NavList:

Hi John,
I'm sure Robin will reply to you, but I spent some time on his two papers a while ago, and maybe can answer your question with enough clarity.

First, Z* is just notation for a point "diametrically opposite" Z. The asterisk in this case does not mean complex conjugation. Z-bar is the complex conjugate of Z.

Second, your use of pole will be (is!) confusing, because he talks about poles of great circles in section 4. There is point on the sphere that maps to the complex plane at point Z. Let's call the point on the sphere S. S maps to Z. Similarly, S* maps to Z*.

Z is a complex number, that in polar form, has magnitude r and angle phi, or in your notation Z= r exp(i phi). I think in terms of vectors, Z is r distant from the origin, angle phi counter-clockwise from the real axis. Z* is in the opposite direction, or phi+pi (or phi-pi). I think you are assuming the opposite direction of phi is -phi. But it's easy to see the bug: plug in zero for phi and you get zero for the angle of both Z and Z*

Here's my tedious derivation of Robin's result:

let S be a point on the sphere. It has coords (A,D) (A= hour angle, D=declination)
S* will be at (A+pi,-D)

Using eq 2.1 and 2.2, we find point Z with r1= tan(D/2+pi/4), phi1=A. Z= r1 exp(i A)
Z* has r2= tan(pi/4-D/2) and phi2= A+pi. Z* = r2 exp(i (A+pi))

I convinced myself (r1)(r2)=1 using tan(a +/- b) = (tan a +/- tan b)/(1 -/+ tan a tan b)

so r2= 1/r1.

and Z2= (1/r1) exp(i (A+pi))= (1/r1) exp(i pi) exp(i A)
= (-1/r1) exp(i A) = -1/(r1 exp(-i A)) = -1/conj(Z)

Robert Bernecky

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[Sent from the message boards by: Robert Bernecky]
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