QuestionFind the value of $i^{99}$ .

Studdy Solution

STEP 1

Assumptions1. We are dealing with complex numbers where $i$ is the imaginary unit with the property that $i^ = -1$ .

STEP 2

We know that the powers of $i$ repeat every4th power. This is because $i^1 = i$ $i^2 = -1$ $i^ = -i$ $i^4 =1$ After this, the pattern repeats.

STEP 3

To find $i^{99}$ , we can use the fact that the powers of $i$ repeat everyth power. So, we can divide99 by to find the remainder.
99 \mod

STEP 4

Calculate the remainder.
$99 \mod4 =3$

STEP 5

The remainder is3, so $i^{99}$ is equivalent to $i^3$ .
$i^{99} = i^3$

STEP 6

We know that $i^3 = -i$ .
$i^{99} = -i$ So, $i^{99}$ equals $-i$ .

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QuestionFind the value of $i^{99}$ .

Studdy Solution

STEP 1

Assumptions1. We are dealing with complex numbers where $i$ is the imaginary unit with the property that $i^ = -1$ .

STEP 2

We know that the powers of $i$ repeat every4th power. This is because $i^1 = i$ $i^2 = -1$ $i^ = -i$ $i^4 =1$ After this, the pattern repeats.

STEP 3

To find $i^{99}$ , we can use the fact that the powers of $i$ repeat everyth power. So, we can divide99 by to find the remainder.
99 \mod

STEP 4

Calculate the remainder.
$99 \mod4 =3$

STEP 5

The remainder is3, so $i^{99}$ is equivalent to $i^3$ .
$i^{99} = i^3$

STEP 6

We know that $i^3 = -i$ .
$i^{99} = -i$ So, $i^{99}$ equals $-i$ .

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QuestionFind the value of i99i^{99}i99.

Studdy Solution

STEP 1

Assumptions1. We are dealing with complex numbers where iii is the imaginary unit with the property that i=−1i^ = -1i=−1.

STEP 2

We know that the powers of iii repeat every4th power. This is becausei1=ii^1 = ii1=ii2=−1i^2 = -1i2=−1i=−ii^ = -ii=−ii4=1i^4 =1i4=1After this, the pattern repeats.

STEP 3

To find i99i^{99}i99, we can use the fact that the powers of iii repeat everyth power. So, we can divide99 by to find the remainder.99 \mod

STEP 4

Calculate the remainder.99mod 4=399 \mod4 =399mod4=3

STEP 5

The remainder is3, so i99i^{99}i99 is equivalent to i3i^3i3.i99=i3i^{99} = i^3i99=i3

STEP 6

We know that i3=−ii^3 = -ii3=−i.i99=−ii^{99} = -ii99=−iSo, i99i^{99}i99 equals −i-i−i.

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QuestionFind the value of $i^{99}$ .

Assumptions1. We are dealing with complex numbers where $i$ is the imaginary unit with the property that $i^ = -1$ .

We know that the powers of $i$ repeat every4th power. This is because $i^1 = i$ $i^2 = -1$ $i^ = -i$ $i^4 =1$ After this, the pattern repeats.

To find $i^{99}$ , we can use the fact that the powers of $i$ repeat everyth power. So, we can divide99 by to find the remainder.
99 \mod

Calculate the remainder.
$99 \mod4 =3$

The remainder is3, so $i^{99}$ is equivalent to $i^3$ .
$i^{99} = i^3$

We know that $i^3 = -i$ .
$i^{99} = -i$ So, $i^{99}$ equals $-i$ .