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Math Help - Studying differentiability of modulus(z)^2

  1. #1
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    Studying differentiability of modulus(z)^2

    Hello mathematicians,

    I am asked to study the differentiability of lzl2 (complex analysis) and don't know how to do it :_(

    I've read that it is only differentiable at 0, but how to reach that conclusion I ignore.

    The two ways that I know for checking if a function is differentiable are

    (1) By checking if the limit definition of derivative exists

    Studying differentiability of modulus(z)^2-img103.jpg
    But I don't know how to work with the squared modulus of numbers adding (numerator of the limit), and I don't think that's the way I'm supposed to do it...

    (2) Checking where C-R equations hold. However, modulus(z)^2 doesn't have imaginary part, so I don't have u and v.

    How can I study the differentiability of this function then?? I don't know to do this for f(z)=Re(z) either.

    Thank you!
    Last edited by Phyba; April 21st 2014 at 12:22 PM.
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  2. #2
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    Re: Studying differentiability of modulus(z)^2

    A function f:\Bbb{C} \to \Bbb{R} is differentiable if and only if there exists a linear function J:\Bbb{C} \to \Bbb{R} such that \lim_{\Delta z \to 0} \dfrac{f(z+\Delta z) - f(z) - J(\Delta z)}{|\Delta z|} = 0.

    Edit: for the function f:\Bbb{C} \to \Bbb{R} defined by f(z) = |z|^2, define J_z(\Delta z) = J_z(\Delta x + i\Delta y) = 2x\Delta x + 2y\Delta y where z = x+iy. Then: \begin{align*}\lim_{\Delta z \to 0} \dfrac{|z+\Delta z|^2 - |z|^2 - J_z(\Delta z)}{|\Delta z|} & = \lim_{\Delta z \to 0} \dfrac{(x^2+2x\Delta x + (\Delta x)^2 + y^2+2y\Delta y + (\Delta y)^2) - (x^2+y^2) - (2x\Delta x + 2y\Delta y)}{|\Delta z|} \\ & = \lim_{\Delta z \to 0} \dfrac{|\Delta z|^2}{|\Delta z|} \\ & = \lim_{\Delta z \to 0} |\Delta z| = 0\end{align*}
    Last edited by SlipEternal; April 21st 2014 at 01:15 PM.
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  3. #3
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    Re: Studying differentiability of modulus(z)^2

    Quote Originally Posted by Phyba View Post
    Hello mathematicians,

    I am asked to study the differentiability of lzl2 (complex analysis) and don't know how to do it :_(

    <snip>

    (2) Checking where C-R equations hold. However, modulus(z)^2 doesn't have imaginary part, so I don't have u and v.

    How can I study the differentiability of this function then?? I don't know to do this for f(z)=Re(z) either.

    Thank you!
    $w=u+\imath v=|z| = |x+\imath y| = \sqrt{x^2(t)+y^2(t)}$

    $u=\sqrt{x^2(t)+y^2(t)}$

    $v=0$

    $u_x=\dfrac{x}{\sqrt{x^2+y^2}}$

    $u_y=\dfrac{y}{\sqrt{x^2+y^2}}$

    $v_x = v_y = 0$

    clearly the CR equations are not satisfied anywhere but at $z=0$
    Thanks from Phyba and HallsofIvy
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  4. #4
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    Re: Studying differentiability of modulus(z)^2

    For f(z) = \text{Re}(z), you can do the same thing. Breaking it down component-wise, you have f(x+iy) = x, so \dfrac{\partial f}{\partial x} = 1 and \dfrac{\partial f}{\partial y} = 0. Then:

    \lim_{\Delta z \to 0} \dfrac{f(z+\Delta z) - f(z) - \begin{bmatrix}1 & 0\end{bmatrix}\begin{bmatrix}\Delta x \\ \Delta y\end{bmatrix} }{|\Delta z|} = \lim_{\Delta z \to 0} \dfrac{x+\Delta x - x - \Delta x}{|\Delta z|} = 0
    Thanks from Phyba
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  5. #5
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    Re: Studying differentiability of modulus(z)^2

    Thank you very much!
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