Results 1 to 4 of 4

Thread: nth roots of unit proofs question

  1. #1
    Junior Member
    Joined
    Mar 2014
    From
    United Kingdom
    Posts
    44

    nth roots of unit proofs question




    For 4i)

    i get to

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

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

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

    ii)

    for even n

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

    but for odd n

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

    for the sum, i got zero

    for the product:

    for even n i got

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

    for odd n

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

    please help with the last one!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,800
    Thanks
    2829
    Awards
    1

    Re: nth roots of unit proofs question

    Quote Originally Posted by Applestrudle View Post
    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: $\rho\cdot\xi^k,~k=0,\cdots,k-1$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2014
    From
    United Kingdom
    Posts
    44

    Re: nth roots of unit proofs question

    Quote Originally Posted by Plato View Post
    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 $\displaystyle {e}^{\frac{i2\pi k}{n}}$ and i got $\displaystyle \frac{1-{e}^{\frac{2 \pi k}{n}}}{1 - {e}^{\frac{i2 \pi}{n}}}$ 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 $\displaystyle {z}_{k} = {e}^{\frac{2 \pi k}{n}}$

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

    for the nth roots of i I got

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

    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 $\displaystyle {e}^{i(\frac{\pi}{2n} - 2\pi)}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2014
    From
    United Kingdom
    Posts
    44

    Re: nth roots of unit proofs question

    Quote Originally Posted by Plato View Post
    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

    $\displaystyle 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?




    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Feb 15th 2013, 01:01 PM
  2. Replies: 0
    Last Post: Jul 5th 2012, 02:18 PM
  3. Roots of unit complex numbers
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Sep 4th 2010, 01:03 AM
  4. Replies: 3
    Last Post: Sep 17th 2009, 07:35 PM
  5. exp & log unit question
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Dec 13th 2008, 09:28 AM

Search Tags


/mathhelpforum @mathhelpforum