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MathGroup Archive 2004

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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




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