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five possible fifth roots of 8-15i

Guest Feb 22, 2015

Best Answer 

 #1
avatar+92777 
+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
avatar+92777 
+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

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