The value of `(i^(23)+((1)/(i))^(29))^(2) is`

Options

A. `4`
B. `1`
C. `-2`
D. `-4`

Explanation

To solve the expression \((i^{23} + \left(\frac{1}{i}\right)^{29})^2\), we will follow these steps:

### Step 1: Simplify \(i^{23}\)
We know that the powers of \(i\) cycle every 4:
- \(i^1 = i\)
- \(i^2 = -1\)
- \(i^3 = -i\)
- \(i^4 = 1\)

To find \(i^{23}\), we can reduce the exponent modulo 4:
\[ 23 \mod 4 = 3 \]
Thus,
\[ i^{23} = i^3 = -i \]

### Step 2: Simplify \(\left(\frac{1}{i}\right)^{29}\)
We can rewrite \(\frac{1}{i}\) as:
\[ \frac{1}{i} = -i \quad \text{(by multiplying numerator and denominator by } i\text{)} \]
Now, we need to calculate \((-i)^{29}\). Again, we reduce the exponent modulo 4:
\[ 29 \mod 4 = 1 \]
Thus,
\[ (-i)^{29} = -i \]

### Step 3: Combine the results
Now we can substitute back into the original expression:
\[ i^{23} + \left(\frac{1}{i}\right)^{29} = -i + (-i) = -2i \]

### Step 4: Square the result
Now we square the result:
\[ (-2i)^2 = 4i^2 \]
And since \(i^2 = -1\):
\[ 4i^2 = 4(-1) = -4 \]

### Conclusion
Thus, the value of \((i^{23} + \left(\frac{1}{i}\right)^{29})^2\) is:
\[ \boxed{-4} \]

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