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Complex Differentiable Function watch

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    Well I need to find when a complex function is complex differentiable however this question is a little more complicated.



    I'm just stuck as surely there are infinitely many solutions? Or have I made a silly mistake...
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    (Original post by TomLeigh)
    Well I need to find when a complex function is complex differentiable however this question is a little more complicated.



    I'm just stuck as surely there are infinitely many solutions? Or have I made a silly mistake...
    u_y=-v_x
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    (Original post by ztibor)
    u_y=-v_x
    -.- I remember that from the lectures now, just read it wrong :P
    So this surely gives us y = 3x^2 - 2 so what are the actual points in the complex field in which the function is complex differentiable?
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    The complex numbers z = x+iy where y = 3x^2 - 2 are the only points that are complex differentiable.
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    Thanks, I wasn't sure if the answer was as simple as that (just explaining) or if there were more steps required. I'll pos rep you tomorrow
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    (Original post by Creole)
    The complex numbers z = x+iy where y = 3x^2 - 2 are the only points that are complex differentiable.
    The question then asks for which points in complex field is the function analytic. What does this mean exactly? Most examples use a simple sentence but I'm not sure what they mean eg:

    Since the imaginary axis contains no non-empty open disc, f is analytic nowhere

    for another question.
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    I believe it is nowhere analytic, although I found the notes quite difficult to understand (at this point I will reveal that I am doing the same module as you :p:).

    The key being that there is no open disc upon which the function is differentiable in the complex plane.
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    (Original post by TomLeigh)
    -.- I remember that from the lectures now, just read it wrong :P
    So this surely gives us y = 3x^2 - 2 so what are the actual points in the complex field in which the function is complex differentiable?

    The Cauchy Riemann equations are a nesacery not sufficient condition for differentiabilty at a point. You can't use them to prove a function is differentiable at at point, only that it is not differentaible at a point if the CR euqations don't hold.

    There is a partial converse which says that
    Let G \subset \mathbb{C} be open and  z \in G . Then if u and v have continuous first order partial derivitives in G and that they exist satisfy the CR equations at z, then f ii differentiable at z.

    a function is analytic in an open set G if it is differentiable at all points in G
 
 
 
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