1. ## Complex number question

Hi there, I'm having trouble with the following question. As usual, I'm probably missing something basic...

If $\displaystyle z=1+\sqrt{3}i$ is a solution to the equation $\displaystyle z^3-a=0,\ a\in\ c$ then the other two solutions are?

I know one of them must be $\displaystyle z=1-\sqrt{3}i$ as the coefficient of $\displaystyle z^3$ is real. But I'm not sure how to find the last solution.
From the options (mc question) it seems that it's either $\displaystyle z=2$ or $\displaystyle z=-2$.

2. Originally Posted by Stroodle
Hi there, I'm having trouble with the following question. As usual, I'm probably missing something basic...

If $\displaystyle z=1+\sqrt{3}i$ is a solution to the equation $\displaystyle z^3-a=0,\ a\in\ c$ then the other two solutions are?

I know one of them must be $\displaystyle z=1-\sqrt{3}i$ as the coefficient of $\displaystyle z^3$ is real. But I'm not sure how to find the last solution.
From the options (mc question) it seems that it's either $\displaystyle z=2$ or $\displaystyle z=-2$.

Complex solutions always appear as conjugates.

So if $\displaystyle z = 1 + \sqrt{3}i$ is a solution, so is $\displaystyle \overline{z} = 1 - \sqrt{3}i$.

That means $\displaystyle z - (1 + \sqrt{3}i)$ and $\displaystyle z - (1 - \sqrt{3}i)$ are both factors, and so is

$\displaystyle [z - (1 + \sqrt{3}i)][z - (1 - \sqrt{3}i)]$

$\displaystyle =z^2 - z(1 - \sqrt{3}i) - z(1 + \sqrt{3}i) + (1 - \sqrt{3}i)(1 + \sqrt{3}i)$

$\displaystyle = z^2 - z + \sqrt{3}iz - z - \sqrt{3}iz + 1 + 3$

$\displaystyle = z^2 - 2z + 4$.

Since $\displaystyle z^2 - 2z + 4$ is a factor, long divide $\displaystyle z^3 - a$ by $\displaystyle z^2 - 2z + 4$ and you will have the third factor. You should note that the third root will need to be real.

Once you have the third factor, you can find $\displaystyle a$.