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):
So you see, it really depends on how you approach 0.
\( e^{(a+bi)\log(z)} \) 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.
