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Need some help with basic complex analysis (no proofs)

  1. Feb 17, 2005 #1
    need some urgent help with basic complex analysis (no proofs)

    This forum is probably more appropriate. please forgive me for double posting.


    Can someone give me examples of the following? (no proofs needed) (C is the complex set)

    1. a non-zero complex number z such that Arg(z^2) is NOT equal to 2 Arg z
    2. a region in C which is not a domain
    3. a non-empty subset of C which has no accumulation points
    4. a continuous function f: C -> C which is not differentiable anywhere
    5. an entire function which is not a polynomial.

    thanks a lot in advance!!
     
  2. jcsd
  3. Feb 17, 2005 #2

    cepheid

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    2. Apply the definition. A domain is an open, connected set. Surely you can think of an example of a region that fails one or both criteria? Examples are even given in the text I'm using, anyway.

    3. Admittedly, I have no idea what this means. (Hey, I'm taking the course as we speak! Gimmie a break. :rofl: )

    4. What happens if f(z) does anything involving the complex conjugate of z?

    5. C'mon! Trigonometric, exponential, you ought to know these, at least. Again, crack open the textbook. I think those would be good ones to look at the proofs of.
     
  4. Feb 17, 2005 #3

    matt grime

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    4. Do you mean holomorphic/analytic (ie complex differentiable) or not? If so it's easy, if you just mean differentiable then that's harder.

    1. Arg has values in what range?

    3. Look up the definition of accumulation point.
     
  5. Feb 17, 2005 #4
    Thanks to both of you!! I looked up the definition of accumulation point and it says:
    A point z0 is said to be an accumulation point of a set S if each deleted neighborhood of z0 contains at least one point of S.
    i'm not sure if i understand that. any ideas??
     
  6. Feb 17, 2005 #5

    Galileo

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    An open neighbourhood of a point [itex]z \in \mathbb{C}[/itex] is an open subset of [itex]\mathbb{C}[/itex] containing the point [itex]z[/itex].
    Often only the 'open circles' around z are considered neighbourhoods though and are always open, so they just call them neighbourhoods, but this doesn't matter, since in any open subset containing z you can draw a circles about z which is contained in that set.
    So with the latter common definition, the set:

    [tex]N(z_0,\epsilon)=\{z: |z-z_0|<\epsilon\}[/tex]
    is called a neighbourhood of [itex]z_0[/itex].

    A deleted neighbourhood of [itex]z_0[/itex] is [itex]N(z_0,\epsilon)\setminus \{z_0\}[/itex].

    So z0 is an accumulation point of S is every neighbourhood of z0 contains a point in S unequal to z0.
     
  7. Feb 17, 2005 #6

    cepheid

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    So..." not differentiable" would mean not diffentiable at any single point, right?
    Which is much more stringent than not analytic anywhere, right? Because you could have a function differentiable at a bunch of isolated points (but not in any neighbourhood of any of those points), that qualifies as not analytic anywhere? Am I getting it?

    Ah, I see...so does this hint you gave him suggest that any number that crosses the "branch cut" (negative real axis) when squared would work?
     
  8. Feb 17, 2005 #7

    matt grime

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    One can create (forgetting the complex bit) continuous functions from R to R that are almost nowhere differentiable, I seem to recall. Possibly even nowhere differentiable, but I can't think of any easy examples that match the rest of the standard of this question, hence I presume it means analytic.

    So, f(z) = |z| is nowhere analytic, but is certainly continuous and considered as a function from R^2 to R it is real differentiable everywhere.
     
  9. Feb 17, 2005 #8
    i think you were right. not differentiable is more stringent than no analytic anywhere... i think...
    i understand the first one now. i just need something that once it's squared, it'll be outside the range of -pi to pi.
    the second one is much easier than i thought ^_^
    the third one i chose {1}
    i have the answers to the last 2 now as well.

    thanks so much to all of you for helping!! you guys rule!! d=^_^=b
     
  10. Feb 17, 2005 #9
    Actually, AFAIK it doesn't have to be a bunch of isolated points. For example, a function that is differentiable on a line (or finite union of lines) is not analytic anywhere, because the lines do not form a domain.

    Basically, yeah. You can analyze the problem by switching from Arg(z) to arg(z), doing the problem and then figuring out with the primary value would be of arg(z^2).
     
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