac{1+i}{1-i} = \dfrac{(1+i)(1+i)}{(1-i)(1+i)} = \dfrac{1 + 2i + i^2}{1^2 - i^2} = \dfrac{1 + 2i - 1}{1 + 1} = \dfrac{2i}{2} = i$.
Simplify the second term by multiplying the numerator and denominator by the conjugate of the denominator $(1-i)$:
$\dfrac{1-i}{1+i} = \dfrac{(1-i)(1-i)}{(1+i)(1-i)} = \dfrac{1 - 2i + i^2}{1^2 - i^2} = \dfrac{1 - 2i - 1}{1 + 1} = \dfrac{-2i}{2} = -i$.
Substitute these values back into the expression for $z$:
$z = i - (-i) = i + i = 2i$.
The modulus of $z$ is $|z| = |2i| = \sqrt{0^2 + 2^2} = 2$.
Answer: 2
Hence, the correct answer is option: 'A'.