<s>[INST] Using substitution in complex equation $x^{10}-2x^5+\frac{4}{3}=0$
I substituted
$t=x^5$
Then I got with the quadratic formula the result
$t_1=1+\frac{\sqrt\frac{4}{3}}{2}i$
$t_2=1-\frac{\sqrt\frac{4}{3}}{2}i$  
How do I calculate the roots now?
When I now resubstitute with  
$t=x^5$  
and form the polar coordinates do I get the roots for the whole equation?
 [/INST] You now need to solve $$x^5 = \left(1 \pm \frac{i}{\sqrt{3}}\right) = \frac{2}{\sqrt{3}}\exp\left({i\left(\frac{\pi}{6} + 2\pi k\right)}\right).$$
This should net you ten solutions, $5$ from each - which is intuitively right since you were originally solving a polynomial of degree $10$.
So $$x = \left(\frac{2}{\sqrt{3}}\right)^{1/5}\exp{\left(i\left(\frac{\pi}{30} + \frac{2\pi k}{5}\right)\right)}$$
Then taking various values of $k$ will get you all your roots. 
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