RE: Test for pure real complex quotient
- To: mathgroup at smc.vnet.net
- Subject: [mg49294] RE: [mg49277] Test for pure real complex quotient
- From: "David Park" <djmp at earthlink.net>
- Date: Mon, 12 Jul 2004 02:11:35 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
y = Sqrt[1 - x^2]; Im[Log[(1 + y)/(1 - y)]/y] // ComplexExpand; Simplify[%, x > 1] 0 Im[Log[(1 + y)/(1 - y)]/y] // ComplexExpand; Simplify[%, 0 < x < 1] 0 But when you put x == 1 you get an indeterminate expression. Im[Log[(1 + y)/(1 - y)]/y] // ComplexExpand; % /. x -> 1 Indeterminate David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ From: Narasimham G.L. [mailto:mathma18 at hotmail.com] To: mathgroup at smc.vnet.net We know that a quotiented complex number Z =(a + I b)/(c + I d) can be pure real if ratios of quotients (angle argument of each complex number,ArcTan[b/a]) are equal, i.e., (b/a=d/c). In general complex variable function theory,is there condition or a method to know if Im[Z1/Z2]=0 ? One way in Mathematica is to ParametricPlot3D [{Re[Z],Im[Z]},{t,,}] and verify a plane, while ignoring error messages ( not machine-size real number ) when result is contradictory. Is there a simple test for pure real (or for that matter, pure imaginary) numbers? This came last week while discussing area/volume relations of an oblate ellipsoid. http://mathforum.org/discuss/sci.math/a/m/614934/617098 y=Sqrt[1-x^2]; Plot [{Log[(1+y)/(1-y)],y,Log[(1+y)/(1-y)]/y},{x,0,2}]; It was not expected the third function would be real for x > 1 . TIA