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Going over complex analysis past exam

  1. Oct 11, 2011 #1
    Hey,

    I've been going through a few past papers for an upcoming exam on complex analysis, I found this T/F question with a few parts i'm not confident on, I'll explain the whole lot of my work and show.


    [PLAIN]http://img404.imageshack.us/img404/2069/asdasdsu.jpg [Broken]



    a) |2+3i|=|2-3i| so false

    b) True since when the coefficients are real the roots come in complex conjugate pairs

    c) Using triangle inequality, 1/|z^2+1|=> 1/|z^2| +2 = 1/ (x^2+y^2+2)
    1/ (x^2+y^2+2)<=1/x^2+2<=1/2 im not sure if these inequality opperations are 100%

    d)False, cannot be every f since f must be analytic within the domain and curve region

    e) False, i think, by Cauchys integral formula C must be a simple closed curve enclosing Zo, so as C in this question is just a line, False

    f) I think this is true, but i'm not sure if you can use cauchys theorem backwards. So true as int f(z)dz=0 if f is analytic in C. So since int f(z)dz=0, f must be analytic inside and on C.
    Is the converse of the cauchy theorem true?

    h) I'm not sure how to approach this, but I was thinking you could just let an be somthing like 1/sqrt(3)^n for the h), then it would work, true

    g) i cant see a way to do somthing similar, so i think the answer is false, since if you make an "an" such that converges for 2-i, then it will also converge for 1+i,
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 12, 2011 #2
    for g,h i think

    ratio test gets a|z|<1 so a < 1/|z|,
    so for g,
    1/sqrt2<a<1/sqrt3 which doesnt make sense so false.

    h,

    1/sqrt3<a<1/sqrt2, so a series exists as required between that inequality, so true
     
  4. Oct 12, 2011 #3

    HallsofIvy

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    Yes, this is false but not because of the equality of the absolute values. There is NO inequality defined on the complex numbers so "2+ 3i< 2- 3i" simply has no meaning.

    No, this is false. You are thinking the wrong way around. Yes, IF the coefficients are real, the roots come in complex conugate parts but you are NOT given that the coefficients are real so that is irrelevant. Look at the equation [itex](x- 2i)(x+ 2i)(x- i)= 0[/itex] which is a cubic polynomial equation with roots 2i, -2i, and i.

    Did you consider z= .9i?

    Yes, false but again your reason is wrong because you are going the wrong way- if f is analytic then the integral around a closed curve is 0 but the converse is not necessarily true. Instead use a specific example: [itex]f(z)= f(re^{i\theta})= \theta[/itex].

    Yes, this is correct.

    Yes, it is called "Morera's theorem":http://en.wikipedia.org/wiki/Morera's_theorem

    No, this is false. A power series always has a "radius of convergence" within which it converges and outside of which it diverges. Here the series is centered on z= 0. The distance from 0 to 1+ i is [itex]\sqrt{2}[/itex] so the radius of convergence must be less than [itex]\sqrt{2}[/itex]. The distance from 0 to 2- i is [itex]\sqrt{5}> \sqrt{2}[/itex] so any such power series that diverges at 1+ i cannot converge for at 2- i.

    Just the reverse. This is true. We can have a power series that converges at at 1+ i but not at 2- i. Take any function that has a pole at a distance between [itex]\sqrt{2}[/itex] and [itex]\sqrt{5}[/itex], for example, the power series expansion of [itex]1/(z- 2)[/itex] converges for all |z|< 2, diverges for all |z|> 2.
     
    Last edited by a moderator: May 5, 2017
  5. Oct 12, 2011 #4
    Thanks alot HallsofIvy,

    I would of gotten dominated if I walked into a test like that without studying
     
  6. Oct 12, 2011 #5

    HallsofIvy

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    So study!
     
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