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..
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
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?
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]
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
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 /)
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?
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
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!