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Math Help - Analyticity of z

  1. #1
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    Analyticity of z

    I was reading through my text, and it stated that z^n (for some integer n) was not analytic at n = 0. Why is this the case? Usually I'd substitute z = x + iy and then use the Cauchy-Riemann equations to check, but (x + iy)^n makes it difficult to separate the real and imaginary parts.

    Thanks.
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  2. #2
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    Quote Originally Posted by Glitch View Post
    I was reading through my text, and it stated that z^n (for some integer n) was not analytic at n = 0. Why is this the case? Usually I'd substitute z = x + iy and then use the Cauchy-Riemann equations to check, but (x + iy)^n makes it difficult to separate the real and imaginary parts.

    Thanks.
    Do you mean when z=0 not n?
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  3. #3
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    Sorry, I should probably provide some context.

    The question I got this from is the following:

    Let C be the circle of radius R centre the origin, taken anticlockwise. Find \int_C{z^n \dz} for integer values of n.
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    Quote Originally Posted by Glitch View Post
    Sorry, I should probably provide some context.

    The question I got this from is the following:

    Let C be the circle of radius R centre the origin, taken anticlockwise. Find \int_C{z^n \dz} for integer values of n.
    Well you can parametrize the circle at the origin by

    z=Re^{i\theta},\theta \in [0,2\pi] \implies dz=Rie^{i\theta}d\theta

    When

    n=0 \implies z^n=1 so you get the integral

    \oint dz=Ri\int_{0}^{2\pi}e^{i\theta}d\theta=0

    In fact you can use this for every integer n but be careful with the case n=-1
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  5. #5
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    I just don't understand why it is not analytic at n = 0 (at least, according to this text).
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  6. #6
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    Quote Originally Posted by Glitch View Post
    I just don't understand why it is not analytic at n = 0 (at least, according to this text).
    It should be when n=0 we get

    z^0=\begin{cases}1, z \ne 0 \\ \text{undefined},z=0 \end{cases}

    Technically 0^0 is undefined, but if we check the limit along any path of

    \lim_{z \to 0}z^0=1 So by the Riemann removable singularity theorem the function can be analytically extended to all of the complex plane

    Removable singularity - Wikipedia, the free encyclopedia

    Note that holomorphic function is a function that is analytic on all of the complex plane.
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