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Thread: Complex Analysis Confusion!

  1. #1
    Nov 2007

    Exclamation Complex Analysis Confusion!


    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!

    Find all values of the following:
    $\displaystyle \log\imath$ 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 $\displaystyle {\imath}^i$=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!

    PS I'm trying with the latex but its slow so sorry about that!
    Last edited by musicmental85; Dec 4th 2007 at 06:08 AM. Reason: Error
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  2. #2
    Global Moderator

    Nov 2005
    New York City
    Quote Originally Posted by musicmental85 View Post
    Find all values of the following:
    $\displaystyle \log\imath$ and log (i+1)
    $\displaystyle \log i = \log |i| + i \arg (i)$ and $\displaystyle \log (i+1) = \log |1+i|+i\arg (1+i)$
    Can you finish?
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  3. #3
    Senior Member
    Dec 2007
    Anchorage, AK

    Complex Analysis

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

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

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

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

    I hope this helps point you in the right direction.

    --Kevin C.
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  4. #4
    Nov 2007
    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; Dec 4th 2007 at 09:39 AM.
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