1. ## complex numbers

how do i solve (x+2i)^3=2

2. Ok so we have $(x + 2i)^{3} = 2$

Take the cubed root of both sides

$x + 2i = \sqrt[3]{2}$

Now subtract 2i from the left and the right

$x = \sqrt[3]{2} - 2i$

3. Originally Posted by SnipedYou
Ok so we have $(x + 2i)^{3} = 2$

Take the cubed root of both sides

$x + 2i = \sqrt[3]{2}$

Now subtract 2i from the left and the right

$x = \sqrt[3]{2} - 2i$

$(x+2i)^3 = 2$, then all the solutions are:
$x+2i = \sqrt[3]{2},\sqrt[3]{2}\zeta,\sqrt[3]{2}\zeta^2$ where $\zeta$ is primitive root of unity.
That means,
$x = -2i + \sqrt[3]{2} \zeta^k$ for $k=0,1,2$.

4. I'm searching for them, but I saw that one right when I started trying to solve it.

I'm stuck here $x^{3} + 6x^{2}i - 12x - i - 2 = 0$ Could I use synthetic division somehow to simplify that knowing that $\sqrt[3]{2}-2i$ is a factor? Or is there another way to factor it?

5. Originally Posted by SnipedYou
I'm searching for them, but I saw that one right when I started trying to solve it.

I'm stuck here $x^{3} + 6x^{2}i - 12x - i - 2 = 0$ Could I use synthetic division somehow to simplify that knowing that $\sqrt[3]{2}-2i$ is a factor? Or is there another way to factor it?
No! Do not fact.

The equation,
$x^3 = a^3$
So,
$x^3 - a^3 = 0$
$(x-a)(x^2+xa+a^2)=0$
So, $x-a=0 \mbox{ and }x^2+xa+a^2 = 0$
So, $x = a, x = \frac{-a\pm \sqrt{a^2 - 4a^2}}{2}$

So in the equation,
$(x+2i)^3 = a^3$ where $a=\sqrt[3]{2}$.
Now use the result above.