Which of the following is equivalent to the complex number i^(54) ? Choose 1 answer: (A) 1 (B) i (C) -1 (D) -i

Definition of i: To solve for i^(54), 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^(54).Cycle of powers of i: The powers of i repeat in a cycle: i, -1, -i, 1, and then back to i. This cycle repeats every 4 powers. So, to simplify i^(54), we can divide 54 by 4 and look at the remainder to determine where we are in the cycle.Determining the position in the cycle: Dividing 54 by 4 gives us 13 with a remainder of 2. This means that i^(54) is equivalent to i^(4*13 + 2). Since i^4 = 1 (and any integer power of it), we can ignore the 4*13 part because it will just be 1 to any power, which is still 1.Simplifying i^(54) using i^2: Now we only need to consider i^2. Since i^2 = -1, i^(54) simplifies to -1.

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Which of the following is equivalent to the complex number
i^(54) ?
Choose 1 answer:
(A) 1
(B)
i
(C) -1
(D)
-i

Which of the following is equivalent to the complex number $i^{54}$ ? $\newline$ 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

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

i^{54}

\newline

Choose

1

answer:

\newline

(A)

1

\newline

(B)

i

\newline

(C)

-1

\newline

(D)

-i

Definition of i: To solve for $i^{54}$ , 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 $i^{54}$ .
Cycle of powers of $i$ : The powers of $i$ repeat in a cycle: $i$ , $-1$ , $-i$ , $1$ , and then back to $i$ . This cycle repeats every $4$ powers. So, to simplify $i^{$ $54$ }, we can divide $54$ by $4$ and look at the remainder to determine where we are in the cycle.
Determining the position in the cycle: Dividing $54$ by $4$ gives us $13$ with a remainder of $2$ . This means that $i^{54}$ is equivalent to $i^{(4\cdot 13 + 2)}$ . Since $i^4 = 1$ (and any integer power of it), we can ignore the $4\cdot 13$ part because it will just be $1$ to any power, which is still $1$ .
Simplifying $i^{54}$ using $i^2$ : Now we only need to consider $i^2$ . Since $i^2 = -1$ , $i^{54}$ simplifies to $-1$ .