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Analyticity problem

  1. Sep 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine if the function is analytic and sketch where it's analytic.

    a) [tex] f(x) = \frac {e^z}{z^2 + 4} [/tex]

    b) [tex] f(z) = \frac {conj(z)}{|z|^2} [/tex]

    c) [tex] \sum_{n=0}^\infty \frac {e^z}{3^n} (2z-4)^n [/tex]


    2. Relevant equations



    3. The attempt at a solution

    a) [tex] e^z [/tex] is analytic everywhere so [tex] f(x) = \frac {e^z}{z^2 + 4} [/tex] is analytic everywhere except at

    [tex] z^2 = (x + yi)^2 = -4 [/tex]

    I tried separating the function into [tex] f(x,y) = u(x,y) + i*v(x,y) [/tex] but get a very complex polynomial when I try to get rid of the imaginary part in the denominator for example:

    [tex] \frac {e^x (cos(y) + i*sin(y)}{(x+yi)^2 + 4} * \frac{(x-yi)^2 + 4}{(x-yi)^2 + 4} [/tex]

    does not work for me

    b)

    [tex] f(z) = \frac{conj(z)}{ |z|^2} [/tex]

    [tex] f(z) = \frac{1}{z} [/tex]

    [tex] \frac{1}{z} = \frac{1}{ x+yi}[/tex]

    [tex] \frac{x - yi} {x^2 + y ^2}[/tex]

    [tex] = \frac{x}{x^2 + y ^2} - \frac {yi}{x^2 + y ^2}[/tex]

    [tex] since \frac{du}{dx} u(x,y) = \frac{dv}{dy} u(x,y)[/tex]

    and [tex] \frac{dv}{dx} u(x,y) = - \frac{dv}{dx} u(x,y) [/tex]

    the function is analytic everywhere except at the origin.

    d) After using the ratio test, the result I get is

    [tex] |\frac{1}{3} (2z - 4) | [/tex]

    [tex] \frac {1}{3} |2z - 4| [/tex]

    so the function is analytic on the disk [tex] 0 < |2z - 4| < 6 [/tex] for all values n
     
    Last edited: Sep 4, 2009
  2. jcsd
  3. Sep 4, 2009 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Re: Analyticity

    Right, so [itex]f(z)[/itex] is analytic everywhere except at [itex]z=\pm2i[/itex]

    I'm not sure why you'd want to do it this way, since you've already found the answer using a much easier method, but if you want to be a masochist about it, this method should work fine as well.

    [tex]\left[(x+yi)^2 + 4\right]\left[(x-yi)^2 + 4\right]=(x+iy)^2(x-iy)^2+4(x+iy)^2+4(x-iy)^2+16=(x^2+y^2)^2+8(x^2-y^2)+16[/tex]

    and so on...

    You mean, "since [itex]\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}[/itex] and [itex]\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}[/itex] except at the origin, where [itex]u[/itex], [itex]v[/itex] and their partial derivatives do not exist, the function is analytic everywhere except at the origin", right?


    Where is the '6' coming from?
     
    Last edited: Sep 4, 2009
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