Question

which of the following is equivalent to the complex number $i^{38}$? 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 property: \(i^1 = i\), \(i^2 = -1\), \(i^3 = i^2 \cdot i=-i\), \(i^4 = (i^2)^2 = (-1)^2 = 1\), and then the cycle repeats every 4 powers. So we can use the fact that \(i^{4k}=1\), \(i^{4k + 1}=i\), \(i^{4k+2}=-1\), \(i^{4k + 3}=-i\) for any integer \(k\).

Step2: Divide the exponent by 4

We need to find the equivalent of \(i^{38}\). Divide 38 by 4: \(38\div4 = 9\) with a remainder of 2. So we can write \(38=4\times9 + 2\).

Step3: Use the cyclic property

Using the property of exponents \(a^{m + n}=a^m\times a^n\), we have \(i^{38}=i^{4\times9+2}=(i^{4})^9\times i^{2}\). Since \(i^{4} = 1\) and \(i^{2}=-1\), we substitute these values in: \((1)^9\times(-1)=1\times(-1)=-1\).

Answer:

C. \(-1\)