nth roots of unit proofs question

• Apr 19th 2014, 04:14 AM
Applestrudle
nth roots of unit proofs question
http://i57.tinypic.com/1678fv9.png

For 4i)

i get to

$\theta = ± k \frac{2 \pi}{n}$ k is an integer

$k = ±1 {z}_{1} = {e}^{i2\pi)} , {z}_{2} = {e}^{-i2\pi}$

I just proved it like that, not sure if its right

ii)

for even n

${(-z)}^{n} = {z}^{n} = 1$

but for odd n

${(-z)}^{n} = -{z}^{n} = -1$

for the sum, i got zero

for the product:

for even n i got

$(-1)\frac{n}{2} . (1)\frac{n}{2} = -\frac{{n}^{2}}{4}$ since you have (-1) half the time and 1 half the time

for odd n

$(1)\frac{n}{2} . (-1)(\frac{n}{2} +1) = -(\frac{{n}^{2}}{4} + \frac{n}{2})$ since you have one more -1 than 1

• Apr 19th 2014, 07:12 AM
Plato
Re: nth roots of unit proofs question
Quote:

Originally Posted by Applestrudle

Here is a suggestion on notation that can simplify this answer. Use the exponentiation notation.
$e^z=\exp(z)=|z|(\cos(\theta)+i~\sin(\theta))$ where $\theta=\text{Arg}(z))$
This notation makes conjugate notation easy: $\overline{\exp(z)}=\exp(\overline{z} )$

For the first part of this question let $\theta=\dfrac{2\pi}{n}~\&~\rho=\exp(i\theta)$
Now it is easy to list the roots: $\rho^k:~k=0,1,\cdots, k-1$, and $\overline{~\rho~}=\exp(-i\theta)$.

Note that $\prod\limits_{k = 0}^{n - 1} {{\rho ^k}} = 1$.

For the second part the n nth roots of $i$: define $\rho=\exp\left(\dfrac{\pi i}{2}\right)~\&~\xi=\exp\left(\dfrac{2\pi i}{n}\right)$
Then those roots can be listed as: $\rho\cdot\xi^k,~k=0,\cdots,k-1$.
• Apr 25th 2014, 08:53 AM
Applestrudle
Re: nth roots of unit proofs question
Quote:

Originally Posted by Plato
Here is a suggestion on notation that can simplify this answer. Use the exponentiation notation.
$e^z=\exp(z)=|z|(\cos(\theta)+i~\sin(\theta))$ where $\theta=\text{Arg}(z))$
This notation makes conjugate notation easy: $\overline{\exp(z)}=\exp(\overline{z} )$

For the first part of this question let $\theta=\dfrac{2\pi}{n}~\&~\rho=\exp(i\theta)$
Now it is easy to list the roots: $\rho^k:~k=0,1,\cdots, k-1$, and $\overline{~\rho~}=\exp(-i\theta)$.

Note that $\prod\limits_{k = 0}^{n - 1} {{\rho ^k}} = 1$.

For the second part the n nth roots of $i$: define $\rho=\exp\left(\dfrac{\pi i}{2}\right)~\&~\xi=\exp\left(\dfrac{2\pi i}{n}\right)$
Then those roots can be listed as: $\rho\cdot\xi^k,~k=0,\cdots,k-1$.

for the sum z1 +z2 + z3+ z4+ .... i did the sum from k =1 to k=n of ${e}^{\frac{i2\pi k}{n}}$ and i got $\frac{1-{e}^{\frac{2 \pi k}{n}}}{1 - {e}^{\frac{i2 \pi}{n}}}$ i used geometric series

x + x^2 and x^3 + x^4 .... x^n = (1-x^n)/(1-x)

for the product z1.z2.z3.z4.z5 ... I used the geometric series in the exponential since ${z}_{k} = {e}^{\frac{2 \pi k}{n}}$

and i got the sum of the product as equal to ${e}^{i2\pi \frac{1-n}{n}}$

for the nth roots of i I got

for the product ${w}_{k} = {e}^{i(\frac{pi}{2n}+\frac{2\pik}{n})}$

so the product is e^(ipi/2n + sum of 2pik/n) for k from 1 to n

for the sum, using the geometric series, I got (2pi/n)(-n), giving -2pi ?

then the final answer for the product is ${e}^{i(\frac{\pi}{2n} - 2\pi)}$
• Apr 27th 2014, 12:13 AM
Applestrudle
Re: nth roots of unit proofs question
Quote:

Originally Posted by Plato
Here is a suggestion on notation that can simplify this answer. Use the exponentiation notation.
$e^z=\exp(z)=|z|(\cos(\theta)+i~\sin(\theta))$ where $\theta=\text{Arg}(z))$
This notation makes conjugate notation easy: $\overline{\exp(z)}=\exp(\overline{z} )$

For the first part of this question let $\theta=\dfrac{2\pi}{n}~\&~\rho=\exp(i\theta)$
Now it is easy to list the roots: $\rho^k:~k=0,1,\cdots, k-1$, and $\overline{~\rho~}=\exp(-i\theta)$.

Note that $\prod\limits_{k = 0}^{n - 1} {{\rho ^k}} = 1$.

For the second part the n nth roots of $i$: define $\rho=\exp\left(\dfrac{\pi i}{2}\right)~\&~\xi=\exp\left(\dfrac{2\pi i}{n}\right)$
Then those roots can be listed as: $\rho\cdot\xi^k,~k=0,\cdots,k-1$.

for iv) i got the product P equals

P = w1.w2.w3.w4.w5.w6....wn (the 1,2,3...n are subs)

since -w* is also a solution, you can say wk = -wk* so they repeat themselves (is this logic correct?) and

P = (w1)^2 (w2)^2 (w3)^2....(w n/2)^2

$P = {e}^{i( \pi\(frac{n}{2}) + 4\pi 8\pi +12\pi ....}$

in the exponent there is the sum from k =1 to k = (n/2) of 4k.pi

but i don't know how to simplify the sum, do i just leave it as a sum, is this even correct?