Question

If \(i = \sqrt{-1}\) then \([i^{18}+(\frac{1}{i})^{25}]^3 =\)

• A. \(4+4i\)
• B. \(2+2i\)
• C. \(4-4i\)
• D. \(2-2i\)

Hint:

Reduce the powers of i using the cycle of four, then cube.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Powers of \(i\) repeat in a cycle of four: \(i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1\).

Step 2: First term
\(18=4\times4+2\), so \(i^{18}=i^2=-1\).

Step 3: Second term
\(\dfrac1i=-i\). So \(\left(\dfrac1i\right)^{25}=(-i)^{25}=-i^{25}\). Since \(25=4\times6+1\), \(i^{25}=i\). So the term is \(-i\).

Step 4: Bracket
\[ i^{18}+\left(\frac1i\right)^{25}=-1-i=-(1+i) \]

Step 5: Cube
\[ (1+i)^2=2i,\quad (1+i)^3=2i(1+i)=2i-2 \]
\[ [-(1+i)]^3=-(2i-2)=2-2i \]
This is option (D).

Final Answer:
The bracket equals \(-1-i\), and its cube is \(2-2i\), option (D). \[ \boxed{2-2i} \]