Precalculus Examples

$i^{- 17}$

Step 1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{i^{17}}$

Step 2

Simplify the denominator.

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Step 2.1

Rewrite $i^{17}$ as ${(i^{4})}^{4} i$ .

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Step 2.1.1

Factor out $i^{16}$ .

$\frac{1}{i^{16} i}$

Step 2.1.2

Rewrite $i^{16}$ as ${(i^{4})}^{4}$ .

$\frac{1}{{(i^{4})}^{4} i}$

Step 2.2

Rewrite $i^{4}$ as $1$ .

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Step 2.2.1

Rewrite $i^{4}$ as ${(i^{2})}^{2}$ .

$\frac{1}{{({(i^{2})}^{2})}^{4} i}$

Step 2.2.2

Rewrite $i^{2}$ as $- 1$ .

$\frac{1}{{({(- 1)}^{2})}^{4} i}$

Step 2.2.3

Raise $- 1$ to the power of $2$ .

$\frac{1}{1^{4} i}$

Step 2.3

One to any power is one.

$\frac{1}{1 i}$

Step 2.4

Multiply $i$ by $1$ .

$\frac{1}{i}$

Step 3

Multiply the numerator and denominator of $\frac{1}{i}$ by the conjugate of $i$ to make the denominator real.

$\frac{1}{i} \cdot \frac{i}{i}$

Step 4

Multiply.

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Step 4.1

Combine.

$\frac{1 i}{i i}$

Step 4.2

Multiply $i$ by $1$ .

$\frac{i}{i i}$

Step 4.3

Simplify the denominator.

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Step 4.3.1

Raise $i$ to the power of $1$ .

$\frac{i}{i^{1} i}$

Step 4.3.2

Raise $i$ to the power of $1$ .

$\frac{i}{i^{1} i^{1}}$

Step 4.3.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{i}{i^{1 + 1}}$

Step 4.3.4

Add $1$ and $1$ .

$\frac{i}{i^{2}}$

Step 4.3.5

Rewrite $i^{2}$ as $- 1$ .

$\frac{i}{- 1}$

Step 5

Move the negative one from the denominator of $\frac{i}{- 1}$ .

$- 1 \cdot i$

Step 6

Rewrite $- 1 \cdot i$ as $- i$ .

$- i$

Step 7

This is the trigonometric form of a complex number where $| z |$ is the modulus and $θ$ is the angle created on the complex plane.

$z = a + b i = | z | (\cos (θ) + i \sin (θ))$

Step 8

The modulus of a complex number is the distance from the origin on the complex plane.

$| z | = \sqrt{a^{2} + b^{2}}$ where $z = a + b i$

Step 9

Substitute the actual values of $a = 0$ and $b = - 1$ .

$| z | = \sqrt{{(- 1)}^{2}}$

Step 10

Find $| z |$ .

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Step 10.1

Raise $- 1$ to the power of $2$ .

$| z | = \sqrt{1}$

Step 10.2

Any root of $1$ is $1$ .

$| z | = 1$

Step 11

The angle of the point on the complex plane is the inverse tangent of the complex portion over the real portion.

$θ = \arctan (\frac{- 1}{0})$

Step 12

Since the argument is undefined and $b$ is negative, the angle of the point on the complex plane is $\frac{3 π}{2}$ .

$θ = \frac{3 π}{2}$

Step 13

Substitute the values of $θ = \frac{3 π}{2}$ and $| z | = 1$ .

$\cos (\frac{3 π}{2}) + i \sin (\frac{3 π}{2})$