Question:
Evaluate
\[
\left(\frac{1-i}{1+i}\right)^{2022} + \left(\frac{1+i}{1-i}\right)^{2021}
\]
Evaluate
\[
\left(\frac{1-i}{1+i}\right)^{2022} + \left(\frac{1+i}{1-i}\right)^{2021}
\]
Show Hint
Convert complex fractions to polar form before exponentiation; then use properties \(i^4=1\) and \((-i)^2=-1\) to simplify powers.
Updated On: Jul 18, 2026
- \(-i\)
- \(i\)
- \(i+1\)
- \(i-1\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Express in polar form.
\[ \frac{1-i}{1+i} = \frac{\sqrt{2} e^{-i\pi/4}}{\sqrt{2} e^{i\pi/4}} = e^{-i\pi/2} = -i \] \[ \frac{1+i}{1-i} = \frac{\sqrt{2} e^{i\pi/4}}{\sqrt{2} e^{-i\pi/4}} = e^{i\pi/2} = i \]
Step 2: Raise to powers.
\[ (-i)^{2022} = [(-i)^2]^{1011} = (-1)^{1011} = -1 \] \[ (i)^{2021} = i^{2020} \cdot i = (i^4)^{505} \cdot i = 1 \cdot i = i \]
Step 3: Add results.
\[ (-i)^{2022} + (i)^{2021} = -1 + i = i - 1 \]
Step 4: Final conclusion.
\[ \boxed{i - 1} \]
\[ \frac{1-i}{1+i} = \frac{\sqrt{2} e^{-i\pi/4}}{\sqrt{2} e^{i\pi/4}} = e^{-i\pi/2} = -i \] \[ \frac{1+i}{1-i} = \frac{\sqrt{2} e^{i\pi/4}}{\sqrt{2} e^{-i\pi/4}} = e^{i\pi/2} = i \]
Step 2: Raise to powers.
\[ (-i)^{2022} = [(-i)^2]^{1011} = (-1)^{1011} = -1 \] \[ (i)^{2021} = i^{2020} \cdot i = (i^4)^{505} \cdot i = 1 \cdot i = i \]
Step 3: Add results.
\[ (-i)^{2022} + (i)^{2021} = -1 + i = i - 1 \]
Step 4: Final conclusion.
\[ \boxed{i - 1} \]
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