Show that `i^(15)+i^(17)+i^(19)+i^(21)+i^(24)` is a real number.

Answer

To show that \( i^{15} + i^{17} + i^{19} + i^{21} + i^{24} \) is a real number, we will simplify each term using the properties of the imaginary unit \( i \), where \( i = \sqrt{-1} \) and \( i^2 = -1 \).

### Step-by-step Solution:

1. **Identify the powers of \( i \)**:
The powers of \( i \) cycle every 4 terms:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
- \( i^5 = i \) (and the cycle repeats)

2. **Calculate each term**:
- \( i^{15} = i^{4 \cdot 3 + 3} = (i^4)^3 \cdot i^3 = 1^3 \cdot (-i) = -i \)
- \( i^{17} = i^{4 \cdot 4 + 1} = (i^4)^4 \cdot i^1 = 1^4 \cdot i = i \)
- \( i^{19} = i^{4 \cdot 4 + 3} = (i^4)^4 \cdot i^3 = 1^4 \cdot (-i) = -i \)
- \( i^{21} = i^{4 \cdot 5 + 1} = (i^4)^5 \cdot i^1 = 1^5 \cdot i = i \)
- \( i^{24} = i^{4 \cdot 6} = (i^4)^6 = 1^6 = 1 \)

3. **Combine the results**:
Now, substituting back into the expression:
\[ i^{15} + i^{17} + i^{19} + i^{21} + i^{24} = -i + i - i + i + 1 \]

4. **Simplify the expression**:
Combine like terms:
\[ (-i + i - i + i) + 1 = 0 + 1 = 1 \]

5. **Conclusion**:
The final result is \( 1 \), which is a real number.

### Final Answer:
Thus, \( i^{15} + i^{17} + i^{19} + i^{21} + i^{24} = 1 \), and it is confirmed to be a real number.