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Thread: complex nos multiplicity

  1. #1
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    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)
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  2. #2
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    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)
    $\displaystyle
    (z^4-1) \sin(\pi\ z)=0
    $

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

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

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

    The roots $\displaystyle 1,\ -1$ 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
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  3. #3
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    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
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  4. #4
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    Quote Originally Posted by edgar davids View Post
    (c) 1 + exp(2z)
    You want,
    $\displaystyle 1+e^{2z}=0$
    Thus,
    $\displaystyle e^{2z}=-1$
    By Euler's formula,
    $\displaystyle e^{\pi i}=0$
    Thus,
    $\displaystyle e^{\pi i+2\pi k}=0$
    Thus,
    $\displaystyle z=2\pi k+\pi i, k\in \mathbb{Z}$
    No two are equal thus, each has multiplicity one 1
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    You want,
    $\displaystyle 1+e^{2z}=0$
    Thus,
    $\displaystyle e^{2z}=-1$
    By Euler's formula,
    $\displaystyle e^{\pi i}=0$
    Typo should be -1

    Thus,
    $\displaystyle e^{\pi i+2\pi k}=0$
    Thus,
    $\displaystyle z=2\pi k+\pi i, k\in \mathbb{Z}$
    No two are equal thus, each has multiplicity one 1
    RonL
    Last edited by CaptainBlack; Nov 22nd 2006 at 10:19 AM.
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  6. #6
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    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 $\displaystyle z_0$ be a root of a function $\displaystyle f(z)$ analytic in some neighbourhood of $\displaystyle z_0$. Expand $\displaystyle f(z)$ as a Taylor series about $\displaystyle z_0$ then if $\displaystyle k$ is the smallest exponent of $\displaystyle z-z_0$ for which the coefficient is non-zero then $\displaystyle k$ 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 $\displaystyle z_0$ which is non-zero.

    Applying this method to (b) shows that each of the roots $\displaystyle z=1/n \pi,\ n=\pm 1,\ \pm 2,\ ...$ has multiplicity 3 as the first and second derivatives of $\displaystyle \sin^3(1/z)$ 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; Nov 22nd 2006 at 10:25 AM.
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  7. #7
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    The second problem is $\displaystyle z\not = 0$
    We have,
    $\displaystyle \sin^3 (1/z)=0$
    Using the field property of zero,
    If and only if,
    $\displaystyle \sin (1/z)=0$

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