Question

question 5
1 pts
i⁻⁵³⁷ =
○ 1
○ i
○ -1
○ -i

Explanation:

Step1: Recall the cycle of imaginary unit

The powers of the imaginary unit \(i\) follow a cycle: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and then the cycle repeats every 4 powers. So we can find the remainder when the exponent is divided by 4 to simplify \(i^{537}\).

Step2: Divide 537 by 4

Calculate \(537 \div 4\). The quotient is 134 and the remainder is \(537 - 4\times134 = 537 - 536 = 1\). So \(i^{537}=i^{4\times134 + 1}\).

Step3: Use the property of exponents

Using the property \(a^{m+n}=a^m\times a^n\), we have \(i^{4\times134 + 1}=(i^4)^{134}\times i^1\). Since \(i^4 = 1\), this becomes \(1^{134}\times i = i\).

Answer:

B. i (assuming the option with \(i\) is labeled as B, based on the visual where the second option is \(i\))