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Math Help - complex analysis

  1. #1
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    complex analysis

    have asked this q to many people and noone is able to help would be much appreciated if anyone could solve this.

    thanx loadz edgar

    (a) Verify that Im(z) and z do not satisfy the Cauchy-Riemann equations
    at any point (so neither function is differentiable anywhere).

    (b) Consider the function f(x+iy) = p|xy|, where x, y ∈ R. Show that
    f satisfies the Cauchy-Riemann equations at the origin, yet f is not
    holomorphic at 0.
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  2. #2
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    Quote Originally Posted by edgar davids View Post
    (a) Verify that Im(z) and z do not satisfy the Cauchy-Riemann equations
    at any point (so neither function is differentiable anywhere).
    The complex function,
    f(z)=z
    Can also be expressed in terms of real functions,
    f(x+iy)=x+iy
    We note the functions x,y are differenciable (Good).
    Next we check Cauchy-Riemann Equations:
    The partial derivative of x along x is 1
    The partail derivative of y along y is 1
    So we have,
    1=1 (Good)
    Next we check the second equation,
    The partial derivative of x along y zero,
    The negative partial derivative of y along x is zero,
    So we have,
    0=0 (Good)
    Thus, the function,
    f(z)=z is homolophric on the entire complex plane. I do not see how you can say it is not.
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  3. #3
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    Quote Originally Posted by edgar davids View Post
    have asked this q to many people and noone is able to help would be much appreciated if anyone could solve this.

    thanx loadz edgar

    (a) Verify that Im(z) and z do not satisfy the Cauchy-Riemann equations
    at any point (so neither function is differentiable anywhere).
    Should these be Im(z) and Re(z)?

    RonL
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  4. #4
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    Quote Originally Posted by edgar davids View Post
    have asked this q to many people and noone is able to help would be much appreciated if anyone could solve this.

    thanx loadz edgar

    (a) Verify that Im(z) and z do not satisfy the Cauchy-Riemann equations
    at any point (so neither function is differentiable anywhere).
    The Cauchy Riemann equations are:

    for a complex valued function of a complex variable:

    f(z) = u(x,y) + i v(x,y)

    where z=x + i y:

    du/dx = dv/dy,
    du/dy = -dv/dx,

    where d here denotes the curly d of partial differentiation.

    Now let f(z) = Im(z) = y, then we have u(x,y)=y, v(x,y)=0,

    so:

    du/dy=1, dv/dx=0,

    so du/dy != -dv/dx,

    and the Cauchy Reimann equations are not satisfied.

    A similar argument can be applied to Re(z).

    RonL
    Last edited by CaptainBlack; October 17th 2006 at 09:14 PM.
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  5. #5
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    Quote Originally Posted by edgar davids View Post
    (b) Consider the function f(x+iy) = p|xy|, where x, y ∈ R. Show that
    f satisfies the Cauchy-Riemann equations at the origin, yet f is not
    holomorphic at 0.
    (again d denotes the partial differential)

    In this case u(x,y)=p|xy| and v(x,y)=0

    now

    du/dx(0,0) = lim(h->0) [p |h.0| - 0]/h = 0 =dv/dy

    and similarly:

    du/dy(0,0) = 0 = -dv/dx,

    so f satisfies the Cauchy-Riemann equations.

    Now f(z) holomorphic at 0 means that f(z) is complex differentiable
    in some neighbourhood of 0. A necessary condition to be complex
    differentiable at z0, is that f satisfy the Cauchy-Riemann equations
    at z0. But our f(z) does not satisfy the Cauchy-Riemann equations
    at any point not on the real or imaginary axes (to be shown below),
    so f(z) cannot be holomorphic at 0 (because it is not complex differentiable
    at some point in every neighbourhood of 0).

    ------------------------------------------------------------

    To show that f(z) does not satisfy the Cauchy-Riemann equations
    at a point not on the axes, we can restrict attention to the first
    quadrant (the others will follow a similar argument)

    Let x>0, y>0, then:

    f(x,y)=p x.y

    so u(x,y)=p x.y, v(x,y)=0

    Then du/dy= px, and dv/dx=0, that is:

    du/dy != -dv/dx

    So f(z) does not satisfy the Cauchy-Riemann equations at any point z
    in C with Re(z), Im(z) >0

    RonL
    Last edited by CaptainBlack; October 17th 2006 at 09:16 PM.
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