# nth roots of unity - Help!Watch

This discussion is closed.
#1
w = e^(2iPi/5).

Express the 5th roots of unity in terms of w.

I'm puzzled.

I thought: 5th root (w) = e^{2iPi/25 +/- 2Pi/5}

Is that right? And if so, how do I get the roots in terms of w and 5 of them?
0
13 years ago
#2
the fifth root of w is e^[(2iPi/5 + 2kpi)/5]. which is as u said "I thought: 5th root (w) = e^{2iPi/25 +/- 2Pi/5}" but u forgot the k.

so:
k=0 ==> 5th root (w) = e^{2iPi/25}
k = 1 ==> 5th root (w) = e^{2iPi/25 + 2Pi/5}
k= 2 ==> 5th root (w) = e^{2iPi/25 + 4Pi/5}
k = 3 ==> 5th root (w) = e^{2iPi/25 + 6Pi/5}
k = 4 ==> 5th root (w) = e^{2iPi/25 + 8Pi/5}
0
13 years ago
#3
cheers, ur welcome
0
13 years ago
#4
well i dont know if those are right and good answers:
let roots be z1 z2 z3 z4 z5:
z1² + z2² + z3² + z4² + z5² = 5 w

i dont know if THE FOLLOWING is right:
z1.z2.z3.z4.z5 = w

i donno, is the product of ALL the roots equal to w? i donno.

btw this post is just to escape from ur question :P i jut donno what the answer is
0
13 years ago
#5
There's slight confusion. Doesnt the question ask to find the 5th roots of unit? You seem to be trying to find the fith roots of w.

The fith roots of unit are

e(2kiPi/5) ... k = {0,1,2,3,4}

which is basically

e(2iPi/5)k = wk

so the answer is just w0, w1, w2, w3, w4.
0
13 years ago
#6
??? i dont understand what u mean.
the fifth root of w is e^[(2iPi/5 + 2kpi)/5]
where k = 0,1,2,3,4
0
13 years ago
#7
(Original post by yazan_l)
??? i dont understand what u mean.
the fifth root of w is e^[(2iPi/5 + 2kipi)/5]
where k = 0,1,2,3,4
Yes, you are right. But the question says that w = this thing, now express the fith roots of unity in terms of w.
0
#8
Yes that's right, we're wanting the 5th roots of 1 (unity) and once we get these values, express them in terms of w.

Also another thing, just been reading through my notes:

5th root (w) = e^[i(2Pi/5 + 2kpi)/5] = e^[i(2Pi/25 + 2kPi/5)] = e^[2ipi/25 + 2ikPi/5]

Yes? Just wanting to make sure that the "i" is outside the bracket and thus when you multiply through you get 2ikPi/5 inside. In the above posts I didn't do that.
0
13 years ago
#9
(Original post by Nima)
Yes that's right, we're wanting the 5th roots of 1 (unity) and once we get these values, express them in terms of w.

Also another thing, just been reading through my notes:

5th root (w) = e^[i(2Pi/5 + 2kpi)/5] = e^[i(2Pi/25 + 2kPi/5)] = e^[2ipi/25 + 2ikPi/5]

Yes? Just wanting to make sure that the "i" is outside the bracket and thus when you multiply through you get 2ikPi/5 inside. In the above posts I didn't do that.
You could do it that way too.

Just remember it this way, sin and cos are 2Pi periodic. Also when you have a solution, multiplying it by 1 will not change the solution.

1k = e2iPik

so say w is a solution, then 1.w is a solution too. ie

1.w = 1kw = w.e2kiPi.

remember, its always two-key-pi-i. [remember it like that ]

dont forget the "i".
0
13 years ago
#10
(Original post by Nima)
Just noticed from your edited post galois about you adding i in red - So I was right about the i being outside the bracket.

Cheers.
No problem.
0
13 years ago
#11
well sorry for forgetting the i, but i'm used to use the cos + isin form, so that's why i didnt include the i
i just dont like to use the e^ form
0
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