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For something like: $$ z^4 + 8z^2 + 3 $$ how can I find all the complex roots using the quadratic equation, or is there a better method?

I tried a u substitution using $u = z^2$, but then when I applied the quadratic formula, I was getting real roots and there was no where to sub u back in.

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up vote 5 down vote accepted

Presumably you got $u=z^2=-4\pm\sqrt{13}$. Now you need to solve for $z$. Both of our values of $z^2$ are negative. The solutions of $z^2=-4+\sqrt{13}$ are $z=\pm i \sqrt{4-\sqrt{13}}$, and the solutions of $z^2=-4-\sqrt{13}$ are $z=\pm i\sqrt{4+\sqrt{13}}$.

Remark: Things get somewhat more complicated when you want to find the square roots of a general complex number. For example, suppose that we want to find the square roots of $3+4i$. One usual way is to first rewrite $z$ as $$\sqrt{3^2+4^2} \left( \frac{3}{\sqrt{3^2+4^2}}+ \frac{4}{\sqrt{3^2+4^2}}i\right).$$ Note that $\sqrt{3^2+4^2}=5$. Let $\theta$ be an angle whose cosine is $\frac{3}{5}$ and whose sine is $\frac{4}{5}$. Then the square roots of $z$ are $$\pm \sqrt{5}\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}i\right).$$

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