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 a curious limit bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 04/16/2011, 08:19 PM (This post was last modified: 04/16/2011, 08:31 PM by bo198214.) But if we are going back to the original question, and I set now $a+bi=1-e^{vi}$, whether $e^{(a+bi)\log(z)}$ converges, for $z\to 0$, one would derive: $\left|e^{a\log( r)-b\phi + i(a\phi+b\log( r))}\right|=r^a e^{-b\phi}$ This implies several things (assuming a>0): b=0, i.e. $v=\pi (2 k+1)$ then you have the limit 0 if b>0, and you wind around anti-clockwise (incresing $\phi$) approaching 0, then you have limit 0. Note that you must put the log-cut accordingly (spiralling) that it allows increasing $\phi$ if b<0, then as above but clockwise (decreasing $\phi$) if b>0 and you wind around slow enough but clockwise, you may also have a limit. I.e. $r\to 0$ faster than $e^{-b\phi}\to\infty$ opposite of the previous So you see, it really depends on how you approach 0. « Next Oldest | Next Newest »

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