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Need some help with basic complex variables (no proofs)

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

    can someone give me examples of the following? (no proofs needed)

    1. a non-zero complex number z such that Arg(z^2) "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
    Arg z is defined to be the angle between -pi and pi which is equivalent to the actual argument of z. So if [tex]z=e^{i\pi}\implies z^2=e^{2i\pi}[/tex] then Arg z = arg z = pi, but arg z² = 2pi so Arg z² = 0.

    If I recall correctly a domain has to be connected, so something like [tex]\left\{z\in\mathbb{C}:|z-2|<1\right\}\cup\left\{z\in\mathbb{C}:|z+2|<1\right\}[/tex] (two discs of radius 1 centred at 2 and -2 on the real axis) would qualify.

    I think an accumulation point only makes sense with respect to a sequence... Off the top of my head I can't think of a set on which no sequence could have an accumulation point, since there is always the possibility of a constant sequence. I could be wrong though...

    Technically f(z)=Re(z) qualifies, although it is only onto [tex]\mathbb{R}[/tex], not [tex]\mathbb{C}[/tex]. But since [tex]\mathbb{R}\subset\mathbb{C}[/tex] it can be thought of as a function f:C->C which is just not surjective.

    I'm pretty sure that [tex]e^z[/tex] is entire.
  4. Feb 17, 2005 #3
    wow thank you!
    i still have no idea how to do the accumulation one. 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?
  5. Feb 17, 2005 #4

    matt grime

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    A z is an accumulation point of S if every open disc centred on z contains some element of S. Every point of S is an accumulation point of S. S is closed if it contains all its accumulation points
  6. Feb 17, 2005 #5


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    nocheesie has a different definition- the 'deleted disc' part is important or the question is false.

    Equivalent definition (worth proving equivalence if you want a better handle on accumulation points): z is an accumulation point of S if and only if there is a sequence in S minus z that converges to z.

    I can't think of a good hint without giving it away, think very simple sets.
  7. Feb 17, 2005 #6

    matt grime

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    So it corrects to "contains some element of S distinct from z".
  8. Feb 17, 2005 #7
    so would the set of say, {1} work? i don't know if i'm understanding this correctly... if you have {1} then there would be no accumulation points right? since there's only 1 itself?
  9. Feb 17, 2005 #8

    matt grime

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    Any disc centred on 1, would not contain any of the other points of S, since there is none. And if z is any other point not equal to 1, then |z-1| is strictly positive, say it equals r. Then the disc of radius r/2 about z does not contain 1 (the only point in S) so it is not an accumulation point of S. Thus S has no accumulation points. Is that what you were thinking?
  10. Feb 17, 2005 #9
    Yes!! that's what i was thinking =) since there's only 1 then it can't have any points in the set other than 1, thus no accumulation points! i think i'm finally getting this... thanks everyone very very much!!
  11. Feb 17, 2005 #10


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