# Complex number question

• May 25th 2010, 09:00 PM
Stroodle
Complex number question
Hi there, I'm having trouble with the following question. As usual, I'm probably missing something basic...

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

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

• May 25th 2010, 09:50 PM
Prove It
Quote:

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

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

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

Complex solutions always appear as conjugates.

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

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

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

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

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

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

Since $z^2 - 2z + 4$ is a factor, long divide $z^3 - a$ by $z^2 - 2z + 4$ 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 $a$.