Precalculus Examples

$\sqrt[3]{- 2 i}$

Step 1

Rewrite $- 2 i$ as ${({(- 1)}^{3})}^{3} \cdot (2 i)$ .

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Step 1.1

Rewrite $- 2$ as $- 1 (2)$ .

$\sqrt[3]{- 1 (2) i}$

Step 1.2

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$\sqrt[3]{{(- 1)}^{3} \cdot 2 i}$

Step 1.3

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$\sqrt[3]{{({(- 1)}^{3})}^{3} \cdot 2 i}$

Step 1.4

Add parentheses.

$\sqrt[3]{{({(- 1)}^{3})}^{3} \cdot (2 i)}$

Step 2

Pull terms out from under the radical.

${(- 1)}^{3} \sqrt[3]{2 i}$

Step 3

Simplify the expression.

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Step 3.1

Raise $- 1$ to the power of $3$ .

$- 1 \sqrt[3]{2 i}$

Step 3.2

Rewrite $- 1 \sqrt[3]{2 i}$ as $- \sqrt[3]{2 i}$ .

$- \sqrt[3]{2 i}$

Step 4

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 5

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 6

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

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

Step 7

Find $| z |$ .

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Step 7.1

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

$| z | = \sqrt{1}$

Step 7.2

Any root of $1$ is $1$ .

$| z | = 1$

Step 8

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 9

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 10

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

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