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Complex Numbers

  1. Aug 29, 2011 #1
    I am reading a chapter on Complex Functions, Laplace Transforms & Cauchy Riemann (as part of Control theory)

    And I don't understand how they arrive at a particular part.
    [ I tried to type it out in tex, but it takes way too much time so uploaded a screenshot to flickr]

    [PLAIN]http://www.flickr.com/photos/66943862@N06/6093176535/ [Broken]

    Here is a http://www.flickr.com/photos/66943862@N06/6093176535/" [Broken]

    I understand how you get to Eqn1 & Eqn2.
    But how does it add up to Equation3?

    Can someone explain?

    Also, I don't understand why it's not analytic at s = -1?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 29, 2011 #2

    micromass

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    Hi phiby! :smile:

    \begin{eqnarray*}
    \frac{\partial G_x}{\partial \sigma}+j\frac{\partial G_y}{\partial \sigma}
    & = & \frac{\omega^2-(\sigma+1)^2+2j\omega(\sigma+1)}{[(\sigma+1)^2+\omega^2]^2}\\
    & = & \frac{\omega^2+2j\omega(\sigma+1)+j^2(\sigma+1)^2}{[\omega^2-j^2(\sigma+1)^2]^2}\\
    & = & \frac{[\omega+j(\sigma+1)]^2}{[(\omega-j(\sigma+1))(\omega+j(\sigma+1))]^2}\\
    & = & \frac{1}{(\omega-j(\sigma+1))^2}\\
    & = & \frac{1}{(-j)^2(\sigma+1+j\omega)^2}\\
    & = & -\frac{1}{\sigma+j\omega+1}
    \end{eqnarray*}

    The function G is not analytic in -1 since it doesn't exist there. Indeed, G(-1) is undefined and is a pole. (so it's not even a removable singularity)
     
  4. Aug 29, 2011 #3

    Awesome. Thanks a lot. I got what you did (the simplification of the equation), but didn't get how you knew you had to do that to simplify the original stuff.

    I studied a lot of engineering math 20 years ago & I am getting back to it after 20 years (almost did none of this in the 20 years). So it's taking me a little time to get this.

    In the original page (my flickr link), I first didn't get how it was separated into Gx & Gy, so I went back & did a review of partial fractions & then it became simple.

    So my question is - what part of math should I review to do what you did above?
     
  5. Aug 29, 2011 #4

    micromass

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    Well, the things I used where the equations

    [tex](a+b)^2=a^2+2ab+b^2[/tex]

    and

    [tex](a+b)(a-b)=a^2-b^2[/tex]

    If you know these very well, then you can find the above solution.
     
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