Powers of the Imaginary Unit 1

Question 1 of 5

Simplify

$i^{15}$ $i^(15)$

1. $1$ $1$
2. $i$ $i$
3. $- i$ $-i$
4. $0$ $0$

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Imaginary numbers have the properties

i^{0} = 1

$i^0=1$ ,

i^{1} = i

$i^1=i$ ,

i^{2} = - 1

$i^2=-1$ and

i^{3} = - i

$i^3=-i$ .

Use the formula

i^{4 n + r}

$i^(4n+r)$ to simplify the powers of the imaginary unit.

	$i^{15}$ $i^(15)$	Divide the power $15$ $15$ by $4$ $4$ . This gives $3$ $3$ with a remainder of $3$ $3$ .
$=$ $=$	$i^{4 \times 3 + 3}$ $i^(4times3+3)$	Rewrite using the formula $i^{4 n + r}$ $i^(4n+r)$ where the $n = 3$ $n=3$ and the $r = 3$ $r=3$ .
$=$ $=$	$i^{12 + 3}$ $i^(12+\color(red)(3))$	Simplify

		$i^{12 + 3}$ $i^(12+\color(red)(3))$	The $3$ $3$ means that this number is the same as $i^{3} = - i$ $i^3=-i$ .
	$=$ $=$	$- i$ $-i$

- i

$-i$

Question 2 of 5

Simplify

$i^{50}$ $i^(50)$

1. $1$ $1$
2. $i$ $i$
3. $- 1$ $-1$
4. $- i$ $-i$

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Imaginary numbers have the properties

i^{0} = 1

$i^0=1$ ,

i^{1} = i

$i^1=i$ ,

i^{2} = - 1

$i^2=-1$ and

i^{3} = - i

$i^3=-i$ .

Use the formula

i^{4 n + r}

$i^(4n+r)$ to simplify the powers of the imaginary unit.

	$i^{50}$ $i^(50)$	Divide the power $50$ $50$ by $4$ $4$ . This gives $12$ $12$ with a remainder of $2$ $2$ .
$=$ $=$	$i^{4 \times 12 + 2}$ $i^(4times12+2)$	Rewrite using the formula $i^{4 n + r}$ $i^(4n+r)$ where the $n = 12$ $n=12$ and the $r = 2$ $r=2$ .
$=$ $=$	$i^{48 + 2}$ $i^(48+\color(red)(2))$	Simplify

		$i^{48 + 2}$ $i^(48+\color(red)(2))$	The $2$ $2$ means that this number is the same as $i^{2} = - 1$ $i^2=-1$ .
	$=$ $=$	$- 1$ $-1$

- 1

$-1$

Question 3 of 5

Simplify

$i^{149}$ $i^(149)$

1. $1$ $1$
2. $i$ $i$
3. $- 1$ $-1$
4. $- i$ $-i$

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Imaginary numbers have the properties

i^{0} = 1

$i^0=1$ ,

i^{1} = i

$i^1=i$ ,

i^{2} = - 1

$i^2=-1$ and

i^{3} = - i

$i^3=-i$ .

Use the formula

i^{4 n + r}

$i^(4n+r)$ to simplify the powers of the imaginary unit.

	$i^{149}$ $i^(149)$	Divide the power $149$ $149$ by $4$ $4$ . This gives $37$ $37$ with a remainder of $1$ $1$ .
$=$ $=$	$i^{4 \times 37 + 1}$ $i^(4times37+1)$	Rewrite using the formula $i^{4 n + r}$ $i^(4n+r)$ where the $n = 37$ $n=37$ and the $r = 1$ $r=1$ .
$=$ $=$	$i^{138 + 1}$ $i^(138+\color(red)(1))$	Simplify

		$i^{138 + 1}$ $i^(138+\color(red)(1))$	The $1$ $1$ means that this number is the same as $i^{1} = i$ $i^1=i$ .
	$=$ $=$	$i$ $i$

i

$i$

Question 4 of 5

Simplify

$i^{20}$ $i^(20)$

1. $- i$ $-i$
2. $i$ $i$
3. $- 1$ $-1$
4. $1$ $1$

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Imaginary numbers have the properties

i^{0} = 1

$i^0=1$ ,

i^{1} = i

$i^1=i$ ,

i^{2} = - 1

$i^2=-1$ and

i^{3} = - i

$i^3=-i$ .

Use the formula

i^{4 n + r}

$i^(4n+r)$ to simplify the powers of the imaginary unit.

	$i^{20}$ $i^(20)$	Divide the power $20$ $20$ by $4$ $4$ . This gives $5$ $5$ with a remainder of $0$ $0$ .
$=$ $=$	$i^{4 \times 5 + 0}$ $i^(4times5+0)$	Rewrite using the formula $i^{4 n + r}$ $i^(4n+r)$ where the $n = 5$ $n=5$ and the $r = 0$ $r=0$ .
$=$ $=$	$i^{20 + 0}$ $i^(20+\color(red)(0))$	Simplify

		$i^{20 + 0}$ $i^(20+\color(red)(0))$	The $0$ $0$ means that this number is the same as $i^{0} = 1$ $i^0=1$ .
	$=$ $=$	$1$ $1$

1

$1$

Question 5 of 5

Simplify

$i^{37}$ $i^(37)$

1. $1$ $1$
2. $i$ $i$
3. $- i$ $-i$
4. $0$ $0$

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Imaginary numbers have the properties

i^{0} = 1

$i^0=1$ ,

i^{1} = i

$i^1=i$ ,

i^{2} = - 1

$i^2=-1$ and

i^{3} = - i

$i^3=-i$ .

Use the formula

i^{4 n + r}

$i^(4n+r)$ to simplify the powers of the imaginary unit.

	$i^{37}$ $i^(37)$	Divide the power $37$ $37$ by $4$ $4$ . This gives $9$ $9$ with a remainder of $1$ $1$ .
$=$ $=$	$i^{4 \times 9 + 1}$ $i^(4times9+1)$	Rewrite using the formula $i^{4 n + r}$ $i^(4n+r)$ where the $n = 37$ $n=37$ and the $r = 1$ $r=1$ .
$=$ $=$	$i^{36 + 1}$ $i^(36+\color(red)(1))$	Simplify

		$i^{36 + 1}$ $i^(36+\color(red)(1))$	The $1$ $1$ means that this number is the same as $i^{1} = i$ $i^1=i$ .
	$=$ $=$	$i$ $i$

i

$i$

Powers of the Imaginary Unit 1

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