Algebra Examples

$x^{3} = - i$

Step 1

Substitute $u$ for $x^{3}$ .

$u^{3} = - i$

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 $b = - 1$ .

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

Step 5

Find $| z |$ .

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Step 5.1

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

$| z | = \sqrt{1}$

Step 5.2

Any root of $1$ is $1$ .

$| z | = 1$

Step 6

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

$θ = \arctan (\frac{- 1}{0})$

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 $| z | = 1$ .

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

Step 9

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

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

Step 10

Use De Moivre's Theorem to find an equation for $u$ .

$r^{3} (c o s (3 θ) + i s i n (3 θ)) = - i = \cos (\frac{3 π}{2}) + i \sin (\frac{3 π}{2})$

Step 11

Equate the modulus of the trigonometric form to $r^{3}$ to find the value of $r$ .

$r^{3} = 1$

Step 12

Solve the equation for $r$ .

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Step 12.1

Subtract $1$ from both sides of the equation.

$r^{3} - 1 = 0$

Step 12.2

Factor the left side of the equation.

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Step 12.2.1

Rewrite $1$ as $1^{3}$ .

$r^{3} - 1^{3} = 0$

Step 12.2.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = r$ and $b = 1$ .

$(r - 1) (r^{2} + r \cdot 1 + 1^{2}) = 0$

Step 12.2.3

Simplify.

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Step 12.2.3.1

Multiply $r$ by $1$ .

$(r - 1) (r^{2} + r + 1^{2}) = 0$

Step 12.2.3.2

One to any power is one.

$(r - 1) (r^{2} + r + 1) = 0$

Step 12.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$r - 1 = 0$

$r^{2} + r + 1 = 0$

Step 12.4

Set $r - 1$ equal to $0$ and solve for $r$ .

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Step 12.4.1

Set $r - 1$ equal to $0$ .

$r - 1 = 0$

Step 12.4.2

Add $1$ to both sides of the equation.

$r = 1$

Step 12.5

Set $r^{2} + r + 1$ equal to $0$ and solve for $r$ .

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Step 12.5.1

Set $r^{2} + r + 1$ equal to $0$ .

$r^{2} + r + 1 = 0$

Step 12.5.2

Solve $r^{2} + r + 1 = 0$ for $r$ .

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Step 12.5.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 12.5.2.2

Substitute the values $a = 1$ , $b = 1$ , and $c = 1$ into the quadratic formula and solve for $r$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 12.5.2.3

Simplify.

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Step 12.5.2.3.1

Simplify the numerator.

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Step 12.5.2.3.1.1

One to any power is one.

$r = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 12.5.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 12.5.2.3.1.2.1

Multiply $- 4$ by $1$ .

$r = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 12.5.2.3.1.2.2

Multiply $- 4$ by $1$ .

$r = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 12.5.2.3.1.3

Subtract $4$ from $1$ .

$r = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 12.5.2.3.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$r = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 12.5.2.3.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$r = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 12.5.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$r = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 12.5.2.3.2

Multiply $2$ by $1$ .

$r = \frac{- 1 \pm i \sqrt{3}}{2}$

Step 12.5.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 12.5.2.4.1

Simplify the numerator.

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Step 12.5.2.4.1.1

One to any power is one.

$r = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 12.5.2.4.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 12.5.2.4.1.2.1

Multiply $- 4$ by $1$ .

$r = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 12.5.2.4.1.2.2

Multiply $- 4$ by $1$ .

$r = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 12.5.2.4.1.3

Subtract $4$ from $1$ .

$r = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 12.5.2.4.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$r = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 12.5.2.4.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$r = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 12.5.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$r = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 12.5.2.4.2

Multiply $2$ by $1$ .

$r = \frac{- 1 \pm i \sqrt{3}}{2}$

Step 12.5.2.4.3

Change the $\pm$ to $+$ .

$r = \frac{- 1 + i \sqrt{3}}{2}$

Step 12.5.2.4.4

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

$r = \frac{- 1 \cdot 1 + i \sqrt{3}}{2}$

Step 12.5.2.4.5

Factor $- 1$ out of $i \sqrt{3}$ .

$r = \frac{- 1 \cdot 1 - (- i \sqrt{3})}{2}$

Step 12.5.2.4.6

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{3})$ .

$r = \frac{- 1 (1 - i \sqrt{3})}{2}$

Step 12.5.2.4.7

Move the negative in front of the fraction.

$r = - \frac{1 - i \sqrt{3}}{2}$

Step 12.5.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 12.5.2.5.1

Simplify the numerator.

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Step 12.5.2.5.1.1

One to any power is one.

$r = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 12.5.2.5.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 12.5.2.5.1.2.1

Multiply $- 4$ by $1$ .

$r = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 12.5.2.5.1.2.2

Multiply $- 4$ by $1$ .

$r = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 12.5.2.5.1.3

Subtract $4$ from $1$ .

$r = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 12.5.2.5.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$r = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 12.5.2.5.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$r = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 12.5.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$r = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 12.5.2.5.2

Multiply $2$ by $1$ .

$r = \frac{- 1 \pm i \sqrt{3}}{2}$

Step 12.5.2.5.3

Change the $\pm$ to $-$ .

$r = \frac{- 1 - i \sqrt{3}}{2}$

Step 12.5.2.5.4

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

$r = \frac{- 1 \cdot 1 - i \sqrt{3}}{2}$

Step 12.5.2.5.5

Factor $- 1$ out of $- i \sqrt{3}$ .

$r = \frac{- 1 \cdot 1 - (i \sqrt{3})}{2}$

Step 12.5.2.5.6

Factor $- 1$ out of $- 1 (1) - (i \sqrt{3})$ .

$r = \frac{- 1 (1 + i \sqrt{3})}{2}$

Step 12.5.2.5.7

Move the negative in front of the fraction.

$r = - \frac{1 + i \sqrt{3}}{2}$

Step 12.5.2.6

The final answer is the combination of both solutions.

$r = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Step 12.6

The final solution is all the values that make $(r - 1) (r^{2} + r + 1) = 0$ true.

$r = 1, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Step 13

Find the approximate value of $r$ .

$r = 1$

Step 14

Find the possible values of $θ$ .

$c o s (3 θ) = c o s (2 π n)$ and $s i n (3 θ) = s i n (2 π n)$

Step 15

Finding all the possible values of $θ$ leads to the equation $3 θ = 2 π n$ .

$3 θ = 2 π n$

Step 16

Find the value of $θ$ for $r = 0$ .

$3 θ = 2 π (0)$

Step 17

Solve the equation for $θ$ .

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Step 17.1

Multiply $2 π \cdot 0$ .

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Step 17.1.1

Multiply $0$ by $2$ .

$3 θ = 0 π$

Step 17.1.2

Multiply $0$ by $π$ .

$3 θ = 0$

Step 17.2

Divide each term in $3 θ = 0$ by $3$ and simplify.

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Step 17.2.1

Divide each term in $3 θ = 0$ by $3$ .

$\frac{3 θ}{3} = \frac{0}{3}$

Step 17.2.2

Simplify the left side.

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Step 17.2.2.1

Cancel the common factor of $3$ .

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Step 17.2.2.1.1

Cancel the common factor.

$\frac{3 θ}{3} = \frac{0}{3}$

Step 17.2.2.1.2

Divide $θ$ by $1$ .

$θ = \frac{0}{3}$

Step 17.2.3

Simplify the right side.

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Step 17.2.3.1

Divide $0$ by $3$ .

$θ = 0$

Step 18

Use the values of $θ$ and $r$ to find a solution to the equation $u^{3} = - i$ .

$u_{0} = 1 (\cos (0) + i \sin (0))$

Step 19

Convert the solution to rectangular form.

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Step 19.1

Multiply $\cos (0) + i \sin (0)$ by $1$ .

$u_{0} = \cos (0) + i \sin (0)$

Step 19.2

Simplify each term.

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Step 19.2.1

The exact value of $\cos (0)$ is $1$ .

$u_{0} = 1 + i \sin (0)$

Step 19.2.2

The exact value of $\sin (0)$ is $0$ .

$u_{0} = 1 + i \cdot 0$

Step 19.2.3

Multiply $i$ by $0$ .

$u_{0} = 1 + 0$

Step 19.3

Add $1$ and $0$ .

$u_{0} = 1$

Step 20

Substitute $x^{3}$ for $u$ to calculate the value of $z$ after the right shift.

$z_{0} = 0 + 1$

Step 21

Find the value of $θ$ for $r = 1$ .

$3 θ = 2 π (1)$

Step 22

Solve the equation for $θ$ .

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Step 22.1

Multiply $2$ by $1$ .

$3 θ = 2 π$

Step 22.2

Divide each term in $3 θ = 2 π$ by $3$ and simplify.

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Step 22.2.1

Divide each term in $3 θ = 2 π$ by $3$ .

$\frac{3 θ}{3} = \frac{2 π}{3}$

Step 22.2.2

Simplify the left side.

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Step 22.2.2.1

Cancel the common factor of $3$ .

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Step 22.2.2.1.1

Cancel the common factor.

$\frac{3 θ}{3} = \frac{2 π}{3}$

Step 22.2.2.1.2

Divide $θ$ by $1$ .

$θ = \frac{2 π}{3}$

Step 23

Use the values of $θ$ and $r$ to find a solution to the equation $u^{3} = - i$ .

$u_{1} = 1 (\cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3}))$

Step 24

Convert the solution to rectangular form.

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Step 24.1

Multiply $\cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3})$ by $1$ .

$u_{1} = \cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3})$

Step 24.2

Simplify each term.

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Step 24.2.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$u_{1} = - \cos (\frac{π}{3}) + i \sin (\frac{2 π}{3})$

Step 24.2.2

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$u_{1} = - \frac{1}{2} + i \sin (\frac{2 π}{3})$

Step 24.2.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$u_{1} = - \frac{1}{2} + i \sin (\frac{π}{3})$

Step 24.2.4

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

$u_{1} = - \frac{1}{2} + i (\frac{\sqrt{3}}{2})$

Step 24.2.5

Combine $i$ and $\frac{\sqrt{3}}{2}$ .

$u_{1} = - \frac{1}{2} + \frac{i \sqrt{3}}{2}$

Step 25

Substitute $x^{3}$ for $u$ to calculate the value of $z$ after the right shift.

$z_{1} = 0 - \frac{1}{2} + \frac{i \sqrt{3}}{2}$

Step 26

Find the value of $θ$ for $r = 2$ .

$3 θ = 2 π (2)$

Step 27

Solve the equation for $θ$ .

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Step 27.1

Multiply $2$ by $2$ .

$3 θ = 4 π$

Step 27.2

Divide each term in $3 θ = 4 π$ by $3$ and simplify.

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Step 27.2.1

Divide each term in $3 θ = 4 π$ by $3$ .

$\frac{3 θ}{3} = \frac{4 π}{3}$

Step 27.2.2

Simplify the left side.

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Step 27.2.2.1

Cancel the common factor of $3$ .

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Step 27.2.2.1.1

Cancel the common factor.

$\frac{3 θ}{3} = \frac{4 π}{3}$

Step 27.2.2.1.2

Divide $θ$ by $1$ .

$θ = \frac{4 π}{3}$

Step 28

Use the values of $θ$ and $r$ to find a solution to the equation $u^{3} = - i$ .

$u_{2} = 1 (\cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3}))$

Step 29

Convert the solution to rectangular form.

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Step 29.1

Multiply $\cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3})$ by $1$ .

$u_{2} = \cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3})$

Step 29.2

Simplify each term.

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Step 29.2.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the third quadrant.

$u_{2} = - \cos (\frac{π}{3}) + i \sin (\frac{4 π}{3})$

Step 29.2.2

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$u_{2} = - \frac{1}{2} + i \sin (\frac{4 π}{3})$

Step 29.2.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the third quadrant.

$u_{2} = - \frac{1}{2} + i (- \sin (\frac{π}{3}))$

Step 29.2.4

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

$u_{2} = - \frac{1}{2} + i (- \frac{\sqrt{3}}{2})$

Step 29.2.5

Combine $i$ and $\frac{\sqrt{3}}{2}$ .

$u_{2} = - \frac{1}{2} - \frac{i \sqrt{3}}{2}$

Step 30

Substitute $x^{3}$ for $u$ to calculate the value of $z$ after the right shift.

$z_{2} = 0 - \frac{1}{2} - \frac{i \sqrt{3}}{2}$

Step 31

These are the complex solutions to $u^{3} = - i$ .

$z_{0} = 1$

$z_{1} = - \frac{1}{2} + \frac{i \sqrt{3}}{2}$

$z_{2} = - \frac{1}{2} - \frac{i \sqrt{3}}{2}$