Which of the following is equivalent to the complex number i^(22) ? Choose 1 answer: (A) 1 (B) i (C) -1 (D) -i

Definition of i: To solve for i^(22), we need to remember that i is the imaginary unit, which is defined by the property that i^2 = -1. We can use this property to simplify i^(22) by breaking it down into powers of i^2.Expressing 22 as a multiple of 4: First, we express 22 as a multiple of 4 plus a remainder because the powers of i repeat every 4th power: i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, and then the cycle repeats. 22 = 4 * 5 + 2, so i^(22) = (i^4)^5 * i^2.Simplifying (i^4)^5: Next, we simplify (i^4)^5. Since i^4 = 1, any power of i^4 is also 1. Therefore, (i^4)^5 = 1^5 = 1.Multiplying by i^2: Now we multiply this result by i^2. We already know that i^2 = -1, so we have: 1 * i^2 = 1 * -1 = -1.Final result: Therefore, i^(22) is equivalent to -1.

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Which of the following is equivalent to the complex number
i^(22) ?
Choose 1 answer:
(A) 1
(B)
i
(C) -1
(D)
-i

Which of the following is equivalent to the complex number $i^{22}$ ? $\newline$ 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^{22}

\newline

Choose

1

answer:

\newline

(A)

1

\newline

(B)

i

\newline

(C)

-1

\newline

(D)

-i

Definition of i: To solve for $i^{22}$ , we need to remember that $i$ is the imaginary unit, which is defined by the property that $i^2 = -1$ . We can use this property to simplify $i^{22}$ by breaking it down into powers of $i^2$ .
Expressing $22$ as a multiple of $4$ : First, we express $22$ as a multiple of $4$ plus a remainder because the powers of $i$ repeat every $4$ th power: $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$ , and then the cycle repeats. $\newline$ $22 = 4 \times 5 + 2$ , so $i^{22} = (i^4)^5 \times i^2$ .
Simplifying $(i^4)^5$ : Next, we simplify $(i^4)^5$ . Since $i^4 = 1$ , any power of $i^4$ is also $1$ . Therefore, $(i^4)^5 = 1^5 = 1$ .
Multiplying by $i^2$ : Now we multiply this result by $i^2$ . We already know that $i^2 = -1$ , so we have: $\newline$ $1 \cdot i^2 = 1 \cdot -1 = -1$ .
Final result: Therefore, $i^{22}$ is equivalent to $-1$ .