Question

Let \( i^2 = -1 \). Then \( \left(i^{10} - \frac{1}{i^{11}}\right) + \left(i^{11} - \frac{1}{i^{12}}\right) + \left(i^{12} - \frac{1}{i^{13}}\right) + \left(i^{13} - \frac{1}{i^{14}}\right) + \left(i^{14} + \frac{1}{i^{15}}\right) \) is equal to

• A. \( -1 + i \)
• B. \( -1 - i \)
• C. \( 1 + i \)
• D. \( -i \)
• E. \( i \)

Hint:

Always reduce powers of \(i\) using modulo 4 and convert reciprocals into negative powers.

Answer 1

The Correct Option is A

Concept: Powers of \(i\) repeat in cycle of 4: \[ i^1 = i,\quad i^2 = -1,\quad i^3 = -i,\quad i^4 = 1 \] Also, \[ \frac{1}{i^n} = i^{-n} \]

Step 1: Reduce powers modulo 4
\[ i^{10} = i^2 = -1,\quad i^{11} = i^3 = -i,\quad i^{12} = 1 \] \[ i^{13} = i,\quad i^{14} = -1,\quad i^{15} = -i \]

Step 2: Convert reciprocals
\[ \frac{1}{i^{11}} = i^{-11} = i^1 = i \] \[ \frac{1}{i^{12}} = 1 \] \[ \frac{1}{i^{13}} = i^{-13} = i^3 = -i \] \[ \frac{1}{i^{14}} = -1 \] \[ \frac{1}{i^{15}} = i^{-15} = i \]

Step 3: Substitute all values
\[ (-1 - i) + (-i - 1) + (1 + i) + (i + 1) + (-1 + i) \]

Step 4: Combine real and imaginary parts
Real part: \[ -1 -1 +1 +1 -1 = -1 \] Imaginary part: \[ - i - i + i + i + i = i \]

Step 5: Final result
\[ \boxed{-1 + i} \]