I have checked this by hand. rectform(y10) gives the
correct result with (-1)^(1/6) = cos(%pi)^(1/6) =
%i/2+sqrt(3)/2 = (%i + sqrt(3)/2.
(%i89) display2d:false$
here is the starting expression
(%i90) y10;
(%o90) sqrt(2)*( sqrt(3)*%i / 2 - 1/2 )/( 2*(-1)^(1/6) )
+(-1)^(1/6)*(-sqrt(3)*%i / 2 - 1/2 ) / sqrt(2)
here is rectform answer:
(%i91) y10re : rectform(y10);
(%o91) 0
here is ratsimp answer
(%i92) y10ra : ratsimp(y10);
(%o92) sqrt(3)*%i/((-1)^(1/6)*sqrt(2))
(%i93) realpart(y10ra);
(%o93) sqrt(3)/(2*sqrt(2))
(%i94) imagpart(y10ra);
(%o94) 3/(2*sqrt(2))
(%i95) rectform(y10ra);
(%o95) 3*%i/(2*sqrt(2))+sqrt(3)/(2*sqrt(2))
so ratsimp returns nonzero answer which has
both a real and imaginary part.
If we simply use our replacement for
( -1 )^(1/6) in y10, automatic simplification
gives zero:
(%i96) z : rectform( (-1)^(1/6) );
(%o96) %i/2+sqrt(3)/2
(%i97) ratsubst(z,(-1)^(1/6),y10);
(%o97) 0
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So what am I doing wrong?
Ted Woollett
win xp, 5.18.1