Question

1. $i^{86}$

Explanation:

Step1: Recall the cycle of \(i\)

The imaginary unit \(i\) has a cyclic property: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and then the cycle repeats every 4 powers. So we can find the remainder when the exponent is divided by 4 to simplify \(i^{n}\).

Step2: Divide 86 by 4

Calculate \(86\div4\). The quotient is 21 and the remainder is \(86 - 4\times21 = 86 - 84 = 2\). So \(i^{86}=i^{4\times21 + 2}\).

Step3: Use the property of exponents

Using the property \(a^{m + n}=a^m\times a^n\), we have \(i^{4\times21 + 2}=(i^4)^{21}\times i^2\). Since \(i^4 = 1\), then \((i^4)^{21}=1^{21}=1\). And we know \(i^2=-1\). So \((i^4)^{21}\times i^2 = 1\times(-1)=-1\).

Answer:

\(-1\)