Question

simplify the complex number $i^{47}$ as much as possible.

Explanation:

Step1: Recall \(i\) powers cycle

The powers of \(i\) cycle every 4: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), then repeats.

Step2: Divide exponent by 4

Divide 47 by 4: \(47 \div 4 = 11\) with a remainder of 3 (since \(4\times11 = 44\), \(47 - 44 = 3\)).

Step3: Simplify using remainder

So \(i^{47}=i^{4\times11 + 3}=(i^4)^{11}\times i^3\). Since \(i^4 = 1\), this becomes \(1^{11}\times i^3\). And \(i^3=-i\).

Answer:

\(-i\)