Results 1 to 7 of 7

Math Help - complex nos multiplicity

  1. #1
    Junior Member
    Joined
    Oct 2006
    Posts
    33

    complex nos multiplicity

    can someone help me with this question pls think i undersdtand trhe concepts but cant quite get the answer thanks

    Edgar

    Determine the multiplicity of the roots of the following functions:
    (a) (z^4 − 1) sin(pi.z)
    (b) sin^3(1/z), z does not equal 0
    (c) 1 + exp(2z)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by edgar davids View Post
    can someone help me with this question pls think i undersdtand trhe concepts but cant quite get the answer thanks

    Edgar

    Determine the multiplicity of the roots of the following functions:
    (a) (z^4 − 1) sin(pi.z)
    <br />
(z^4-1) \sin(\pi\ z)=0<br />

    has roots when z^4-1=0 and/or \sin(\pi\ z)=0

    The roots of z^4-1=0 are 1,\ -1,\ i,\ -i all of multiplicity 1

    The roots of \sin(\pi z) are 0,\ \pm1,\ \pm2,\ ... again each of multiplicity 1.

    The roots 1,\ -1 occur in both terms so they are roots of multiplicity 2 for the original function, all the other roots are roots of multiplicity 1 of the original function.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2006
    Posts
    33

    thanks and a question

    firstly thanks captain black for that

    second: can yuo pls help me with parts b and c and if poss =give me somehints as to the method

    thanks in advance
    Edgar
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by edgar davids View Post
    (c) 1 + exp(2z)
    You want,
    1+e^{2z}=0
    Thus,
    e^{2z}=-1
    By Euler's formula,
    e^{\pi i}=0
    Thus,
    e^{\pi i+2\pi k}=0
    Thus,
    z=2\pi k+\pi i, k\in \mathbb{Z}
    No two are equal thus, each has multiplicity one 1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ThePerfectHacker View Post
    You want,
    1+e^{2z}=0
    Thus,
    e^{2z}=-1
    By Euler's formula,
    e^{\pi i}=0
    Typo should be -1

    Thus,
    e^{\pi i+2\pi k}=0
    Thus,
    z=2\pi k+\pi i, k\in \mathbb{Z}
    No two are equal thus, each has multiplicity one 1
    RonL
    Last edited by CaptainBlack; November 22nd 2006 at 11:19 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by edgar davids View Post
    can someone help me with this question pls think i undersdtand trhe concepts but cant quite get the answer thanks

    Edgar

    Determine the multiplicity of the roots of the following functions:
    (a) (z^4 − 1) sin(pi.z)
    (b) sin^3(1/z), z does not equal 0
    (c) 1 + exp(2z)
    A general if laborious method to determine the multiplicity of a root is as follows:

    Let z_0 be a root of a function f(z) analytic in some neighbourhood of z_0. Expand f(z) as a Taylor series about z_0 then if k is the smallest exponent of z-z_0 for which the coefficient is non-zero then k is the multiplicity of the root.

    This of course may be recast as the multiplicity of the root is the order of the lowest order derivative of the function at z_0 which is non-zero.

    Applying this method to (b) shows that each of the roots z=1/n \pi,\ n=\pm 1,\ \pm 2,\ ... has multiplicity 3 as the first and second derivatives of \sin^3(1/z) are zero at these points, but the third derivative is non-zero at these points.

    (This method may also be applied to (c), which will confirm ImPerfectHacker conclusion)

    RonL
    Last edited by CaptainBlack; November 22nd 2006 at 11:25 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    The second problem is z\not = 0
    We have,
    \sin^3 (1/z)=0
    Using the field property of zero,
    If and only if,
    \sin (1/z)=0

    We know that,
    \sin (1/z)=\frac{e^{i/z}-e^{-i/z}}{2i}
    We require that,
    e^{i/z}-e^{-i/z}=0
    Thus,
    e^{i/z}=e^{-i/z}
    Since the complex exponential is no-where zero we can divde through by it obtaining a biconditional statement,
    e^{2i/z}=1
    By Euler's Formula,
    e^{2\pi i}=1
    Thus,
    2i/z=2\pi i+2\pi k
    Thus,
    1/z=-\pi+\pi i k
    Thus,
    z=\frac{1}{\pi i k-\pi }
    It seems all solutions are of multiplicity of order 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Determine multiplicity
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: November 22nd 2011, 12:30 PM
  2. eigenvalues --- multiplicity
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 14th 2011, 04:31 PM
  3. Eigenvalue with multiplicity of 2?
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 15th 2011, 06:08 PM
  4. geometric multiplicity
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: November 22nd 2008, 08:47 PM
  5. Multiplicity of roots
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: March 26th 2007, 11:15 AM

Search Tags


/mathhelpforum @mathhelpforum