Question

compute $i^{139}$.

Explanation:

Step1: Recall the cycle of \(i\)

The imaginary unit \(i\) has a cyclic property: \(i^1 = i\), \(i^2=-1\), \(i^3 = i^2\times i=-i\), \(i^4=(i^2)^2 = (-1)^2 = 1\), and then the cycle repeats every 4 powers. So we can find the remainder when the exponent is divided by 4 to simplify \(i^{139}\).

Step2: Divide 139 by 4

We perform the division \(139\div4\). \(4\times34 = 136\), so the remainder is \(139 - 136=3\). This means \(i^{139}=i^{4\times34 + 3}\).

Step3: Use the property of exponents

Using the property \(a^{m + n}=a^m\times a^n\), we can rewrite \(i^{4\times34+3}\) as \((i^4)^{34}\times i^3\). Since \(i^4 = 1\), \((i^4)^{34}=1^{34}=1\). And we know from the cycle that \(i^3=-i\). So \((i^4)^{34}\times i^3=1\times(-i)=-i\).

Answer:

\(-i\)