Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Log(z) : singularity, series and pole

  1. Apr 30, 2012 #1
    Firstly does log(z) have any singularity other than z=0?
    Secondly, z=0 is a pole of what order for log(z)? What is the Laurent series expansion for log(z)
    at z=0?
     
  2. jcsd
  3. Apr 30, 2012 #2
    You got [itex]\sqrt{z}[/itex] straight? How many singular points that one has? Two right. Zero and infinity. That is, there exists a neighborhood of zero and infinity such that one complete circuit around them carries a branch of [itex]\sqrt{z}[/itex] into another branch. There's one at infinity cus' any very large circle around the origin is a circle around the point at infinity and that circle carries the value of [itex]\sqrt{z}[/itex] into another branch. For example a circle starting at 4 starts at [itex]\sqrt{4}=2[/itex], then after one full circuit along an analytically continuous path over the function, goes to [itex]-2[/itex]. After a [itex]4\pi[/itex] circuit, it goes back to 2. And the order of a branch-point is the number of times for this minus one so [itex]\sqrt{z}[/itex] has an algebraic branch point of order one at both the origin and infinity. If it were [itex]\sqrt[3]{z}[/itex], same dif except the order is 2. Those singular point are not poles because they are not issolated. That is, there is no neighborhood around them where the function is analytic and single-valued. Obviously since [itex]\sqrt{z}[/itex] will be different after one full circuit around the neighborhood so it's not single-valued.

    But the Laurent series expansion is a power series around poles so therefore [itex]\sqrt[n]{z}[/itex] does not have a Laurent expansion about it's singular points.

    Now what about log(z)? That one has singular points called "logarithmic branch points" of infinite order since circuits around the origin or infinity around the function never return to the starting value. These too are not poles since they also are not issolated and so therefore do not have Laurent series expansions about their singular points.
     
    Last edited: Apr 30, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Log(z) : singularity, series and pole
  1. Series of cos(exp(-z)) (Replies: 3)

Loading...