Status: Accepted
Owner: asmeurer
Labels: Type-Defect Priority-Medium Simplify
New issue 2327 by asmeurer: sqrt(I) should return sqrt(2)/2 + sqrt(2)/2*I
by default
http://code.google.com/p/sympy/issues/detail?id=2327
In [31]: sqrt(I)
Out[31]:
4
╲╱ -1
In [32]: sqrt(I).expand(complex=True)
Comment #6 on issue 2327 by pr...@goodok.ru: sqrt(I) should return
sqrt(2)/2 + sqrt(2)/2*I by default
http://code.google.com/p/sympy/issues/detail?id=2327
This operation is related with simplification and complex expansion.
The original expression save information about multiplicity of solut
Updates:
Status: undefin
Comment #7 on issue 2327 by asmeu...@gmail.com: sqrt(I) should return
sqrt(2)/2 + sqrt(2)/2*I by default
http://code.google.com/p/sympy/issues/detail?id=2327
Yeah, it was a bad idea.
--
You received this message because you are subscribed to the Google Groups
Updates:
Status: WontFix
Comment #8 on issue 2327 by asmeu...@gmail.com: sqrt(I) should return
sqrt(2)/2 + sqrt(2)/2*I by default
http://code.google.com/p/sympy/issues/detail?id=2327
(No comment was entered for this change.)
--
You received this message because you are subscribed to
Comment #1 on issue 2327 by asmeurer: sqrt(I) should return sqrt(2)/2 +
sqrt(2)/2*I by default
http://code.google.com/p/sympy/issues/detail?id=2327
It seems like this should happen automatically too:
In [45]: 1/(-sqrt(2)/2 - sqrt(2)/2*I)
Out[45]:
1
─
⎽⎽⎽ ⎽⎽⎽
Comment #2 on issue 2327 by sapta.ii...@gmail.com: sqrt(I) should return
sqrt(2)/2 + sqrt(2)/2*I by default
http://code.google.com/p/sympy/issues/detail?id=2327
I don't think this is a good idea. Trying to accomplish this will require
changes to how a Pow/Expr type is initialized. Basically
Comment #3 on issue 2327 by asmeurer: sqrt(I) should return sqrt(2)/2 +
sqrt(2)/2*I by default
http://code.google.com/p/sympy/issues/detail?id=2327
So you are *probably* right. It's ironic, since I am usually the stickler
about automatic evaluation around here. But I was a little annoyed
Comment #4 on issue 2327 by smi...@gmail.com: sqrt(I) should return
sqrt(2)/2 + sqrt(2)/2*I by default
http://code.google.com/p/sympy/issues/detail?id=2327
I could see trying to automatically remove the complex number from the
denominator as we do with radicals so 1/(1+I) goes to 1/2 - I/2
Updates:
Status: NeedsDecision
Comment #5 on issue 2327 by asmeurer: sqrt(I) should return sqrt(2)/2 +
sqrt(2)/2*I by default
http://code.google.com/p/sympy/issues/detail?id=2327
(No comment was entered for this change.)
--
You received this message because you are subscribed to the