# Complex Numbers making Real Numbers

As a point of curiosity, I found myself wondering when a complex number to an integer power creates a real number. For the sake of completeness, I also looked at when the result is a fully imaginary number.

More rigorously, define $$\mathbb{C}^*$$ as the set of complex numbers $$a + bi$$ where both $$a$$ and $$b$$ are non-zero, and $$k\in\mathbb{N}^*$$. For what values $$x \in \mathbb{C}^*$$ is $$x^k \in \mathbb{R} \lor \mathbb{I}$$?

### k = 1

This is the trivial case. Since $$x^1 = x$$, there are no real or imaginary cases.

### k = 2

$$(a + bi)^2 = a^2 + 2abi + b^2i^2 = a^2 + 2abi – b^2 = (a^2 – b^2) + 2abi$$.

This is real when the coefficient of $$i$$, i.e., $$2ab$$ is equal to 0. This is only true when $$a = 0$$ or $$b = 0$$, so it is not true for any element of  $$\mathbb{C}^*$$.

This is imaginary when $$a^2 – b^2 = 0 \Rightarrow a^2 = b^2$$. That is, $$(a \pm ai)^2 \in \mathbb{C}^*$$.

### k = 3

$$(a + bi)^3 = a^2 + 3a^2bi – 3ab^2 – b^3i = a(a^2 – 3b^2) + b(3a^2 – b^2)i$$.

This is real when $$b(3a^2 – b^2) = 0$$. If $$b = 0, x\notin\mathbb{C}^*$$, so we ignore that case. $$3a^2 – b^2 = 0 \Rightarrow 3a^2 = b^2$$, so $$b = \pm\sqrt{3} a$$. Hence, if $$x\in\mathbb{C}^*$$, $$x = a \pm a\sqrt{3}i$$.

Similarly, this is imaginary when $$a(a^2 – 3b^2) = 0$$, so $$b = \pm\frac{\sqrt{3}}{3} a$$ and $$x = a \pm\frac{a\sqrt{3}}{3}i$$.

### k = 4

Using similar methods, we get: $$(a + bi)^4 = (a^4 – 6a^2b^2 + b^4) + 4ab(a^2 – b^2)i$$.

This is real when $$a^2 = b^2$$, so $$x = a \pm ai$$.

Replacing $$a^2 = c$$ and $$b^2 = d$$ gives us the real part of $$d^2 – 6dc + c^2$$. Applying the quadratic formula with $$a = 1$$, $$b = -6c$$, and $$d = c^2$$ yields $$c = \frac{6c \pm \sqrt{36c^2 – 4c^2}}{2} = 3c \pm c\sqrt{8} = (3 \pm 2\sqrt{2}) c$$. Note that since the quadratic is symmetrical, that means that $$3 + 2\sqrt{2} = \frac{1}{3 – 2\sqrt{2}}$$.

Since $$a^2 = c$$ and $$b^2 = d = (3 \pm 2\sqrt{2})c$$, $$a = \pm\sqrt{c}$$ and $$b = \pm(3 \pm 2\sqrt{2})\sqrt{c} = \pm(3 \pm 2\sqrt{2})a$$.

This is imaginary, then, when $$x = a \pm(3 \pm 2\sqrt{2}) ai$$.

### k = 5

This time, we get: $$(a + bi)^5 = a(a^4 – 10a^2b^2 + 5b^4) + b(5a^4 – 10a^2b^2 + b^4)i$$

This is real when $$5a^4 – 10a^2b^2 + b^4 = 5c^2 – 10cd + d^2 = 0$$.

$$d = \frac{10c \pm\sqrt{100c^2 – 20c^2}}{10} = \frac{10c \pm 4c\sqrt{5}}{2} = 5 \pm 2\sqrt{5} c$$, so $$b = \pm\sqrt{5 \pm 2\sqrt{5}}a$$.

This is real, then, when $$x = a \pm \sqrt{5 \pm 2\sqrt{5}}ai$$.

This is imaginary when $$a^4 – 10a^2b^2 + 5b^4 = c^2 – 10cd + 5d^2 = 0$$.

$$d = \frac{10c \pm \sqrt{100c^2 – 20c^2}}{10} = \frac{10c \pm 4c\sqrt{5}}{10} = (1 \pm \frac{2\sqrt{5}}{5})c$$, so $$b = \pm\sqrt{1\pm\frac{2\sqrt{5}}{5}}a$$.

This is imaginary, then, when $$x = a \pm\sqrt{1\pm\frac{2\sqrt{5}}{5}}ai$$.

### k = 6

This time, we get: $$(a + bi)^6 = (a^6 – 15a^4b^2 + 15a^2b^4 – b^6) + 2ab(3a^4 – 10a^2b^2 + 3b^4)i$$

This is real when $$3a^4 – 10a^2b^2 + 3b^4 = 3c^2 – 10cd + 3d^2 = 0$$.

$$d = \frac{10c \pm \sqrt{100c^2 – 36c^2}}{6} = \frac{10c \pm 8c}{6} = \frac{5\pm 4}{3}c$$, so $$b = \pm\sqrt{\frac{5\pm 4}{3}}a$$.

This is real, then, when $$x = a \pm\sqrt{\frac{5\pm 4}{3}}ai$$.

This is imaginary when $$a^6 – 15a^4b^2 + 15a^2b^4 – b^6 = c^3 – 15c^2d + 15cd^2 – d^3 = 0$$.

We can factor $$c^3 – 15c^2d + 15cd^2 – d^3 = (c – d)(c^2 – 14cd + d^2)$$. $$c – d = 0$$ when $$c = d$$, so $$b = \pm a$$ is one possibility. Let’s look at the other case.

$$c^2 – 14cd + d^2 = 0$$ when $$d = \frac{14c \pm\sqrt{196c^2 – 4c^2}}{2} = 7c \pm\sqrt{48}c = (7 \pm 4\sqrt{3})c$$, so $$b = \pm\sqrt{7 \pm 4\sqrt{3}}a$$.

This is imaginary, then, when $$x \in \{a \pm ai, a \pm\sqrt{7\pm 4\sqrt{3}}ai\}$$.

### Summary

At $$k = 7$$, we have the cubics $$c^3 – 21c^2d + 35cd^2 – 7d^3$$ and $$7c^3 – 35c^2d + 21cd^2 – d^3$$, which do not have factors, so this is a good place to stop.

Factoring out $$a$$, here are the cases where $$x\in\mathbb{C}^*$$ yields $$x^k \in \mathbb{R} \lor \mathbb{I}$$:

 k $$x^k\in\mathbb{R}$$ $$x^k\in\mathbb{I}$$ 1 (None) (None) 2 (None) $$1 \pm i$$ 3 $$1\pm\sqrt{3}i$$ $$1\pm\frac{\sqrt{3}}{3}i$$ 4 $$1\pm i$$ $$1 \pm(3 \pm 2\sqrt{2})i$$ 5 $$1\pm\sqrt{5\pm 2\sqrt{5}}i$$ $$1\pm\sqrt{1\pm\frac{2\sqrt{5}}{5}}i$$ 6 $$1\pm\sqrt{\frac{5\pm 4}{3}}i$$ $$\{1\pm i, 1\pm\sqrt{7\pm 4\sqrt{3}}i\}$$