After spending 3 hours on this problem without any fruition, I looked up like a dozen sites and still don't have an answer. The question is: Show that stereographic projection takes parallels of latitude (phi) on the sphere (unit sphere S2), given as z = sin (phi), into the circles in complex numbers with centre O and radius 1 + tan(phi/2) 1 - tan(phi/2) Geometric, algebraic, trigonometric... I don't care what the answer uses, I'm dying to know...

I haven't done this in a few months now so I really don't have a clue but if I remember correctly, you should change the radii into their polar forms by use of trigonometric identities. Then you should draw out the graph. I can't remember any further but drawing out the graph used to help me a lot in complex numbers.

A notch above what we usually get around here and my usual rule against direct answers applies. I also hate projection with a passion (spent far too long messing with the internals of crystals) but OK. That is a really nastily worded question too. As I understand it. You have a sphere, this sphere would happen to be a unit sphere (r=1) but we will leave that for now. Stereographic projection ( http://mathworld.wolfram.com/StereographicProjection.html ) is a simplification (supposedly) for working with spheres and other 3d items on a 2d plane. Seen as we are supposed to lack computers (my preferred way of working with 3d) a Wulff net it is but some work before then we have to lay it out. The question defines latitude as phi. Although it should not be necessary to use anything any angles are naturally in radians. z=sin(phi) being the "distance" form (phi is the angle form), as it is a unit sphere we do not have to worry about things making it more difficult (I only mention it so you can see how things work). The question is: Stereographic projection can be used to show a sphere on a 2d plane. Parallels of latitude on the sphere (defined above) can be shown as circles in complex numbers with centre O (origin) and radius 1 + tan(phi/2) 1 - tan(phi/2) Show this using the above equations. Trigonometry can quite easily be put into terms of complex numbers but make it easy on yourself and define them normally first. Equation of a circle (x - h)^2 + (y - k)^2 = r^2 The middle of the circle is the origin so x^2+y^2=r^2 parallel lines means if you have an equation for one line (or circle as the case may be) take another parallel line will leave you with a constant. In more simple terms the gradient (differentiated equation of line hence the complex form being hinted at) will be identical. It seems my body decided to be ill today and if I went to work this out I would mess up something so I am going to have to leave this to you. I can already see some simplifications and expansions coming up so brush up on trigonometric identities if you are hazy. "Geometric, algebraic, trigonometric"..... it is a combination of the lot and more.

If we use spherical coordinates, wouldn't be better? Or maybe cylindrical coordinate? The problem is just that or there's more explained, it feels like there's isn't enough information

Wow thanks for the input people, especially FAST6191! The problem is just that, I've typed the exact words (except phi symbol). I originally posted just before 4 a.m. my time and I went to bed right after posting it. I woke up this morning, spent twenty minutes-ish and came up with a (sloppy) solution. 1) As phi -> 90 degrees, radius -> infinity (because you're projecting from the "north pole" to a horizontal cross section of a sphere -a circle- at latitude phi, and phi -> 90 degrees means it's getting closer to the north pole) So here the radius is some number a / (1 - sin(phi)) where a =/= 0. Or we can write a' / (1 - tan(phi/2)) 2) As phi -> -90 degrees, radius -> 0 (because the projection from the north pole to the south pole is a point, or a circle with radius 0). So here the radius is (1 + sin(phi)) / b where b =/= 0. Or, again, we can write (1 + tan(phi/2)) / b'. Since (1 - tan(-90/2)) is not zero and (1 + tan(90/2)) is not zero, we can merge two expressions to come up with radius = (1 + tan(phi/2)) / (1 - tan(phi/2)) This is a sloppy proof because I haven't proved that radius is exactly that for phi between 90 and -90 degrees (although the values are feasible; it's easy to have radius between 0 and infinity =P). But I'm happy enough with this much work; I'm so done with it.