# Help with finding all the roots to $z^6 - 2z^3 + 2 = 0$.

I need help with finding the roots to the equation $z^6 - 2z^3 + 2 = 0$

I start with assigning $x$ as $z^3$. This gives me the equation: $x^2 - 2x + 2 = (x-1)^2 + 1 = 0$.

Further developments:

$(x-1)^2 = -1 = e^{i\pi} \leftrightarrow x = 1 \pm e^{i\pi\frac{1}{2}}$.

This means that $x_1 = 1 + e^{i\pi\frac{1}{2}}$ and $x_2 = 1 - e^{i\pi\frac{1}{2}}$

The problem is that my math book tells me that at this stage the result for $x$ should be $x = 1 \pm i\sqrt{2}e^{\pm i\pi\frac{1}{4}}$.

What have I done wrong?

Thank you kindly for your help!

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You are right. The solution of $x^2-2x+2$ is $1\pm i=1\pm e^{i\frac\pi 2}$ and not $1 \pm i\sqrt{2}e^{\pm i\frac\pi 4}$. – P.. Feb 20 '13 at 19:14
From the answer by @nbubis it appears that your textbook simply is missing an ' $=$ ' between the $i$ and the $\sqrt2$. – half-integer fan Feb 20 '13 at 19:46

What's wrong with using the simple root formula? $$x = \frac{2\pm\sqrt{4-8}}{2} = 1\pm i = 1 \pm i=\sqrt{2}e^{\pm i\pi/4}$$ You then get: $$z = x^{1/3} = 2^{1/6}e^{\pm i\pi/12}e^{2n i\pi/3}$$
$$-1 = e^{\pm i\pi}$$