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Math Help - differentiation

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    Talking differentiation

    Show that f(z) = |z|^2 is differentiable only at z = 0. Hence it is no where analytical.?
    Last edited by mr fantastic; October 19th 2009 at 11:25 PM. Reason: I assume what I added in latex is what was meant.
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    Quote Originally Posted by SNESAM View Post
    Show that f(z) = |z|^2 is differentiable only at z = 0. Hence it is no where analytical.?
    |z|^2=z\cdot \bar{z}=(x+iy)\cdot (x-iy)=x^2+y^2

    Now as f(z)=u(x,y)+iv(x,y)

    This gives u(x,y)=x^2+y^2; v(x,y)=0

    Using the cauchy Riemann equaitons

    u_x=2x only equals v_y=0 at the point x=0

    A similar argument shows the same for y so the function is diff at (0,0).
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