` i^(57) + 1// i^(125)` , when simplified has the value

Options

A. 0
B. 2i
C. `-2i`
D. 2

Answer

To simplify the expression \( i^{57} + \frac{1}{i^{125}} \), we can follow these steps:

### Step 1: Simplify \( i^{57} \)
We know that the powers of \( i \) (the imaginary unit) repeat every 4:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)

To simplify \( i^{57} \), we can find the remainder when 57 is divided by 4:
\[ 57 \div 4 = 14 \quad \text{(remainder 1)} \]
Thus, \( i^{57} = i^{4 \cdot 14 + 1} = (i^4)^{14} \cdot i^1 = 1^{14} \cdot i = i \).

### Step 2: Simplify \( \frac{1}{i^{125}} \)
Next, we simplify \( i^{125} \) in a similar way. We find the remainder when 125 is divided by 4:
\[ 125 \div 4 = 31 \quad \text{(remainder 1)} \]
Thus, \( i^{125} = i^{4 \cdot 31 + 1} = (i^4)^{31} \cdot i^1 = 1^{31} \cdot i = i \).

Now, we can find \( \frac{1}{i^{125}} \):
\[ \frac{1}{i^{125}} = \frac{1}{i} \]
To simplify \( \frac{1}{i} \), we multiply the numerator and denominator by \( -i \):
\[ \frac{1}{i} = \frac{-i}{i^2} = \frac{-i}{-1} = i. \]

### Step 3: Combine the results
Now we can substitute back into the original expression:
\[ i^{57} + \frac{1}{i^{125}} = i + i = 2i. \]

### Final Result
The simplified value of \( i^{57} + \frac{1}{i^{125}} \) is:
\[ \boxed{2i}. \]

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