#1
Dec1303, 01:27 PM

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I'm trying to find the: 4 4th roots of [itex] {\sqrt{3}} [/tex] + i .
So I made a Cartesian plane and graphed radical 3 and 1.. but these numbers can be in 2 quadrants, 1st and 3rd. r=2 ==> 2([itex] {\sqrt{3}} [/tex] + i) ===> 2(cos 30+isin 30) =4rad2(cos 7.5+isin 7.5) Would I also use 2(cos 210+isin210) ??? Thanks.. 



#2
Dec1303, 01:30 PM

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Recall that, for instance, [itex]\cos 30 = \cos 390[/itex]. Does that give you any ideas about how to proceed?
P.S., the symbols you're looking for are: [itex]\sqrt{3}[/itex] (click it to see the LaTeX source) and "& radic ; 3" (the stuff between the quotes, with no spaces), gives √3 


#3
Dec1303, 01:58 PM

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kinda. i know that to get the 4 4th roots of this equation are
30/4 = 7.5 ===> (cos 7.5 + isin 7.5) +30 (cos 37.5 + isin 37.5) +30 (cos 67.5 + isin 67.5) +30 (cos 97.5 + isin 97.5) but there is also a second set which you could start from 210. 210/4 = 52.5 ===> (cos 52.5 + isin 52.5) +210 (cos 262.5 + isin 262.5) +210 (cos 472.5 + isin 472.5) +210 (cos 682.5 + isin 682.5) Would there be 2 sets of answers to this root? 



#4
Dec1303, 02:09 PM

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Roots of Complex Numbers, Help!(For the record, I agree with the last one as well, but not the middle two) Also, why would you use 210 anyplace? [tex]2(\cos 210 + i \sin 210) \neq 2(\cos 30 + i \sin 30)[/tex] However, note that [tex]2(\cos 390 + i \sin 390) = 2(\cos 30 + i \sin 30)[/tex] 


#5
Dec1303, 02:53 PM

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you use 30 to find all of the roots of that particular equation.
substituting k =0,1,2,3..etc. k=0 for general solution (cos 30 + isin30) =2[1/2 + i[itex]/sqrt{3}[/itex] /2] = [itex] /sqrt{3} [/itex] + i k=1 you add 30 because 30 is the reference angle. k=2 add 30 k=3 add 30. haha im going to stop im confusing myself 



#6
Dec1303, 02:57 PM

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Can you write out what the general solution is supposed to be?
(P.S. You're using the wrong slash in your LaTeX; it should be \ not /) 


#7
Dec1303, 03:12 PM

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i wish my teacher would go by the book, he solves it differently than how the book does, which is very confusing. this is homework so i don't have the solution yet.
but i think i understand your point. i use k=90, because 360/4 = 90. if i wanted the 6th roots, i would use 60 as k? 



#8
Dec1303, 03:25 PM

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Right (I think):
You know that: [tex]\sqrt{3} + i = 2(\cos (30 + 360n) + i \sin(30 + 360n)) \quad n \in \mathbb{Z}[/tex] So the fourth roots would be [itex] 2 ( \cos \frac{30 + 360n}{4} + i \sin \frac{30 + 360n}{4}) = 2 (\cos (7.5 + 90n) + i \sin (7.5 + 90)) [/itex] edit: trying to get the formats right 


#9
Dec1303, 03:31 PM

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ok thanks. so i would use k=0,1,2,3.
so the 210 angle has nothing to do with the equation? in essence would it be the same as [tex]2(\cos \frac{30 + 360n}{4} + i \sin \frac{30 + 360n}{4}) = 2 (\cos (7.5 + 90n} + i \sin (7.5 + 90))[/tex] Thanks for the help I appreciate it much! 


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