# The cube root of a complex number

Warning! This proof is work in progress, and it probably contains false statements.

Consider the cube root of a complex number, $\sqrt[3]{a + bi}$. Now write this in the standard form: $c + di$, so you get:

$$\sqrt[3]{a + bi} = c + di$$

Cube both sides and expand the brackets.

\begin{aligned}a + bi &= (c + di)^3 \newline &= c^3 + 3c^2di + 3c(di)^2 + (di)^3 \newline &= c^3 + 3c^2di - 3cd^2 - d^3i \newline &= c^3 - 3cd^2 + i(3c^2d - d^3)\end{aligned}

Since $a, b, c, d \in \R$, the real parts and the imaginary parts must match, so you get this systems of equations:

$$a = c^3 - 3cd^2$$

$$b = 3c^2d - d^3$$

From the first equation, isolate $d^2$:

$$3cd^2 = c^3 - a$$

$$d^2 = \frac{c^3 - a}{3c}$$

From the second equation, divide every term by $d$ and substitute $d^2$.

$$3c^2 - d^2 = \frac{b}{d}$$

$$3c^2 - \frac{c^3 - a}{3c} = \frac{b}{d}$$

Split the fraction and add like terms.

$$3c^2 - \frac{1}{3}c^2 + \frac{a}{3c} = \frac{b}{d}$$

$$\frac{8}{3}c^2 + \frac{a}{3c} = \frac{b}{d}$$

Square both sides and expand the brackets.

$$\left(\frac{8}{3}c^2 + \frac{a}{3c}\right)^2 = \left(\frac{b}{d}\right)^2$$

$$\frac{64}{9}c^4 + \frac{16}{9}ac + \frac{a^2}{9c^2} = \frac{b^2}{d^2}$$

Substitute $d^2$ and multiply both sides by $c^2$.

$$\frac{64}{9}c^6 + \frac{16}{9}ac^3 + \frac{a^2}{9} = \frac{b^2c^2}{\frac{c^3 - a}{3c}} = \frac{3b^2c^3}{c^3 - a}$$

Multiply both sides by $c^3 - a$.

$$\left(\tfrac{64}{9}c^6 + \tfrac{16}{9}ac^3 + \tfrac{1}{9}a^2 \right)(c^3 - a) = 3b^2c^3$$

$$\tfrac{64}{9}c^9 + \tfrac{16}{9}ac^6 + \tfrac{1}{9}a^2c^3 - \tfrac{64}{9}ac^6 - \tfrac{16}{9}a^2c^3 - \tfrac{1}{9}a^3 - 3b^2c^3 = 0$$

Factor out powers of $c$.

$$\tfrac{64}{9}c^9 + \left(\tfrac{16}{9}a - \tfrac{64}{9}a\right)c^6 + \left(\tfrac{1}{9}a^2 - \tfrac{16}{9}a^2 - 3b^2\right)c^3 - \tfrac{1}{9}a^3 = 0$$

$$\tfrac{64}{9}c^9 - \tfrac{48}{9}ac^6 + \left(- \tfrac{15}{9}a^2 - 3b^2\right)c^3 - \tfrac{1}{9}a^3 = 0$$

Divide each term by $\frac{64}{9}$, so multiply by $9$ and divide by $64$.

$$c^9 - \tfrac{3}{4}ac^6 + \left(-\tfrac{15}{64}a^2 - \tfrac{27}{64}b^2\right)c^3 - \tfrac{1}{64}a^3 = 0$$

Let $c^3 = n + \tfrac{1}{4} a$ and expand the brackets.

$$\left(n + \tfrac{1}{4}a\right)^3 - \tfrac{3}{4}a\left(n + \tfrac{1}{4}a\right)^2 + \left(-\tfrac{15}{64}a^2 - \tfrac{27}{64}b^2\right)\left(n + \tfrac{1}{4}a\right) - \tfrac{1}{64}a^3 = 0$$

$$\bigg(n^3 + 3n^2 \cdot \tfrac{1}{4}a + 3n \cdot \tfrac{1}{16}a^2 + \tfrac{1}{64}a^3 \bigg) - \tfrac{3}{4}a\bigg(n^2 + \tfrac{1}{2}an + \tfrac{1}{16}a^2\bigg) + \bigg(-\tfrac{15}{64}a^2n - \tfrac{15}{256}a^3 - \tfrac{27}{64}b^2n - \tfrac{27}{256}ab^2 \bigg) - \tfrac{1}{64}a^3 = 0$$

$$n^3 + \tfrac{3}{4}an^2 + \tfrac{3}{16}a^2n + \tfrac{1}{64}a^3 - \tfrac{3}{4}an^2 - \tfrac{1}{8}a^2n - \tfrac{3}{64}a^3 - \tfrac{15}{64}a^2n - \tfrac{15}{256}a^3 - \tfrac{27}{64}b^2n - \tfrac{27}{256}ab^2 - \tfrac{1}{64}a^3 = 0$$

Group all terms by $n$ and add like terms.

$$n^3 + \bigg(\tfrac{3}{4}a - \tfrac{3}{4}a\bigg)n^2 + \bigg(\tfrac{3}{16}a^2 - \tfrac{3}{8}a^2 - \tfrac{15}{64}a^2 - \tfrac{27}{64}b^2\bigg)n + \bigg(\tfrac{1}{64}a^3 - \tfrac{3}{64}a^3 - \tfrac{15}{256}a^3 - \tfrac{27}{256}ab^2 - \tfrac{1}{64}a^3\bigg) = 0$$

$$n^3 + \bigg(-\tfrac{27}{64}a^2 - \tfrac{27}{64}b^2\bigg)n + \bigg(-\tfrac{27}{256}a^3 - \tfrac{27}{256}ab^2\bigg) = 0$$

Let $n = u + v$ and $-\tfrac{27}{64}a^2 - \tfrac{27}{64}b^2 = -3uv \implies v = \frac{\frac{27}{64}a^2 + \frac{27}{64}b^2}{3u} = \frac{\frac{9}{64}a^2 + \frac{9}{64}b^2}{u}$.

$$(u + v)^3 - 3uv(u + v) - \tfrac{27}{256}a^3 - \tfrac{27}{256}ab^2 = 0$$

$$u^3 + 3u^2v + 3uv^2 + v^3 - 3u^2v - 3uv^2 - \tfrac{27}{256}a^3 - \tfrac{27}{256}ab^2 = 0$$

$$u^3 + v^3 - \tfrac{27}{256}a^3 - \tfrac{27}{256}ab^2 = 0$$

Substitute $v$ and multiply both sides by $u^3$.

$$u^3 + \left(\frac{\tfrac{9}{64}a^2 + \frac{9}{64}b^2}{u}\right)^3 - \tfrac{27}{256}a^3 - \tfrac{27}{256}ab^2 = 0$$

$$u^3 + \frac{\left(\frac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3}{u^3} - \tfrac{27}{256}a^3 - \tfrac{27}{256}ab^2 = 0$$

$$u^6 + \left(\tfrac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3 + \bigg(-\tfrac{27}{256}a^3 - \tfrac{27}{256}ab^2\bigg)u^3 = 0$$

$$u^6 + \bigg(-\tfrac{27}{256}a^3 - \tfrac{27}{256}ab^2\bigg)u^3 + \left(\tfrac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3 = 0$$

Use the quadratic formula to solve for $u^3$.

$$u^3 = \frac{-\left(-\tfrac{27}{256}a^3 - \tfrac{27}{256}ab^2\right) \pm \sqrt{(-\tfrac{27}{256}a^3 - \tfrac{27}{256}ab^2)^2 - 4 \left(\tfrac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3}}{2}$$

$$u^3 = \tfrac{1}{2}\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right) \pm \tfrac{1}{2} \sqrt{\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right)^2 - 4 \left(\tfrac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3}$$

Take the cube root on both sides to isolate $u$.

$$u = \sqrt[3]{\tfrac{1}{2}\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right) \pm \tfrac{1}{2} \sqrt{\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right)^2 - 4 \left(\tfrac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3}}$$

Now substitute $u$ to solve for $v$.

$$v = \frac{\frac{9}{64}a^2 + \frac{9}{64}b^2}{u} = \frac{\frac{9}{64}a^2 + \frac{9}{64}b^2}{\sqrt[3]{\tfrac{1}{2}\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right) \pm \tfrac{1}{2} \sqrt{\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right)^2 - 4 \left(\tfrac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3}}}$$

Substiute $u$ and $v$ back into $n$:

$$n = u + v = \sqrt[3]{\tfrac{1}{2}\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right) \pm \tfrac{1}{2} \sqrt{\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right)^2 - 4 \left(\tfrac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3}} + \frac{\frac{9}{64}a^2 + \frac{9}{64}b^2}{\sqrt[3]{\tfrac{1}{2}\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right) \pm \tfrac{1}{2} \sqrt{\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right)^2 - 4 \left(\tfrac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3}}}$$

Finally, substitute $n$ back into $c$.

$$c^3 = n + \tfrac{1}{4}a$$

$$c = \sqrt[3]{\sqrt[3]{\tfrac{1}{2}\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right) \pm \tfrac{1}{2} \sqrt{\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right)^2 - 4 \left(\tfrac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3}} + \frac{\frac{9}{64}a^2 + \frac{9}{64}b^2}{\sqrt[3]{\tfrac{1}{2}\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right) \pm \tfrac{1}{2} \sqrt{\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right)^2 - 4 \left(\tfrac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3}}} + \tfrac{1}{4}a}$$

Or, shorter:

$$c = \sqrt[3]{u + \frac{\frac{9}{64}a^2 + \frac{9}{64}b^2}{u} + \tfrac{1}{4}a}, \text{where } u = \sqrt[3]{\tfrac{1}{2}\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right) \pm \tfrac{1}{2} \sqrt{\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right)^2 - 4 \left(\tfrac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3}}$$

From $d^2 = \frac{c^3 - a}{3c}$ follows that:

\begin{aligned}d &= \sqrt{\frac{c^3 - a}{3c}} \newline&= \sqrt{\frac{u + v + \frac{1}{4}a - a}{3c}} \newline&= \sqrt{\frac{u + v - \frac{3}{4}a}{3\sqrt[3]{u + \frac{\frac{9}{64}a^2 + \frac{9}{64}b^2}{u} + \tfrac{1}{4}a}}} \newline&, \text{where } u = \sqrt[3]{\tfrac{1}{2}\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right) \pm \tfrac{1}{2} \sqrt{\left(\tfrac{27}{256}a^3 + \tfrac{27}{256}ab^2\right)^2 - 4 \left(\tfrac{9}{64}a^2 + \tfrac{9}{64}b^2\right)^3}}\end{aligned}