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Math Help - Error in Complex Analysis Proof

  1. #1
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    Error in Complex Analysis Proof

    My Complex Analysis book states the following interesting theorem:

    Let N(c,r) be a open disk in \mathbb{C}. Let f be a complex function whose domain contains the open disk. Let f(x+iy) = u(x,y)+iv(x,y) \mbox{ for }x+iy \in N(c,r). If the parital derivaties u_x,u_y,v_x,v_y all exist and are continous on N(c,r) and at c \in N(c,r) the Cauchy-Riemann equations are satisfied then f is differenciable at c.

    This is a partial converse for the necessay Cauchy-Riemann condition.

    Here is how the proof begins.

    Let c=a+bi and let h,k be so that |h+ik|<r (that is they lie in the open disk).
    Then,
    u(a+b,b+k)-u(a,b) = [u(a+h,b+k)-u(a,b+k)] + [u(a,b+k)-u(a,b)]
    By the Mean Value Theorem,
    =hu_x(a+x_0h,b+k) + kv_y(a,b+y_0k) where x_0,y_0\in (0,1).

    And it goes on....

    My problem is that, they did not use the MVT properly. Do you agree? (The problem is that this is one of those books which likes to skip steps).
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  2. #2
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    Quote Originally Posted by ThePerfectHacker View Post
    My Complex Analysis book states the following interesting theorem:

    Let N(c,r) be a open disk in \mathbb{C}. Let f be a complex function whose domain contains the open disk. Let f(x+iy) = u(x,y)+iv(x,y) \mbox{ for }x+iy \in N(c,r). If the parital derivaties u_x,u_y,v_x,v_y all exist and are continous on N(c,r) and at c \in N(c,r) the Cauchy-Riemann equations are satisfied then f is differenciable at c.

    This is a partial converse for the necessay Cauchy-Riemann condition.

    Here is how the proof begins.

    Let c=a+bi and let h,k be so that |h+ik|<r (that is they lie in the open disk).
    Then,
    u(a+b,b+k)-u(a,b) = [u(a+h,b+k)-u(a,b+k)] + [u(a,b+k)-u(a,b)]
    By the Mean Value Theorem,
    =hu_x(a+x_0h,b+k) + kv_y(a,b+y_0k) where x_0,y_0\in (0,1).

    And it goes on....

    My problem is that, they did not use the MVT properly. Do you agree? (The problem is that this is one of those books which likes to skip steps).
    What exactly is your objection, it looks OK to me.

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlank View Post
    What exactly is your objection, it looks OK to me.
    Tell me then then how you got it?

    I defined a function of f(x)=u(a+hx,b+k): [0,1]\to \mathbb{R}. We can see that this is a well-defined function on this interval and differenciable on (0,1). By mean value theorem there exists x_0\in (0,1) so that, \frac{f(1)-f(0)}{1-0} = f'(x_0). Thus, f(1)-f(0)=f'(x_0). Write out it out fully, u(a+h,b+k)-u(a,b+k)=u_x(a+x_0h,b+k).
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    Tell me then then how you got it?

    I defined a function of f(x)=u(a+hx,b+k): [0,1]\to \mathbb{R}. We can see that this is a well-defined function on this interval and differenciable on (0,1). By mean value theorem there exists x_0\in (0,1) so that, \frac{f(1)-f(0)}{1-0} = f'(x_0). Thus, f(1)-f(0)=f'(x_0). Write out it out fully, u(a+h,b+k)-u(a,b+k)=u_x(a+x_0h,b+k).
    <br />
f(a+h) - f(a) = h f'(u_0)<br />

    u_0 in (0,h), or:

    <br />
f(a+h) - f(a) = h f'(x_0 h)<br />

    x_0 in (0,1)

    RonL
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