Solve $z^3-3z^2+3z+7=0$ and sketch solution set

Question

Solve the equation $$z^3-3z^2+3z+7=0$$ and sketch solution set.

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My work:

Since $z=-1$ is a root of the equation then we proceed by doing Ruffini's Rule we observe that $$z^3-3z^2+3z+7=(z+1)(z^2-4z+7)=(z+1)(z-(2+\sqrt{3}i))(z-(2-\sqrt{3}i))=0.$$ Hence the solution set is $\{-1,2+\sqrt{3}i,2-\sqrt{3}i\}$ and the sketch is:

Is it okay?

Answer 1

yes, your work is correct.

You have factored the polynomial correctly. You have found the roots and plotted the roots on the complex plane.

Answer 2

Yes, it is fine. Here is an alternative way to do it. $$z^3-3z^2+3z+7=0$$ $$\Longrightarrow (z-1)^3=(-2)^3$$ $$\Longrightarrow z=1-2, 1-2\omega, 1-2\omega^2$$ where $\omega$ is a cube root of unity.

Also, now it becomes much easier to plot the roots on argand plane, just plot the cube roots of unity and apply transformation.