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Math Help - Cauchy-Riemann equations

  1. #1
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    Question Cauchy-Riemann equations

    I understand how the equations work but i can't seem to be able to determine whether "1 / (1 + z)" satisfies the cauchy-riemann equations and hence determine whether f ' (z) exists.

    Any help would be greatly apprichiated
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  2. #2
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    Quote Originally Posted by jezzyjez View Post
    I understand how the equations work but i can't seem to be able to determine whether "1 / (1 + z)" satisfies the cauchy-riemann equations and hence determine whether f ' (z) exists.

    Any help would be greatly apprichiated
    If you substitute z = x + iy and multiply numerator and denominator by the complex conjugate of the denominator and simplify, you will get f = \frac{x + 1}{x^2 + 2x + 1 + y^2} - i \frac{y}{x^2 + 2x + 1 + y^2}.
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    Yeh cool i get that now thanks, am i missing a trick with the last part of the question

    "determine whether f ' (x) exists"

    or does it literally exist when the equations satisfy the C-R equations??
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    Quote Originally Posted by jezzyjez View Post
    I understand how the equations work but i can't seem to be able to determine whether "1 / (1 + z)" satisfies the cauchy-riemann equations and hence determine whether f ' (z) exists.

    Any help would be greatly apprichiated

    You can evaluate the partial derivatives with the function as Mr. Fantastic showed you, or else realize (most books have it) that the CR-equations

    are equivalent to \frac{\partial f}{\partial\,\overline{z}}=0 , where we have that \frac{\partial}{\partial\,\overline{z}}=\frac{1}{2  }\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right) , so putting z=x+iy, we get:

    f(z)=f(x,y)=\frac{1}{1+x+iy}\Longrightarrow \frac{\partial f}{\partial x}=-\,\frac{1}{(1+x+iy)^2}\,,\,\,\frac{\partial f}{\partial y}=-\,\frac{i}{(1+x+iy)^2}.

    Check the above and conclude your function is analytic in \mathbb{C}\setminus \{1\}

    Tonio
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    Quote Originally Posted by tonio View Post
    You can evaluate the partial derivatives with the function as Mr. Fantastic showed you, or else realize (most books have it) that the CR-equations

    are equivalent to \frac{\partial f}{\partial\,\overline{z}}=0 , where we have that \frac{\partial}{\partial\,\overline{z}}=\frac{1}{2  }\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right) , so putting z=x+iy, we get:

    f(z)=f(x,y)=\frac{1}{1+x+iy}\Longrightarrow \frac{\partial f}{\partial x}=-\,\frac{1}{(1+x+iy)^2}\,,\,\,\frac{\partial f}{\partial y}=-\,\frac{i}{(1+x+iy)^2}.

    Check the above and conclude your function is analytic in \mathbb{C}\setminus \{{\color{red}-1}\}

    Tonio
    A thread of related interest: http://www.mathhelpforum.com/math-he...erivation.html

    (And a correction of a minor typo in red).
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    Quote Originally Posted by jezzyjez View Post
    or does it literally exist when the equations satisfy the C-R equations??
    The function is analytic at a point if and only if the C-R equations are satisfied in some neighbourhood of the point.
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