Precalculus Examples

$x^{3} + 27 i = 0$

Step 1

Subtract $27 i$ from both sides of the equation.

$x^{3} = - 27 i$

Step 2

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 3

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 4

Substitute the actual values of $a = 0$ and $b = - 27$ .

$| z | = \sqrt{{(- 27)}^{2}}$

Step 5

Find $| z |$ .

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Step 5.1

Raise $- 27$ to the power of $2$ .

$| z | = \sqrt{729}$

Step 5.2

Rewrite $729$ as $27^{2}$ .

$| z | = \sqrt{27^{2}}$

Step 5.3

Pull terms out from under the radical, assuming positive real numbers.

$| z | = 27$

Step 6

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

$θ = \arctan (\frac{- 27}{0})$

Step 7

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 8

Substitute the values of $θ = \frac{3 π}{2}$ and $| z | = 27$ .

$27 (\cos (\frac{3 π}{2}) + i \sin (\frac{3 π}{2}))$

Step 9

Replace the right side of the equation with the trigonometric form.

$x^{3} = 27 (\cos (\frac{3 π}{2}) + i \sin (\frac{3 π}{2}))$