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For what positive real value(s) of k will (2+ki)^3 be a real number?
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I got \[\left(\begin{matrix}+ \\ -\end{matrix}\right)2\sqrt{3}\], just want to know if I did it right
yea, i got the same thing
You could use the cube roots of unity to get it. that is correct though.
if all you want is a real number, any k works
if k isnt plus minus 2sqrt(3), then when you calculate:\[(2+ki)^3\]it will still be complex.
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\[(2+ki)^3=8-6k^2+(12k-k^3)i\]for this to be real, the imaginary component must be zero, so:\[12k-k^3=0\]\[k(12-k^2)=0\]which means we have 3 solutions:\[k=0\]or\[k=\pm\sqrt{12}=\pm2\sqrt{3}\]
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