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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: [mg49289] Re: [mg49277] Test for pure real complex quotient
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Mon, 12 Jul 2004 02:11:30 -0400 (EDT)
  • References: <200407110616.CAA16681@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 11 Jul 2004, at 15:16, Narasimham G.L. wrote:

> *This message was transferred with a trial version of CommuniGate(tm) 
> Pro*
> 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
>
>


I am not sure what do you mean by "simple". There is the obvious 
"simple test":

Simplify[ComplexExpand[
     Im[Z1/Z2], {Z1, Z2},
     TargetFunctions ->
      {Arg, Abs}] == 0,
   {Abs[Z1] > 0, Abs[Z2] > 0}]

Sin[Arg[Z1] - Arg[Z2]] == 0

(One doesn't really need Mathematica for that!)

But it does not mean it will be easy to check this in all cases by 
hand. In fact in the above case a better test is:

y = Sqrt[1 - x^2];
v = Log[(1 + y)/(1 - y)]/y;

FullSimplify[ComplexExpand[Im[v]],x>1]

0


Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/


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