Question

simplify the term.
i^{47}

Explanation:

Step1: Find remainder of exponent mod 4

$47 \div 4 = 11$ remainder $3$, so $47 = 4\times11 + 3$

Step2: Rewrite using $i^4=1$

$i^{47} = i^{4\times11 + 3} = (i^4)^{11} \times i^3$

Step3: Substitute $i^4=1$ and $i^3=-i$

$(1)^{11} \times (-i) = 1 \times (-i)$

Answer:

$-i$