# Roots of Complex Numbers, Help!

by StarkyDee
Tags: complex, numbers, roots
 P: n/a I'm trying to find the: 4 4th roots of ${\sqrt{3}} [/tex] + i . So I made a Cartesian plane and graphed radical 3 and 1.. but these numbers can be in 2 quadrants, 1st and 3rd. r=2 ==> 2([itex] {\sqrt{3}} [/tex] + i) ===> 2(cos 30+isin 30) =4rad2(cos 7.5+isin 7.5) Would I also use 2(cos 210+isin210) ??? Thanks..  Emeritus Sci Advisor PF Gold P: 16,091 Recall that, for instance, [itex]\cos 30 = \cos 390$. Does that give you any ideas about how to proceed? P.S., the symbols you're looking for are: $\sqrt{3}$ (click it to see the LaTeX source) and "& radic ; 3" (the stuff between the quotes, with no spaces), gives √3
 P: n/a 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?
Emeritus
PF Gold
P: 16,091
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?

$$2(\cos 210 + i \sin 210) \neq 2(\cos 30 + i \sin 30)$$

However, note that

$$2(\cos 390 + i \sin 390) = 2(\cos 30 + i \sin 30)$$
 P: n/a 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$/sqrt{3}$ /2] = $/sqrt{3}$ + 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
 Emeritus Sci Advisor PF Gold P: 16,091 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 /)
 P: n/a 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?
 Emeritus Sci Advisor PF Gold P: 16,091 Right (I think): You know that: $$\sqrt{3} + i = 2(\cos (30 + 360n) + i \sin(30 + 360n)) \quad n \in \mathbb{Z}$$ So the fourth roots would be $2 ( \cos \frac{30 + 360n}{4} + i \sin \frac{30 + 360n}{4}) = 2 (\cos (7.5 + 90n) + i \sin (7.5 + 90))$ edit: trying to get the formats right
 P: n/a 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 $$2(\cos \frac{30 + 360n}{4} + i \sin \frac{30 + 360n}{4}) = 2 (\cos (7.5 + 90n} + i \sin (7.5 + 90))$$ Thanks for the help I appreciate it much!

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