Which of the following is equivalent to the complex number i^(38) ? Choose 1 answer: (A) 1 (B) i (C) -1 (D) -i

Step 1: Define i as the imaginary unit: To solve for i^(38), we need to remember that i is the imaginary unit, which is defined by i^2 = -1. We can use the powers of i to simplify i^(38) because the powers of i repeat in a cycle: i, -1, -i, 1, and then back to i. Let's find the remainder when 38 is divided by 4, since the cycle repeats every 4 powers. 38 ÷ 4 = 9 remainder 2Step 2: Simplify i^(38) using the powers of i: Since the remainder is 2, i^(38) is equivalent to i^2. We know that i^2 = -1.Step 3: Find the remainder when 38 is divided by 4: Therefore, i^(38) is equivalent to -1.

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Which of the following is equivalent to the complex number
i^(38) ?
Choose 1 answer:
(A) 1
(B)
i
(C) -1
(D)
-i

Which of the following is equivalent to the complex number $i^{38}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $1$ $\newline$ (B) $i$ $\newline$ (C) $-1$ $\newline$ (D) $-i$

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Math Problems

Algebra 1

Domain and range of quadratic functions: equations

Full solution

Q. Which of the following is equivalent to the complex number

i^{38}

\newline

Choose

1

answer:

\newline

(A)

1

\newline

(B)

i

\newline

(C)

-1

\newline

(D)

-i

Step $1$ : Define $i$ as the imaginary unit: To solve for $i^{38}$ , we need to remember that $i$ is the imaginary unit, which is defined by $i^2 = -1$ . We can use the powers of $i$ to simplify $i^{38}$ because the powers of $i$ repeat in a cycle: $i$ , $-1$ , $-i$ , $i^{38}$ $0$ , and then back to $i$ . Let's find the remainder when $i^{38}$ $2$ is divided by $i^{38}$ $3$ , since the cycle repeats every $i^{38}$ $3$ powers. $\newline$ $i^{38}$ $5$ remainder $i^{38}$ $6$
Step $2$ : Simplify $i^{38}$ using the powers of $i$ : Since the remainder is $2$ , $i^{38}$ is equivalent to $i^2$ . We know that $i^2 = -1$ .
Step $3$ : Find the remainder when $38$ is divided by $4$ : Therefore, $i^{38}$ is equivalent to $-1$ .