Subject: (1.0+%i)^0.5 does not evaluate numerically
From: Richard Fateman
Date: Sat, 04 Jul 2009 07:00:51 -0700
I think this issue is that the original programmers hesitated to make a
single choice of a multi-valued function.
For example, (-1)^(1/3) is simplified to -1. Arguably, wrong.
(-1)^(1/3.0)
rectform(%)
gives 0.86602540378444*%i+0.5
That is another possibility.
If you plunge ahead and replace (-1)^(1/3.0) with 0.866*%i+0.5 you will
undoubtedly get some answers more
directly. Probably you will be closer to what Mathematica does under
similar circumstances.
You will presumably also get inconsistencies, as presumably does
Mathematica.
Note that Mathematica does NOT do (-1)^(1/3) --> -1.
RJF