Question

If \( z = i^9 + i^{19} \), then \( z \) is equal to:

• A. \( 0 + 0i \)
• B. \( 1 + 0i \)
• C. \( 0 + i \)
• D. \( 1 + 2i \)
• E. \( 1 + 3i \)

Hint:

The sum of any four consecutive powers of \( i \) is always zero (\( i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0 \)). For isolated terms, always look for the nearest multiple of 4 in the exponent.

Answer 1

The Correct Option is A

Concept: The powers of the imaginary unit \( i \) repeat in a cycle of four: \( i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1 \). To simplify \( i^n \), divide the exponent \( n \) by 4 and use the remainder \( r \), such that \( i^n = i^r \).

Step 1: Simplify the individual powers of \( i \).
For \( i^9 \): Divide 9 by 4: \( 9 = 4 \times 2 + 1 \). The remainder is 1. \[ i^9 = i^1 = i \] For \( i^{19} \): Divide 19 by 4: \( 19 = 4 \times 4 + 3 \). The remainder is 3. \[ i^{19} = i^3 = -i \]

Step 2: Sum the results.
\[ z = i^9 + i^{19} \] \[ z = i + (-i) = 0 \] In complex number form \( a + bi \), this is represented as \( 0 + 0i \).