Find All Complex Number Solutions z^3=27i

$z^{3} = 27 i$

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

$u^{3} = 27 i$

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

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

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

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

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

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

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

$| z | = 27$

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

$θ = \arctan (\frac{27}{0})$

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

$θ = \frac{π}{2}$

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

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

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

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

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

$r^{3} (c o s (3 θ) + i s i n (3 θ)) = 27 i = 27 (\cos (\frac{π}{2}) + i \sin (\frac{π}{2}))$

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

$r^{3} = 27$

Solve the equation for $r$ .

Subtract $27$ from both sides of the equation.

$r^{3} - 27 = 0$

Factor the left side of the equation.

Rewrite $27$ as $3^{3}$ .

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

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 $b = 3$ .

$(r - 3) (r^{2} + r \cdot 3 + 3^{2}) = 0$

Simplify.

Move $3$ to the left of $r$ .

$(r - 3) (r^{2} + 3 r + 3^{2}) = 0$

Raise $3$ to the power of $2$ .

$(r - 3) (r^{2} + 3 r + 9) = 0$

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

$r - 3 = 0$

$r^{2} + 3 r + 9 = 0$

Set $r - 3$ equal to $0$ and solve for $r$ .

Set $r - 3$ equal to $0$ .

$r - 3 = 0$

Add $3$ to both sides of the equation.

$r = 3$

Set $r^{2} + 3 r + 9$ equal to $0$ and solve for $r$ .

Set $r^{2} + 3 r + 9$ equal to $0$ .

$r^{2} + 3 r + 9 = 0$

Solve $r^{2} + 3 r + 9 = 0$ for $r$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 3$ , and $c = 9$ into the quadratic formula and solve for $r$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$r = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$r = \frac{- 3 \pm \sqrt{9 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$r = \frac{- 3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Subtract $36$ from $9$ .

$r = \frac{- 3 \pm \sqrt{- 27}}{2 \cdot 1}$

Rewrite $- 27$ as $- 1 (27)$ .

$r = \frac{- 3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$r = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

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

$r = \frac{- 3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$r = \frac{- 3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

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

Pull terms out from under the radical.

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

Move $3$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

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

Simplify the numerator.

Raise $3$ to the power of $2$ .

$r = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$r = \frac{- 3 \pm \sqrt{9 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$r = \frac{- 3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Subtract $36$ from $9$ .

$r = \frac{- 3 \pm \sqrt{- 27}}{2 \cdot 1}$

Rewrite $- 27$ as $- 1 (27)$ .

$r = \frac{- 3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$r = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

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

$r = \frac{- 3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$r = \frac{- 3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

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

Pull terms out from under the radical.

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

Move $3$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $+$ .

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

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

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

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

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

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

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

Move the negative in front of the fraction.

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

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

Simplify the numerator.

Raise $3$ to the power of $2$ .

$r = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$r = \frac{- 3 \pm \sqrt{9 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$r = \frac{- 3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Subtract $36$ from $9$ .

$r = \frac{- 3 \pm \sqrt{- 27}}{2 \cdot 1}$

Rewrite $- 27$ as $- 1 (27)$ .

$r = \frac{- 3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$r = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

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

$r = \frac{- 3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$r = \frac{- 3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

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

Pull terms out from under the radical.

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

Move $3$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $-$ .

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

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

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

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

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

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

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

Move the negative in front of the fraction.

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

The final answer is the combination of both solutions.

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

The final solution is all the values that make $(r - 3) (r^{2} + 3 r + 9) = 0$ true.

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

Find the approximate value of $r$ .

$r = 3$

Find the possible values of $θ$ .

$c o s (3 θ) = c o s (\frac{π}{2} + 2 π n)$ and $s i n (3 θ) = s i n (\frac{π}{2} + 2 π n)$

Finding all the possible values of $θ$ leads to the equation $3 θ = \frac{π}{2} + 2 π n$ .

$3 θ = \frac{π}{2} + 2 π n$

Find the value of $θ$ for $r = 0$ .

$3 θ = \frac{π}{2} + 2 π (0)$

Solve the equation for $θ$ .

Simplify.

Multiply $2 π (0)$ .

Multiply $0$ by $2$ .

$3 θ = \frac{π}{2} + 0 π$

Multiply $0$ by $π$ .

$3 θ = \frac{π}{2} + 0$

Add $\frac{π}{2}$ and $0$ .

$3 θ = \frac{π}{2}$

Divide each term in $3 θ = \frac{π}{2}$ by $3$ and simplify.

Divide each term in $3 θ = \frac{π}{2}$ by $3$ .

$\frac{3 θ}{3} = \frac{\frac{π}{2}}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 θ}{3} = \frac{\frac{π}{2}}{3}$

Divide $θ$ by $1$ .

$θ = \frac{\frac{π}{2}}{3}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{π}{2} \cdot \frac{1}{3}$

Multiply $\frac{π}{2} \cdot \frac{1}{3}$ .

Multiply $\frac{π}{2}$ by $\frac{1}{3}$ .

$θ = \frac{π}{2 \cdot 3}$

Multiply $2$ by $3$ .

$θ = \frac{π}{6}$

Use the values of $θ$ and $r$ to find a solution to the equation $u^{3} = 27 i$ .

$u_{0} = 3 (\cos (\frac{π}{6}) + i \sin (\frac{π}{6}))$

Convert the solution to rectangular form.

Simplify each term.

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$u_{0} = 3 (\frac{\sqrt{3}}{2} + i \sin (\frac{π}{6}))$

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$u_{0} = 3 (\frac{\sqrt{3}}{2} + i (\frac{1}{2}))$

Combine $i$ and $\frac{1}{2}$ .

$u_{0} = 3 (\frac{\sqrt{3}}{2} + \frac{i}{2})$

Simplify terms.

Apply the distributive property.

$u_{0} = 3 (\frac{\sqrt{3}}{2}) + 3 (\frac{i}{2})$

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

$u_{0} = \frac{3 \sqrt{3}}{2} + 3 (\frac{i}{2})$

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

$u_{0} = \frac{3 \sqrt{3}}{2} + \frac{3 i}{2}$

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

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

Find the value of $θ$ for $r = 1$ .

$3 θ = \frac{π}{2} + 2 π (1)$

Solve the equation for $θ$ .

Simplify.

Multiply $2$ by $1$ .

$3 θ = \frac{π}{2} + 2 π$

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$3 θ = \frac{π}{2} + 2 π \cdot \frac{2}{2}$

Combine $2 π$ and $\frac{2}{2}$ .

$3 θ = \frac{π}{2} + \frac{2 π \cdot 2}{2}$

Combine the numerators over the common denominator.

$3 θ = \frac{π + 2 π \cdot 2}{2}$

Multiply $2$ by $2$ .

$3 θ = \frac{π + 4 π}{2}$

Add $π$ and $4 π$ .

$3 θ = \frac{5 π}{2}$

Divide each term in $3 θ = \frac{5 π}{2}$ by $3$ and simplify.

Divide each term in $3 θ = \frac{5 π}{2}$ by $3$ .

$\frac{3 θ}{3} = \frac{\frac{5 π}{2}}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 θ}{3} = \frac{\frac{5 π}{2}}{3}$

Divide $θ$ by $1$ .

$θ = \frac{\frac{5 π}{2}}{3}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{5 π}{2} \cdot \frac{1}{3}$

Multiply $\frac{5 π}{2} \cdot \frac{1}{3}$ .

Multiply $\frac{5 π}{2}$ by $\frac{1}{3}$ .

$θ = \frac{5 π}{2 \cdot 3}$

Multiply $2$ by $3$ .

$θ = \frac{5 π}{6}$

Use the values of $θ$ and $r$ to find a solution to the equation $u^{3} = 27 i$ .

$u_{1} = 3 (\cos (\frac{5 π}{6}) + i \sin (\frac{5 π}{6}))$

Convert the solution to rectangular form.

Simplify each term.

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.

$u_{1} = 3 (- \cos (\frac{π}{6}) + i \sin (\frac{5 π}{6}))$

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$u_{1} = 3 (- \frac{\sqrt{3}}{2} + i \sin (\frac{5 π}{6}))$

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

$u_{1} = 3 (- \frac{\sqrt{3}}{2} + i \sin (\frac{π}{6}))$

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

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

Combine $i$ and $\frac{1}{2}$ .

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

Apply the distributive property.

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

Multiply $3 (- \frac{\sqrt{3}}{2})$ .

Multiply $- 1$ by $3$ .

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

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

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

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

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

Move the negative in front of the fraction.

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

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

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

Find the value of $θ$ for $r = 2$ .

$3 θ = \frac{π}{2} + 2 π (2)$

Solve the equation for $θ$ .

Simplify.

Multiply $2$ by $2$ .

$3 θ = \frac{π}{2} + 4 π$

To write $4 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$3 θ = \frac{π}{2} + 4 π \cdot \frac{2}{2}$

Combine $4 π$ and $\frac{2}{2}$ .

$3 θ = \frac{π}{2} + \frac{4 π \cdot 2}{2}$

Combine the numerators over the common denominator.

$3 θ = \frac{π + 4 π \cdot 2}{2}$

Multiply $2$ by $4$ .

$3 θ = \frac{π + 8 π}{2}$

Add $π$ and $8 π$ .

$3 θ = \frac{9 π}{2}$

Divide each term in $3 θ = \frac{9 π}{2}$ by $3$ and simplify.

Divide each term in $3 θ = \frac{9 π}{2}$ by $3$ .

$\frac{3 θ}{3} = \frac{\frac{9 π}{2}}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 θ}{3} = \frac{\frac{9 π}{2}}{3}$

Divide $θ$ by $1$ .

$θ = \frac{\frac{9 π}{2}}{3}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{9 π}{2} \cdot \frac{1}{3}$

Cancel the common factor of $3$ .

Factor $3$ out of $9 π$ .

$θ = \frac{3 (3 π)}{2} \cdot \frac{1}{3}$

Cancel the common factor.

$θ = \frac{3 (3 π)}{2} \cdot \frac{1}{3}$

Rewrite the expression.

$θ = \frac{3 π}{2}$

Use the values of $θ$ and $r$ to find a solution to the equation $u^{3} = 27 i$ .

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

Convert the solution to rectangular form.

Simplify each term.

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

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

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$u_{2} = 3 (0 + i \sin (\frac{3 π}{2}))$

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 fourth quadrant.

$u_{2} = 3 (0 + i (- \sin (\frac{π}{2})))$

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

$u_{2} = 3 (0 + i (- 1 \cdot 1))$

Multiply $- 1$ by $1$ .

$u_{2} = 3 (0 + i \cdot - 1)$

Move $- 1$ to the left of $i$ .

$u_{2} = 3 (0 - 1 \cdot i)$

Rewrite $- 1 i$ as $- i$ .

$u_{2} = 3 (0 - i)$

Simplify the expression.

Subtract $i$ from $0$ .

$u_{2} = 3 (- i)$

Multiply $- 1$ by $3$ .

$u_{2} = - 3 i$

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

$z_{2} = 0 - 3 i$

These are the complex solutions to $u^{3} = 27 i$ .

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

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

$z_{2} = - 3 i$

Find All Complex Number Solutions z^3=27i

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