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Math Help - show that f'(c) exists

  1. #1
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    show that f'(c) exists

    let f(z) = u + iv, z = x + iy
    let u and v be defined as follows:
    u(x,y) = (x^3 - y^3) / (x^2 + y^2) if (x,y)≠(0,0), u(0,0)=0
    v(x,y) = (x^3 + y^3) / (x^2 + y^2) if (x,y)≠(0,0), v(0,0)=0

    solve lim [f(z) - f(c)] / (z-c) as z approaches c. that is, show f'(c) exists

    our instructor told us a hint: solve it component-wise

    the thing is, I do not know how to apply that in such a complex function please help me.. teach me some sorcery

    edit: btw, this limit is defined to be the derivative
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  2. #2
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    Re: show that f'(c) exists

    Quote Originally Posted by StillInUse View Post
    let f(z) = u + iv, z = x + iy
    let u and v be defined as follows:
    u(x,y) = (x^3 - y^3) / (x^2 + y^2) if (x,y)≠(0,0), u(0,0)=0
    v(x,y) = (x^3 + y^3) / (x^2 + y^2) if (x,y)≠(0,0), v(0,0)=0

    solve lim [f(z) - f(c)] / (z-c) as z approaches c. that is, show f'(c) exists

    our instructor told us a hint: solve it component-wise

    the thing is, I do not know how to apply that in such a complex function please help me.. teach me some sorcery

    edit: btw, this limit is defined to be the derivative
    To find the real and imaginary components, rationalise the denominator:

    \frac{f(z)-f(c)}{z-c} = \frac{u+iv}{x+iy} = \frac{(u+iv)(x-iy)}{x^2+y^2}.

    Now substitute the formulas for u and v into the fraction ... .
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  3. #3
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    Re: show that f'(c) exists

    uhm could you kindly explain what happened to f(c)?
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    Re: show that f'(c) exists

    Quote Originally Posted by Opalg View Post
    To find the real and imaginary components, rationalise the denominator:

    \frac{f(z)-f(c)}{z-c} = \frac{u+iv}{x+iy} = \frac{(u+iv)(x-iy)}{x^2+y^2}.

    Now substitute the formulas for u and v into the fraction ... .
    Hello Sir Opalg i take it that what you gave me is a hint..

    after multiplying a conjugate of the denominator and then some further algebraic manipulations I ended up with these bunch of letters

    \frac{(x-a)[u(x,y)-u(a,b)]+(y-b)[v(x,y)-v(a,b)]+i[(x-a)[v(x,y)-v(a,b)]-(y-b)[u(x,y)+u(a,b)]}{(x-a)^2-(y-b)^2}.

    is this, by any chance, going towards the right direction? i still do not know how to take the limit of this function though
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  5. #5
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    Re: show that f'(c) exists

    Quote Originally Posted by StillInUse View Post
    uhm could you kindly explain what happened to f(c)?
    Good point! I also left the c out of the denominator. I must have been subconsciously assuming that c=0. In fact, I think that c probably should be 0, because actually f(z) is not differentiable at any other point.
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    Re: show that f'(c) exists

    Quote Originally Posted by Opalg View Post
    Good point! I also left the c out of the denominator. I must have been subconsciously assuming that c=0. In fact, I think that c probably should be 0, because actually f(z) is not differentiable at any other point.

    f'(c) exists if the Cauchy-Riemann conditions are satisfied at c. Look them up. You have to compute some partial derivatives on u and v.
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    Re: show that f'(c) exists

    Quote Originally Posted by Hartlw View Post
    f'(c) exists if the Cauchy-Riemann conditions are satisfied at c. Look them up. You have to compute some partial derivatives on u and v.
    exactly. now, the dilemma here is how can i prove that f'(c) exist? i mean, does rationalizing the denominator then separating the real and imaginary parts of

    [(f(z)-f(c)][(x-a)+i(y-b)]

    enough of a reason to say that the limit as z approaches to c do exist?
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  8. #8
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    Re: show that f'(c) exists

    Quote Originally Posted by Opalg View Post
    Good point! I also left the c out of the denominator. I must have been subconsciously assuming that c=0. In fact, I think that c probably should be 0, because actually f(z) is not differentiable at any other point.
    By C-R f'(z) doesn't exist anywhere except possibly at z=0.

    If you let x^2 + y^2 = r^2 and note |x|<r and |y|<r, you can show limit of u and v and their derivatives approach 0 as r approaches 0. If you then define u and v and their drivatives = 0 at z=0 they are continuous at z=0 and C-R is satisfied at z=0 so that f'(0)=0.
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