$$\sqrt[5]{8-15i}$$
I saw Alan do one like this recently. I will give it a go.
$$\sqrt{8^2+15^2}=17$$
$$\\8-15i = 17\left(\frac{8}{17}-\frac{15}{17}i\right)=17(cos\theta+isin\theta)=17e^{i\theta}\\\\
cos\theta=\frac{8}{17}\\
sin\theta=-\frac{15}{17}\\
\theta $ must be in the 4th quad $\\
tan\theta=-\frac{15}{8}\\
\theta\approx -61.9275\\\\
\sqrt[5]{8-15i}\\
=(17e^{i\theta})^{1/5}\\
=17^{1/5}e^{(\theta/5)i}\\
=17^{1/5}e^{(-61.9275/5)i}\\
=17^{1/5}e^{(-61.9275/5)i}\\
=1.7623e^{-12.3855i}\\$$
now these angles are in degrees abd 360/5=72degrees
-12.3855+72=59.61
59.61+72=131.61
131.61+72=203.61
203.61+72=275.61
275.61+72=347.61
so the radius of the polar co-ordinates is 1.7623 and $$\theta$$ = 59.61, 131.61, 203.61, 275.61 and 347.61
these polar co-ordinates are graphed on this page
http://www.wolframalpha.com/input/?s=51&_=1424610503524&i=%288-15i%29^%281%2f5%29&fp=1&incTime=true
Now I suppose I should change these polar coordinates back to standard complex numbers but it is very late for me and it is all a bit much for now.
That should be a good start for you anyway.
$$\sqrt[5]{8-15i}$$
I saw Alan do one like this recently. I will give it a go.
$$\sqrt{8^2+15^2}=17$$
$$\\8-15i = 17\left(\frac{8}{17}-\frac{15}{17}i\right)=17(cos\theta+isin\theta)=17e^{i\theta}\\\\
cos\theta=\frac{8}{17}\\
sin\theta=-\frac{15}{17}\\
\theta $ must be in the 4th quad $\\
tan\theta=-\frac{15}{8}\\
\theta\approx -61.9275\\\\
\sqrt[5]{8-15i}\\
=(17e^{i\theta})^{1/5}\\
=17^{1/5}e^{(\theta/5)i}\\
=17^{1/5}e^{(-61.9275/5)i}\\
=17^{1/5}e^{(-61.9275/5)i}\\
=1.7623e^{-12.3855i}\\$$
now these angles are in degrees abd 360/5=72degrees
-12.3855+72=59.61
59.61+72=131.61
131.61+72=203.61
203.61+72=275.61
275.61+72=347.61
so the radius of the polar co-ordinates is 1.7623 and $$\theta$$ = 59.61, 131.61, 203.61, 275.61 and 347.61
these polar co-ordinates are graphed on this page
http://www.wolframalpha.com/input/?s=51&_=1424610503524&i=%288-15i%29^%281%2f5%29&fp=1&incTime=true
Now I suppose I should change these polar coordinates back to standard complex numbers but it is very late for me and it is all a bit much for now.
That should be a good start for you anyway.