Complex number question

Jun 2009
367
122
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 :)
 

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MHF Helper
Aug 2008
12,883
4,999
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 \(\displaystyle a\).
 
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