#1**+5 **

$$\sqrt[5]{8-15i}$$

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.

Melody
Feb 22, 2015

#1**+5 **

Best Answer

$$\sqrt[5]{8-15i}$$

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.

Melody
Feb 22, 2015