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Math Help - Complex Analysis Confusion!

  1. #1
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    Exclamation Complex Analysis Confusion!

    Hi,

    I'm doing revision for an exam in complex analysis at the mo. I've been going through examples and I've found a couple that I've never seen before and have no idea how to do-any help would be much appreciated!

    a)
    Find all values of the following:
    \log\imath and log (i+1)

    (Can't figure it out in Latex, I'm only learning!)


    b) Let D= C\(-oo,0], and suppose log 1=0
    Find (1-i)^4i and [e(-1-ROOT(3)i)/2]^(3PIi)
    (Again if someone could show me this in latex it'd be appreciated, I kept getting 1-i^4 when I tried)

    c)Given D= C\(-oo,0] and {\imath}^i=exp(PI/2), its principle value, show that
    (-1+(ROOT(3))i)^(3/2)= -2ROOT2

    Please help, I'm so confused. I'm thinking it has to do with the section we did on simply connected domains and isolated singularities of an analytic domain
    but I can't find any examples to follow and I'm lost!

    Thanks
    PS I'm trying with the latex but its slow so sorry about that!
    Last edited by musicmental85; December 4th 2007 at 06:08 AM. Reason: Error
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  2. #2
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    Quote Originally Posted by musicmental85 View Post
    a)
    Find all values of the following:
    \log\imath and log (i+1)
    \log i = \log |i| + i \arg (i) and \log (i+1) = \log |1+i|+i\arg (1+i)
    Can you finish?
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  3. #3
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    Complex Analysis

    A. For all values of \log(\imath), what we want is, from the definition of the logarithm, all z such that e^z=\imath.
    This is, of course, z=\imath(\tfrac{\pi}{2}+2n\pi), where n is any integer.
    Similarly, note that |1+\imath|=\sqrt{2} and \arg (1+\imath) \equiv \dfrac{\pi}{4} \pmod{2\pi} , and use these to find all z such that e^z=1+\imath=|1+\imath|e^{\imath\arg (1+\imath)} .


    B. Our branch cut (and log(1)=0) gives us -\pi < \arg z < \pi for any z not on the cut. Thus we use the fact that z = |z|e^{\imath\arg z} for any complex z to get
    z^{a+b\imath} = \left ( |z|e^{\imath\arg z} \right )^{a+b\imath}
      = |z|^{a+b\imath}e^{(\imath\arg z)(a+b\imath)}
      = |z|^a e^{\imath b \log |z|} e^{-b \arg z} e^{\imath a \arg z}

    So you can use this (along with basic exponentiation rules like (ab)^c=a^c b^c and r^{a+b} = r^a r^b) for (1 - \imath)^{a\imath} and for \left [ e \left ( -1-\dfrac{\imath\sqrt{3}}{2} \right ) \right ]^{3\pi\imath}


    C. Are you sure that it's \imath^\imath = e^{\tfrac{\pi}{2}} and not \imath^\imath = e^{-\tfrac{\pi}{2}}? I dont think the former is possible:
    |\imath| = 1, so \imath = e^{\imath \arg \imath} and \imath^\imath = e^{-\arg \imath}.
    Thus \imath^\imath = e^{\tfrac{\pi}{2}} gives \arg \imath = -\dfrac{\pi}{2}, but \arg \imath \equiv \dfrac{\pi}{2} \pmod{2\pi}.


    I hope this helps point you in the right direction.

    --Kevin C.
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  4. #4
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    Ok, just working on the first time (I'm slow at these)
    Do I need the perfect hackers comment?

    log i= log|i|+iarg(i)
    =log 1 +i(PI/2 +2nPI)
    =0 + i(PI/2 +2nPI)

    Can you just use the fact that z=|z|e^(iargz) and state answer as z=e^(i)(PI/2+2nPI) for example in the first case?

    For the second does the final answer become z=[ROOT(2)][e^(i)(7PI/4 + 2nPI)] ?

    In part b)
    Does the answer come out something like (1-i)^4i= e^(-4logROOT(2)-4i(7PI/4 +2nPI)) ?

    and the second comes out as e^(3PI(-1/ROOT3 +2nPI +ilog e^2)?

    In part c) you are right but it is supposed to be (-i)^i=exp(PI/2)
    Thanks so much for the help!
    Last edited by musicmental85; December 4th 2007 at 09:39 AM.
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