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Math Help - limit from sides in complex function..

  1. #1
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    limit from sides in complex function..

    i was told here that there is no +infinity and -infinity in complex functions
    only infinity

    but i have a solved question from a test which is solved exactly like it you said was wrong
    and got full credit

    maybe i am missing something here

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  2. #2
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  3. #3
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    Quote Originally Posted by transgalactic View Post
    i was told here that there is no +infinity and -infinity in complex functions
    only infinity

    but i have a solved question from a test which is solved exactly like it you said was wrong
    and got full credit

    maybe i am missing something here

    Really, it would have sufficed to note that since z^2 has a finite limit, and \sin \left( \frac{z}{z+1} \right) has an essential singularity at -1 then f has an essential singularity at -1, then what you showed (basically) is that the limit doesn't exist. I assume they gave you the credit since your argument vaguely appeals to what I said.

    As an example of what you're doing wrong take z^{-1}. Clearly we can think of it as a real or complex function. Analyze the behaviour in both cases and remember why we can talk about poles.
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  4. #4
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    wrong place
    Last edited by transgalactic; July 12th 2010 at 12:13 PM.
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  5. #5
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    Quote Originally Posted by transgalactic View Post
    so you still say that 0+ and 0- +-infinity method here is wwrong?
    Taking the limits [Math]0^+[/tex] and 0^- is not wrong, you're just taking two different curves to approach the limit (which is always valid), what is wrong is your distinction of infinities. Tell me how would you define -\infty as a limit of a complex function?
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  6. #6
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    so you still say that 0+ and 0- +-infinity method here is wwrong?

    and i did developed the function into a series

    and i saw that there is no z in a negative power

    so it cannot be significant singular point
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  7. #7
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    Quote Originally Posted by transgalactic View Post
    so you still say that 0+ and 0- +-infinity method here is wwrong?

    and i did developed the function into a series

    and i saw that there is no z in a negative power

    so it cannot be significant singular point
    What about the powers of z+1 there, which blow up near -1 whereas the z's remain bounded.

    By the way, please answer the question in my previous post, it will clarify things for you.
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  8. #8
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    if my development is correct how whould you continue to develop it
    so we will get a series with only one z variable

    ??
    i cant make a guess here its all has to be series with only one z in a certain power in order to makea conclution
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  9. #9
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    Frankly I wouldn't know how to turn that expression into a Laurent series, but it's not necessary, that is why one develops propositions like: if z_0\in U\subseteq \mathbb{C} with U open, f,g:U\setminus \{z_0\} \rightarrow \mathbb{C} holomorphic and f has a removable singularity at z_0 and g has an essential singularity at z_0 then fg has an essential singularity at z_0
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  10. #10
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    what is f
    what is g
    in our case
    ?
    could you explain your method more thurely
    i dont know how to apply it here
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  11. #11
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    f(z)=z^2, \ g(z)=\sin \left( \frac{z}{z+1} \right), \ U=\mathbb{C} \setminus \{ -1 \}
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  12. #12
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    ok and how to prove that at -1 the sin expression is significant
    you canceled my +-inf method
    and by developing it into a series is not working either
    so how to prove that at -1 the sin expression is significant?
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  13. #13
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    Quote Originally Posted by transgalactic View Post
    ok and how to prove that at -1 the sin expression is significant
    you canceled my +-inf method
    and by developing it into a series is not working either
    so how to prove that at -1 the sin expression is significant?
    \sin (z) has an essential singularity at infinity, which means \sin \left( \frac{1}{z} \right) has an essential singularity at 0 (this is easy to check since the principal part of the Laurent expansion has infinitely many terms) and this gives that \sin ( f(z)) has an essential sing. wherever f has a pole. Take f(z)= \frac{z}{z+1} and the conclusion follows.
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  14. #14
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    Quote Originally Posted by Jose27 View Post
    \sin (z) has an essential singularity at infinity, which means \sin \left( \frac{1}{z} \right) has an essential singularity at 0 (this is easy to check since the principal part of the Laurent expansion has infinitely many terms) and this gives that \sin ( f(z)) has an essential sing. wherever f has a pole. Take f(z)= \frac{z}{z+1} and the conclusion follows.
    i understand why
    \sin \left( \frac{1}{z} \right) has an essential singularity at 0

    but how we can take any function we want and say sin(f) has significant point
    at the pole of f
    ??
    like here
    \sin ( f(z))
    f(z)= \frac{z}{z+1}
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  15. #15
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    Basically, because near a pole, a function behaves like \frac{1}{(z-z_0)^m} for some integer m. Try to work out the details by yourself (maybe using the multiplication formula for \sin will simplify the argument). For this specific case a simple change of variables (and maybe a little geometric argument) gives the conclusion.
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