Question:
The modulus of \( \frac{1+i}{1-i} - \frac{1-i}{1+i} \) is
The modulus of \( \frac{1+i}{1-i} - \frac{1-i}{1+i} \) is
Show Hint
Always rationalize denominator first when dealing with complex fractions.
Updated On: May 1, 2026
- \(2 \)
- \( \sqrt{2} \)
- \(4 \)
- \(8 \)
- \(10 \)
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The Correct Option is A
Solution and Explanation
Concept:
To find modulus:
\[
|z| = \sqrt{a^2 + b^2}
\]
Also simplify complex fractions using conjugates.
Step 1: Simplify first term.
\[ \frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{1+1} = \frac{1 + 2i + i^2}{2} = \frac{2i}{2} = i \]
Step 2: Simplify second term.
\[ \frac{1-i}{1+i} \cdot \frac{1-i}{1-i} = \frac{(1-i)^2}{2} = \frac{1 -2i + i^2}{2} = \frac{-2i}{2} = -i \]
Step 3: Subtract.
\[ i - (-i) = 2i \]
Step 4: Find modulus.
\[ |2i| = 2 \]
Step 1: Simplify first term.
\[ \frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{1+1} = \frac{1 + 2i + i^2}{2} = \frac{2i}{2} = i \]
Step 2: Simplify second term.
\[ \frac{1-i}{1+i} \cdot \frac{1-i}{1-i} = \frac{(1-i)^2}{2} = \frac{1 -2i + i^2}{2} = \frac{-2i}{2} = -i \]
Step 3: Subtract.
\[ i - (-i) = 2i \]
Step 4: Find modulus.
\[ |2i| = 2 \]
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