Question Find the value of $i^{52}$ , where $i$ is the imaginary unit.

Studdy Solution

STEP 1

Assumptions
1. We need to find the value of $i^{52}$ .
2. Here, $i$ is the imaginary unit, which is defined as $i = \sqrt{-1}$ .
3. The powers of $i$ are cyclic with a period of 4, meaning $i^4 = 1$ , and thus $i^n = i^{n \mod 4}$ for any integer $n$ .

STEP 2

First, we need to find the remainder when 52 is divided by 4 because the powers of $i$ repeat every 4th power.
$52 \mod 4$

STEP 3

Calculate the remainder.
$52 \mod 4 = 0$

STEP 4

Since the remainder is 0, we know that $i^{52}$ is equivalent to $i^0$ because the powers of $i$ are cyclic with a period of 4.
$i^{52} = i^{4 \times 13} = (i^4)^{13}$

STEP 5

Use the fact that $i^4 = 1$ to simplify the expression.
$(i^4)^{13} = 1^{13}$

STEP 6

Any non-zero number to the power of any integer is itself, so $1^{13} = 1$ .
$1^{13} = 1$

STEP 7

Conclude that $i^{52} = 1$ .
$i^{52} = 1$

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Question Find the value of $i^{52}$ , where $i$ is the imaginary unit.

Studdy Solution

STEP 1

Assumptions
1. We need to find the value of $i^{52}$ .
2. Here, $i$ is the imaginary unit, which is defined as $i = \sqrt{-1}$ .
3. The powers of $i$ are cyclic with a period of 4, meaning $i^4 = 1$ , and thus $i^n = i^{n \mod 4}$ for any integer $n$ .

STEP 2

First, we need to find the remainder when 52 is divided by 4 because the powers of $i$ repeat every 4th power.
$52 \mod 4$

STEP 3

Calculate the remainder.
$52 \mod 4 = 0$

STEP 4

Since the remainder is 0, we know that $i^{52}$ is equivalent to $i^0$ because the powers of $i$ are cyclic with a period of 4.
$i^{52} = i^{4 \times 13} = (i^4)^{13}$

STEP 5

Use the fact that $i^4 = 1$ to simplify the expression.
$(i^4)^{13} = 1^{13}$

STEP 6

Any non-zero number to the power of any integer is itself, so $1^{13} = 1$ .
$1^{13} = 1$

STEP 7

Conclude that $i^{52} = 1$ .
$i^{52} = 1$

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Question Find the value of i52i^{52}i52, where iii is the imaginary unit.

Studdy Solution

STEP 1

Assumptions1. We need to find the value of i52i^{52}i52.2. Here, iii is the imaginary unit, which is defined as i=−1i = \sqrt{-1}i=−1​.3. The powers of iii are cyclic with a period of 4, meaning i4=1i^4 = 1i4=1, and thus in=inmod 4i^n = i^{n \mod 4}in=inmod4 for any integer nnn.

STEP 2

First, we need to find the remainder when 52 is divided by 4 because the powers of iii repeat every 4th power.52mod 452 \mod 452mod4

STEP 3

Calculate the remainder.52mod 4=052 \mod 4 = 052mod4=0

STEP 4

Since the remainder is 0, we know that i52i^{52}i52 is equivalent to i0i^0i0 because the powers of iii are cyclic with a period of 4.i52=i4×13=(i4)13i^{52} = i^{4 \times 13} = (i^4)^{13}i52=i4×13=(i4)13

STEP 5

Use the fact that i4=1i^4 = 1i4=1 to simplify the expression.(i4)13=113(i^4)^{13} = 1^{13}(i4)13=113

STEP 6

Any non-zero number to the power of any integer is itself, so 113=11^{13} = 1113=1.113=11^{13} = 1113=1

STEP 7

Conclude that i52=1i^{52} = 1i52=1.i52=1i^{52} = 1i52=1

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Question Find the value of $i^{52}$ , where $i$ is the imaginary unit.

Assumptions
1. We need to find the value of $i^{52}$ .
2. Here, $i$ is the imaginary unit, which is defined as $i = \sqrt{-1}$ .
3. The powers of $i$ are cyclic with a period of 4, meaning $i^4 = 1$ , and thus $i^n = i^{n \mod 4}$ for any integer $n$ .

First, we need to find the remainder when 52 is divided by 4 because the powers of $i$ repeat every 4th power.
$52 \mod 4$

Calculate the remainder.
$52 \mod 4 = 0$

Since the remainder is 0, we know that $i^{52}$ is equivalent to $i^0$ because the powers of $i$ are cyclic with a period of 4.
$i^{52} = i^{4 \times 13} = (i^4)^{13}$

Use the fact that $i^4 = 1$ to simplify the expression.
$(i^4)^{13} = 1^{13}$

Any non-zero number to the power of any integer is itself, so $1^{13} = 1$ .
$1^{13} = 1$

Conclude that $i^{52} = 1$ .
$i^{52} = 1$