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Math Help - Simple Complex Analysis Question

  1. #1
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    Simple Complex Analysis Question

    Hi I would just like for someone to check my proof/point out anything I am missing. Thanks!

    Prove that the function f(z)= \sqrt{|xy|} is not diff at origin.

    Proof:
    If it was differentiable at the origin, then f(z) should have the same derivative no matter how you approach the origin. We consider two paths approaching the origin from the x axis (positive side) and the x=y line from the 1st quadrant.

    Case 1 (x-axis):
    \lim_{(x,0)\to(0,0)}\frac{f(x,0)-f(0,0)}{x} = \lim_{(x,0)\to(0,0)}\frac{0}{x}

    Case 2 (x=y line):
    \lim_{(x,x)\to(0,0)}\frac{f(x,x)-f(0,0)}{(x,x)} = \lim_{(x,x)\to(0,0)}\frac{x-0}{x+ix}=  \lim_{(x,x)\to(0,0)}(1/(1+i))

    Since these two limits are different. The function is not differentiable at the origin.
    Last edited by arsenicbear; April 14th 2010 at 08:15 PM.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by arsenicbear View Post
    Hi I would just like for someone to check my proof/point out anything I am missing. Thanks!

    Prove that the function f(z)= \sqrt{|xy|} is not diff at origin.

    Proof:
    If it was differentiable at the origin, then f(z) should have the same derivative no matter how you approach the origin. We consider two paths approaching the origin from the x axis (positive side) and the x=y line from the 1st quadrant.

    Case 1 (x-axis):
    \lim_{(x,0)\to(0,0)}\frac{f(x,0)-f(0,0)}{x} = \lim_{(x,0)\to(0,0)}\frac{0}{x}

    Case 2 (x=y line):
    \lim_{(x,x)\to(0,0)}\frac{f(x,x)-f(0,0)}{(x,x)} = \lim_{(x,x)\to(0,0)}\frac{x-0}{x+ix}=  \lim_{(x,x)\to(0,0)}(\sqrt{2}/2-\sqrt{2}/2*i)

    Since these two limits are different. The function is not differentiable at the origin.
    Look at that last limit again, but yeah your method works.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by arsenicbear View Post
    Hi I would just like for someone to check my proof/point out anything I am missing. Thanks!

    Prove that the function f(z)= \sqrt{|xy|} is not diff at origin.

    Proof:
    If it was differentiable at the origin, then f(z) should have the same derivative no matter how you approach the origin. We consider two paths approaching the origin from the x axis (positive side) and the x=y line from the 1st quadrant.

    Case 1 (x-axis):
    \lim_{(x,0)\to(0,0)}\frac{f(x,0)-f(0,0)}{x} = \lim_{(x,0)\to(0,0)}\frac{0}{x}

    Case 2 (x=y line):
    \lim_{(x,x)\to(0,0)}\frac{f(x,x)-f(0,0)}{(x,x)} = \lim_{(x,x)\to(0,0)}\frac{x-0}{x+ix}=  \lim_{(x,x)\to(0,0)}(1/(1+i))

    Since these two limits are different. The function is not differentiable at the origin.
    To make it easier on yourself and assume it's diff. and use the chain rule to show it's composition with another function (which will simplify things) is diff. but it won't be.

    Spoiler:
    z^2
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