Question:

Evaluate $\left[i^{22} + \left(\dfrac{1}{i}\right)^{25}\right]^3$

Show Hint

Use modulo 4 pattern for powers of $i$ and simplify step-by-step.
Updated On: Apr 15, 2025
  • $4(i - 1)$
  • $2 - 7i$
  • $i - 1$
  • $2(1 - i)$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation


Recall: $i^4 = 1$, so powers of $i$ repeat every 4 terms.
\[ i^{22} = i^{(4 \times 5 + 2)} = i^2 = -1
\left(\dfrac{1}{i}\right)^{25} = (-i)^{25} = (-1)^{25} i^{25} = -i
\Rightarrow i^{22} + \left(\dfrac{1}{i}\right)^{25} = -1 - i
\text{So, expression becomes: } (-1 - i)^3
= (-1)^3(1 + i)^3 = - (1 + i)^3 = - (1 + 3i - 3 - i) = - (1 + 3i - 3 - i) \text{(Expand)}
= - (2 + 2i) = -2(1 + i) \]
simplify:
\[ (-1 - i)^3 = (-1 - i)(-1 - i)(-1 - i) = (-1 - i)^2 \cdot (-1 - i) = (1 + 2i + i^2)(-1 - i) = (1 + 2i - 1)(-1 - i) = (2i)(-1 - i) = -2i - 2i^2 = -2i + 2 = 2 - 2i \]
Final answer: $(2 - 2i) = 2(1 - i)$
Was this answer helpful?
0
0