zero's of exp^[1/2](x) ?
#7
(12/08/2010, 10:16 AM)sheldonison Wrote:
(12/08/2010, 02:44 AM)mike3 Wrote: Here's a graph of \( \exp^{1/2}(z) \), acquired via the Cauchy integral. Scale is from -5 to +5 on both real and imag axis:
very nice. Thanks Mike! I can see the cut points at L and conj(L). Your particular cutpoint choice don't show the singularity at -0.36237+iPi. The singularity at L is somewhat less intense, in that exp^[1/2](L) is defined and equals L.

-0.36237 + iPi ??

where does this come from ? and what is its closed form ?

regards

tommy1729
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Messages In This Thread
zero's of exp^[1/2](x) ? - by tommy1729 - 12/04/2010, 12:16 AM
RE: zero's of exp^[1/2](x) ? - by JJacquelin - 12/07/2010, 08:53 AM
RE: zero's of exp^[1/2](x) ? - by sheldonison - 12/07/2010, 01:03 PM
RE: zero's of exp^[1/2](x) ? - by mike3 - 12/08/2010, 02:44 AM
RE: zero's of exp^[1/2](x) ? - by sheldonison - 12/08/2010, 10:16 AM
RE: zero's of exp^[1/2](x) ? - by tommy1729 - 12/08/2010, 01:03 PM
RE: zero's of exp^[1/2](x) ? - by sheldonison - 12/08/2010, 01:09 PM
RE: zero's of exp^[1/2](x) ? - by tommy1729 - 12/08/2010, 01:39 PM
RE: zero's of exp^[1/2](x) ? - by sheldonison - 12/08/2010, 02:24 PM
RE: zero's of exp^[1/2](x) ? - by tommy1729 - 12/29/2015, 01:23 PM
RE: zero's of exp^[1/2](x) ? - by tommy1729 - 12/08/2010, 12:32 AM
RE: zero's of exp^[1/2](x) ? - by tommy1729 - 11/14/2012, 05:11 PM



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