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 powers of \( i \).

### Step 1: Identify the pattern of powers of \( i \)
The powers of \( i \) cycle every four terms:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
- \( i^5 = i \) (and so on...)

### Step 2: Calculate each term
Now we will calculate each term in the expression based on the cycle:
- \( i^{15} = i^{4 \cdot 3 + 3} = (i^4)^3 \cdot i^3 = 1^3 \cdot (-i) = -i \)
- \( i^{17} = i^{4 \cdot 4 + 1} = (i^4)^4 \cdot i^1 = 1^4 \cdot i = i \)
- \( i^{19} = i^{4 \cdot 4 + 3} = (i^4)^4 \cdot i^3 = 1^4 \cdot (-i) = -i \)
- \( i^{21} = i^{4 \cdot 5 + 1} = (i^4)^5 \cdot i^1 = 1^5 \cdot i = i \)
- \( i^{24} = i^{4 \cdot 6} = (i^4)^6 = 1^6 = 1 \)

### Step 3: Substitute the calculated values into the expression
Now we can substitute these values back into the original expression:
\[ i^{15} + i^{17} + i^{19} + i^{21} + i^{24} = -i + i - i + i + 1 \]

### Step 4: Simplify the expression
Now, we can simplify the expression:
\[ (-i + i) + (-i + i) + 1 = 0 + 0 + 1 = 1 \]

### Conclusion
Since \( 1 \) is a real number, we conclude that:
\[ i^{15} + i^{17} + i^{19} + i^{21} + i^{24} = 1 \]
Thus, \( i^{15} + i^{17} + i^{19} + i^{21} + i^{24} \) is indeed a real number.

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