Question

which of the following is equivalent to the complex number $i^{41}$? choose 1 answer: a 1 b $i$ c $-1$ d $-i$

Explanation:

Step1: Recall the cycle of \(i\)

The imaginary unit \(i\) has a cyclic pattern: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and then the pattern repeats every 4 powers. So we can find the remainder when the exponent is divided by 4 to simplify \(i^{41}\).

Step2: Divide 41 by 4

Calculate \(41\div4\). The quotient is 10 and the remainder is 1 (since \(4\times10 = 40\) and \(41 - 40 = 1\)). So \(i^{41}=i^{4\times10 + 1}\).

Step3: Use the property of exponents

Using the property \(a^{m + n}=a^m\times a^n\), we can rewrite \(i^{4\times10 + 1}\) as \((i^4)^{10}\times i^1\). We know that \(i^4 = 1\), so \((i^4)^{10}=1^{10}=1\). Then \((i^4)^{10}\times i^1 = 1\times i = i\).

Answer:

B. \(i\)