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Math Help - Prove f is analytic in C, and find f'(i)

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    Prove f is analytic in C, and find f'(i)

    Quote Originally Posted by practice exam
    Let [MATh]z=x+iy[/tex]. Prove that the function f(z)=x^3-3xy^2-3+i(3x^2y-y^3) is analytic in \mathbb{C}, and find f'(i).
    My textbook doesn't cover this, that I'm aware, and I must have missed it in lecture. It's a practice exam question, which means it's meant to be done fairly quickly. But I don't see any easy or quick ways to do this.

    Any help would be much appreciated!
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hatsoff View Post
    My textbook doesn't cover this, that I'm aware, and I must have missed it in lecture. It's a practice exam question, which means it's meant to be done fairly quickly. But I don't see any easy or quick ways to do this.

    Any help would be much appreciated!
    Since your ' u' and ' v' are continuously differentiable you can apply the Cauchy-Riemann equations.
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    u and v are continuously differentiable and satisfy the CR equations implies that f is holomorphic, sure. But how do I go from "holomorphic" to "analytic" ?

    Presumably I need an actual solution so that I can differentiate and plug in i.
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    Is it a fact that U_x=V_y~\&~Uy_=-V_x
    Please answer with your workings.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hatsoff View Post
    u and v are continuously differentiable and satisfy the CR equations implies that f is holomorphic, sure. But how do I go from "holomorphic" to "analytic" ?

    Presumably I need an actual solution so that I can differentiate and plug in i.
    Well, technically holomorphic and analytic are synonymous...but you may not know that yet (assuming what level you're at). But, I just realized this is much easier since a quick check shows that f(z)=z^3-3.
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    Quote Originally Posted by Plato View Post
    Is it a fact that U_x=V_y~\&~Uy_=-V_x
    Please answer with your workings.
    Well, let's see...

    u_x=3x^2-3y_2=v_y and u_y=-6xy=-v_x. So I guess I could do

    f'(z)=u_x+iv_x=(3x^2-3y^2)+i(6xy)=3(x^2+2ixy-y^2)=3(x+iy)^2=3z^2. It follows that f has a Taylor series about any a\in\mathbb{C}, and that f'(i)=-3.

    Is that the easiest way to do this though? I suspect that I knew to jump from f'(z)=(3x^2-3y^2)+i(6xy) to f'(z)=3z^2 because I was expecting it based on Drexel28's observation. But on an exam that might not be quite so obvious.

    Thanks for the help guys!
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    Quote Originally Posted by hatsoff View Post
    u and v are continuously differentiable and satisfy the CR equations implies that f is holomorphic, sure. But how do I go from "holomorphic" to "analytic" ?

    Presumably I need an actual solution so that I can differentiate and plug in i.
    Are you clear on the definition of "holomophic"? A function is "analytic" (or not) at individual points or on a given set. A function that is "holomorphic" is analytic on the entire complex plane.

    Also, it might help to remember why we have the "Cauchy-Riemann" equations.

    Suppose f(z)= u(x,y)+ iv(x,y) is to be differentiated at (x_0, y_0). Of course,
    \displaytype\frac{df}{dz}= \lim_{h\to 0} \frac{f(z+h)- f(z)}{h}
    but because the complex plane is two dimensional, for that limit to exist, we must get tthe same limit taking any path toward (x_0, y_0). In particular, if we approach along a line parallel to the real axis, we have
    \frac{df}{dz}= \lim_{h\to 0}\frac{u(x_0+h, y_0)- u(x_0,y_0)}{h}+ i\lim_{h\to 0}\frac{v(x_0+h,y_0)- v(x_0,y_0)}{h}
    \frac{df}{dz}= \frac{\partial u}{\partial x}+ i \frac{\partial v}{\partial x}

    While if we approach (x_0, y_0) along a path parallel to the imaginary axis
    \frac{df}{dz}= \lim_{h\to 0}\frac{u(x_0, y_0+ h)- u(x_0,y_0)}{hi}+ i\lim_{h\to 0}\frac{v(x_0,y_0+h)- v(x_0,y_0)}{hi}
    \frac{df}{dz}= -i\frac{\partial u}{\partial y}+ \frac{\partial v}{\partial y}

    In order that f be differentiable at (x_0, y_0), those must be the same so, setting real part equal to real part and imaginary part equal to imaginary part, we have
    \frac{\partial u}{\partial x}= \frac{\partial v}{\partial y} and
    \frac{\partial u}{\partial y}= -\frac{\partial v}{\partial x}

    Those also tell us how to calculate \frac{df}{dz}- use either
    \frac{df}{dz}= \frac{\partial u}{\partial x}+ i \frac{\partial v}{\partial x} or

    \frac{df}{dz}= -i\frac{\partial u}{\partial y}+ \frac{\partial v}{\partial y}
    Last edited by HallsofIvy; March 4th 2011 at 06:37 AM.
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