Vladimir Bondarenko wrote:
> evalf(-2*arctan(25/8*2^(1/2))-2*arctan(8/3)+(3+4*I)*Pi);
>
> 4.304187764+12.56637062*I
>
> Ummm... near this, but it sounds like your result
> has some room for further compression :)
>
> Chop[N[Pi+2 I (2 ArcCot[1/2+(-1)^(3/4)]+Log[(Root[73+112 #1 +
> 146 #1^2 + 48 #1^3 + 9 #1^4 &, 3] Root[9 + 60 #1 + 182 #1^2 +
> 188 #1^3 + 73 #1^4 &,1]^(1 + I))/Root[81- 324 #1 - 666 #1^2 +
> 1980 #1^3 + 5329 #1^4 &, 3]^I])]]
>
> 0
>
> ;)
>
> On Mar 24, 7:42 am, Axel Vogt <&nore...@[EMAIL PROTECTED]
> wrote:
>> Vladimir Bondarenko wrote:
>>
>> ...
>>
>>> Pi + 2 I (2 ArcCot[1/2 + (-1)^(3/4)] + Log[(Root[73 + 112 #1 +
>>> 146 #1^2 + 48 #1^3 + 9 #1^4 &, 3] Root[9 + 60 #1 + 182 #1^2 +
>>> 188 #1^3 + 73 #1^4 &,1]^(1 + I))/Root[81 - 324 #1 - 666 #1^2 +
>>> 1980 #1^3 + 5329 #1^4 &, 3]^I])
>>> ?
>> -2*arctan(25/8*2^(1/2))-2*arctan(8/3)+(3+4*I)*Pi
>
"Pi + 2 I (2 ArcCot[1/2 + (-1)^(3/4)] + Log[(Root[73 + 112 #1 +
146 #1^2 + 48 #1^3 + 9 #1^4 &, 3] Root[9 + 60 #1 + 182 #1^2 +
188 #1^3 + 73 #1^4 &,1]^(1 + I))/Root[81 - 324 #1 - 666 #1^2 +
1980 #1^3 + 5329 #1^4 &, 3]^I])":
convert(%,FromMma): lprint(%);
Pi+2*I*(2*Pi-2*arccot(-1/2-(-1)^(3/4))+ln(
RootOf(73+112*_Z+146*_Z^2+48*_Z^3+9*_Z^4,index = 1)*
RootOf(9+60*_Z+182*_Z^2+188*_Z^3+73*_Z^4,index = 3)^(1+I)/
(RootOf(81-324*_Z-666*_Z^2+1980*_Z^3+5329*_Z^4,index = 4)^I)))
evalf[30](%): evalf(%);
4.3041877626100 + 12.566370614359 I
|
