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Math Help - hard complex question

  1. #1
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    hard complex question

    captain black can you or someone else give me a hand to solve this problem. spent hours trying to do it but finding it very very difficult so if someone could help me would be really grateful and would really help my understanding

    thanx in advance


    (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
    Grand Panjandrum
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    Quote Originally Posted by edgar davids View Post
    captain black can you or someone else give me a hand to solve this problem. spent hours trying to do it but finding it very very difficult so if someone could help me would be really grateful and would really help my understanding

    thanx in advance


    (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.
    You have already asked these questions and had them answered in this
    thread.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by edgar davids View Post
    captain black can you or someone else give me a hand to solve this problem. spent hours trying to do it but finding it very very difficult so if someone could help me would be really grateful and would really help my understanding

    thanx in advance


    (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.
    Edgar has emailed me with the question he had meant to post when he
    posted the above by mistake:



    Suppose that f is holomorphic in an open set D. Use the Cauchy-Riemann
    equations to prove that in any of the following cases:
    (a) Re(f) is constant;
    (b) Im(f) is constant;
    (c) |f| is constant;
    one can conclude that f is constant.
    Hint for part (c): calculate the partial derivatives of |f|2 = u2+v2 = const.


    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    Edgar has emailed me with the question he had meant to post when he
    posted the above by mistake:



    Suppose that f is holomorphic in an open set D. Use the Cauchy-Riemann
    equations to prove that in any of the following cases:
    (a) Re(f) is constant;
    (b) Im(f) is constant;
    (a) Let f(x,y) = u(x,y) + i v(x,y),

    then the Cauchy-Riemann equations tell us that:

    <br />
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}<br />
,

    and:

    <br />
\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}<br />

    But we have Re(f) is a constant, so:

    <br />
\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x}=0<br />
,

    and

    <br />
\frac{\partial u}{\partial y} = 0<br />

    So along any curve such that Re(z)=constant; f is a constant.

    But a holomorphic function is determined every where by its values on a
    curve, but g(z)=Re(f(0)) is holomorphic and equal to f(z) on Re(z)=0,
    so f(z)=g(z)=Re(f(0)) every where in \mathbb{C}.

    Part (b) is similar.

    RonL
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  5. #5
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    thanks

    thanx loads for that

    how do you do part c? sorry to keep hassling you and really do appreciate all the help

    all the best
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post

    Suppose that f is holomorphic in an open set D. Use the Cauchy-Riemann
    equations to prove that in any of the following cases:
    :
    :
    (c) |f| is constant;
    one can conclude that f is constant.
    Hint for part (c): calculate the partial derivatives of |f|2 = u2+v2 = const.
    Now this last one is fiddly, and so prone to error, so you will need to
    check this carefully.

     |f| constant implies |f|^2 is a constant, so:

    <br />
\frac{\partial}{\partial x}|f|^2 = 2u \frac{\partial u}{\partial x} + 2 v\frac{\partial v}{\partial x}=0<br />

    and:

    <br />
\frac{\partial}{\partial y}|f|^2 = 2u \frac{\partial u}{\partial y} + 2 v\frac{\partial v}{\partial y}=0<br />

    Hence we have:

    <br />
u \frac{\partial u}{\partial x} = -v\frac{\partial v}{\partial x}<br />
and <br />
u \frac{\partial u}{\partial y} = -v\frac{\partial v}{\partial y}<br />
.

    But u and v also satisfy the Cauchy-Riemann equations:

    <br />
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}<br />
and <br />
\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}<br />
.

    So with a bit of jiggery pokery we have:

    <br />
u \frac{\partial u}{\partial x} = v\frac{\partial u}{\partial y}<br />
and <br />
u \frac{\partial u}{\partial y} = -v\frac{\partial u}{\partial x}<br />
.

    Now multiply the first of these by v and the second by u to get:

    <br />
vu \frac{\partial u}{\partial x} = v^2\frac{\partial u}{\partial y}<br />
and <br />
u^2 \frac{\partial u}{\partial y} = -vu\frac{\partial u}{\partial x}<br />
,

    so:

    <br />
u^2 \frac{\partial u}{\partial y} = -v^2\frac{\partial u}{\partial y}<br />
,

    Now chose a vertical line segment on which f(z) has no zeros (such a segment exists because the zeros of a non-zero holomorphic function are isolated), then on this segment we have {\partial u}/{\partial y} = 0 (as if this were non zero at some point we would have u^2+v^2=0 which would mean that f(z)=0 at that point - a contradiction), so u(z) is a constant on this segment.

    Now I don't have the patience to do this myself, but I would expect a similar argument to show that on this segment {\partial v}/{\partial y} = 0, which implies that v(z) is a constant on the segment, which with our previous result shows that f(z) is constant on this segment, but a holomorphic function is determined everywhere by its values on a curve (of any kind) and so determined by its values on the segment. But on this segment f(z) is equal to a constant function which is also holomorphic so f(z) is constant on \mathbb{C}.

    RonL
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