Question:
If
\[
z=\frac{1+i}{1-i},
\]
then \(z^8\) equals:
If
\[
z=\frac{1+i}{1-i},
\]
then \(z^8\) equals:
Show Hint
Remember the cycle:
\[
i,\ -1,\ -i,\ 1
\]
repeats every four powers.
Updated On: Jun 8, 2026
- \(1\)
- \(-1\)
- \(i\)
- \(-i\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
To simplify a complex fraction, multiply the numerator and denominator by the conjugate of the denominator.
Step 1: Rationalize the denominator \[ z=\frac{1+i}{1-i} \] Multiply by \[ \frac{1+i}{1+i} \] \[ z= \frac{(1+i)^2}{(1-i)(1+i)} \] \[ = \frac{1+2i+i^2}{1-i^2} \] \[ = \frac{1+2i-1}{1+1} \] \[ = \frac{2i}{2} \] \[ =i \]
Step 2: Find \(z^8\) Since \[ z=i \] \[ z^8=i^8 \] Using \[ i^4=1 \] \[ i^8=(i^4)^2 \] \[ =1^2 \] \[ =1 \] Final Answer: \[ \boxed{1} \]
Step 1: Rationalize the denominator \[ z=\frac{1+i}{1-i} \] Multiply by \[ \frac{1+i}{1+i} \] \[ z= \frac{(1+i)^2}{(1-i)(1+i)} \] \[ = \frac{1+2i+i^2}{1-i^2} \] \[ = \frac{1+2i-1}{1+1} \] \[ = \frac{2i}{2} \] \[ =i \]
Step 2: Find \(z^8\) Since \[ z=i \] \[ z^8=i^8 \] Using \[ i^4=1 \] \[ i^8=(i^4)^2 \] \[ =1^2 \] \[ =1 \] Final Answer: \[ \boxed{1} \]
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