Question

Simplify \( i^{57} + \frac{1}{i^{25}} \) and find its value:

• A. 0
• B. \( 2i \)
• C. \( -2i \)
• D. 2

Hint:

When simplifying powers of \( i \), remember that the powers of \( i \) follow a repeating cycle of length 4. This allows you to find the equivalent smaller power by taking the remainder when dividing the exponent by 4.

Answer 1

The Correct Option is A

We are tasked with simplifying \( i^{57} + \frac{1}{i^{25}} \), where \( i \) is the imaginary unit, defined by \( i^2 = -1 \). 
Step 1: Simplifying \( i^{57} \) We know that powers of \( i \) cycle every four terms: \[ i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad \dots \] To simplify \( i^{57} \), we divide 57 by 4 and find the remainder: \[ 57 \div 4 = 14 { remainder } 1 \] Thus: \[ i^{57} = i^1 = i \] 
Step 2: Simplifying \( \frac{1}{i^{25}} \) Next, we simplify \( \frac{1}{i^{25}} \). Again, powers of \( i \) cycle every 4 terms. To simplify \( i^{25} \), we divide 25 by 4 and find the remainder: \[ 25 \div 4 = 6 { remainder } 1 \] Thus: \[ i^{25} = i^1 = i \] 
So: \[ \frac{1}{i^{25}} = \frac{1}{i} \] 
Now, we can multiply the numerator and denominator by \( i \) to get rid of the imaginary unit in the denominator: 
\[ \frac{1}{i} = \frac{1}{i} \times \frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i \] 
Step 3: Final Calculation Now, substitute the results back into the original expression: \[ i^{57} + \frac{1}{i^{25}} = i + (-i) = 0 \] 
Thus, the value of the expression is \( 0 \). 
The correct answer is \( \boxed{(A)} \).