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Roots of Complex Numbers, Help!

by StarkyDee
Tags: complex, numbers, roots
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StarkyDee
#1
Dec13-03, 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..
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Hurkyl
#2
Dec13-03, 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
StarkyDee
#3
Dec13-03, 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?

Hurkyl
#4
Dec13-03, 02:09 PM
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Roots of Complex Numbers, Help!

(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)
I agree with the first of these four. What justification do you have for the other three? It seems you are simply adding 30 degrees... why?

(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]
StarkyDee
#5
Dec13-03, 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
Hurkyl
#6
Dec13-03, 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 /)
StarkyDee
#7
Dec13-03, 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?
Hurkyl
#8
Dec13-03, 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
StarkyDee
#9
Dec13-03, 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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