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Thread: Approaching complex problem.....

  1. #1
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    Approaching complex problem.....

    Lets say we have a function: $\displaystyle f(z) = x+i|y| $.

    We want to find all point that is it differentiable. Right of the bat we know that it is differentiable for $\displaystyle y >0 $. For $\displaystyle f(z) = x+iy $ is a "direct" function of $\displaystyle x+iy $. E.g. we do not have to appeal to CR equations.

    Also it is not differentiable for $\displaystyle y<0 $ since $\displaystyle f(z) = x-iy $ is not a "direct" function of $\displaystyle x+iy $.


    Now for $\displaystyle y = 0 $ can we use the same argument to show that $\displaystyle f(z) = x+i|y| $ is not differentiable there?

    Because then we have $\displaystyle f(z) = x $. And this is not a "direct" function of $\displaystyle x+iy $.
    Last edited by heathrowjohnny; Feb 8th 2009 at 01:41 PM.
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  2. #2
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    I mean analytic takes care of everything. Its stronger then differentiability. Thats what "function of z" is.
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  3. #3
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    Quote Originally Posted by heathrowjohnny View Post
    Lets say we have a function: $\displaystyle f(z) = x+i|y| $.

    We want to find all point that is it differentiable. Right of the bat we know that it is differentiable for $\displaystyle y >0 $. For $\displaystyle f(z) = x+iy $ is a "direct" function of $\displaystyle x+iy $. E.g. we do not have to appeal to CR equations.

    Also it is not differentiable for $\displaystyle y<0 $ since $\displaystyle f(z) = x-iy $ is not a "direct" function of $\displaystyle x+iy $.


    Now for $\displaystyle y = 0 $ can we use the same argument to show that $\displaystyle f(z) = x+i|y| $ is not differentiable there?

    Because then we have $\displaystyle f(z) = x $. And this is not a "direct" function of $\displaystyle x+iy $.
    For $\displaystyle y=0$ you can use the definition of differenciable, $\displaystyle \lim_{z\to z_0} \frac{f(z)-f(z_0)}{z-z_0}$. Now $\displaystyle z=x+iy$ and $\displaystyle z_0 = x$. Therefore, $\displaystyle \lim_{(x,y)\to (x_0,0)}\frac{x+i|y| - x}{iy} = \lim_{y\to 0}\frac{y}{|y|}$.
    And this limit does not exist.
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  4. #4
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    so you can't just say that $\displaystyle f(z) = x $ is not an "analytic function" of $\displaystyle x+iy $?
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