Results 1 to 7 of 7

Thread: Residue

  1. #1
    Newbie
    Joined
    Nov 2006
    Posts
    15

    Residue

    I have a doubt with the singularities and poles of a function, for example in the next complex function:

    $\displaystyle f(x)=\sqrt{x^2+a}$

    with $\displaystyle a$ a constant. The function is zero when $\displaystyle x=\pm{\sqrt{a}}$

    For this value of x, is it a singularity or a pole? If it is a pole, of which order? Is it possible to calculate the residue for $\displaystyle x=\pm{\sqrt{a}}$?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by germana2006 View Post
    I have a doubt with the singularities and poles of a function, for example in the next complex function:

    $\displaystyle f(x)=\sqrt{x^2+a}$

    with $\displaystyle a$ a constant. The function is zero when $\displaystyle x=\pm{\sqrt{a}}$ Mr F says: No. You want $\displaystyle {\color{red}x=\pm i {\sqrt{a}}}$.

    For this value of x, is it a singularity or a pole? If it is a pole, of which order? Is it possible to calculate the residue for $\displaystyle x=\pm{\sqrt{a}}$?
    It's a branch point. A branch point is a special type of singularity.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    This may help

    $\displaystyle f(x)=\sqrt{x^2+a}=e^{ln(\sqrt{(x^2+a)})}=e^{1/2ln(x^2+a)}$
    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 mr fantastic View Post
    It's a branch point. A branch point is a special type of singularity.
    I disagree. To be a singularity we require that the function is analytic in a puntured neighborhood of that point. This is not true for $\displaystyle \sqrt{z^2+a}$ along its branch.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by ThePerfectHacker View Post
    I disagree. To be a singularity we require that the function is analytic in a puntured neighborhood of that point. This is not true for $\displaystyle \sqrt{z^2+a}$ along its branch.
    I quote from Churchill et al Complex Variables and Applications:

    "... a point $\displaystyle z_0$ is called a singular point of a function f if f faails to be analytic at $\displaystyle z_0$ but is analytic at some point in every neighbourhood of $\displaystyle z_0$. A singular point is said to be isolated if, in addition, there is some neighbourhood of $\displaystyle z_0$ throughout which f is analytic except at the point itself."

    A branch point is a singular point - what it's not is an isolated singular point.

    -----------------------------------------------------------------------------------------

    There are two categories of singularity:

    1. Isolated singularity.

    The point $\displaystyle z = z_0$ is an isolated singularity of $\displaystyle f(z)$ if $\displaystyle f(z)$ is not analytic at $\displaystyle z = z_0$ but is analytic in some deleted neighbourhood of $\displaystyle z = z_0$.

    There are three types:

    i. Removable. $\displaystyle \lim_{z \rightarrow z_0} f(z)$ exists and is finite. Alternatively, the Laurent series expansion of $\displaystyle f(z)$ about $\displaystyle z = z_0$ has no principle part.

    Eg. $\displaystyle f(z) = \frac{\sin z}{z}$ has a removable singularity at z = 0.

    ii. Pole. $\displaystyle (z - z_0)^m f(z)$ is analytic at $\displaystyle z = z_0$ for some positive integer m. The smallest such m is called the order of the pole. Alternatively, the principle part of the Laurent series expansion of $\displaystyle f(z)$ about $\displaystyle z = z_0$ terminates after m terms.

    Eg. $\displaystyle f(z) = \frac{1}{z-1}$ has a simple pole at z = 1.

    iii. Essential. It's neither removable nor a pole. Alternatively, the principle part of the Laurent series expansion of $\displaystyle f(z)$ about $\displaystyle z = z_0$ does not terminate. Alternatively, $\displaystyle \lim_{z \rightarrow z_0} f(z)$ does not exist (NB. Picard's Theorem).

    Eg. $\displaystyle f(z) = e^{1/z}$ has an essential singularity at z = 0.

    2. Non-isolated singularity.

    If the singular point $\displaystyle z = z_0$ is not isolated, then it is a non-isolated singularity.

    Eg. The branch point z = 0 of $\displaystyle f(z) = \ln z$ is a non-isolated singularity. In fact, any branch point of a multiple-valued function is a non-isolated singularity.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Okay. I just never seen such terms before. But if you use that terminology then it is not just $\displaystyle \pm i\sqrt{a}$ it then should be $\displaystyle (-\infty,0]$.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by ThePerfectHacker View Post
    Okay. I just never seen such terms before. But if you use that terminology then it is not just $\displaystyle \pm i\sqrt{a}$ it then should be $\displaystyle (-\infty,0]$.
    Yes. But it should be noted that branch cuts are defined by convention ..... You say $\displaystyle (-\infty,0]$, I could just as well define the branch cut to be $\displaystyle [0, \infty)$ in which case the points in $\displaystyle (-\infty,0]$ are not singular .....

    The branch point is (the only point that is) common to all branch cuts .... This makes it fundamentally different from all other points on a branch cut.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. residue
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 23rd 2011, 12:17 PM
  2. residue
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Apr 27th 2011, 04:43 AM
  3. residue
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Dec 3rd 2009, 03:08 PM
  4. Residue
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Mar 9th 2009, 01:30 AM
  5. residue
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Feb 17th 2009, 09:22 PM

Search Tags


/mathhelpforum @mathhelpforum