# Complex number question

#### 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 $$\displaystyle z=-2$$.

Thanks for your help #### Prove It

MHF Helper
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 $$\displaystyle z=-2$$.

Thanks for your help 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 $$\displaystyle a$$.

• Stroodle