Question

which of the following is equivalent to the complex number $i^{19}$? 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 cycle repeats every 4 powers. So we can find the remainder when the exponent is divided by 4 to simplify \(i^{19}\).

Step2: Divide 19 by 4

Divide 19 by 4: \(19\div4 = 4\) with a remainder of 3. This means \(i^{19}=i^{4\times4 + 3}\).

Step3: Use exponent rules

Using the exponent rule \(a^{m + n}=a^m\times a^n\), we can rewrite \(i^{4\times4 + 3}\) as \((i^4)^4\times i^3\). Since \(i^4 = 1\), this becomes \(1^4\times i^3\). And since \(i^3=-i\), we have \(1\times(-i)=-i\).

Answer:

D. \(-i\)