Locus of z(z+c) for |z|=1 (z complex) and an optical phenomenon



I was motivated to consider the following subject by the analysis
of an optical phenomenon I observed by chance and then found
its cause ... see below

What can be said about the image of the unit circle |z|=1 (z complex
number) by the map f: z |-> z * (z+c), c a complex constant ? In fact,
we don't loose soemthing essential by taking c real and >= 0
(indeed, let t be any complex nb. of modulus 1; replacing z by t*z
which is just a reparametrization of the unit circle and then
multiplying f(z) by t^(-2) which is a rotation about the center 0 will
do the same as replacing c by c*t^(-1), and t may be chosen such
that this is real and >= 0)

Considering f(z)+1 = z^2+c*z+1 instead of f(z) actually simplifies
things, this is just a change of origin in the plane. For c=0 the
the said image - I'll just call it 'my curve' - can be considered as
the unit circle "taken twice". For c=2 ( f(z)+1 = (z+1)^2 ) we get
the curve called a cardioid (heart curve) - this I recognized
from its shape and then found in an old book where it is described:
that the general case c>0 gives in fact a type of curve (which
I had forgotten) known as a "limaçon", more precisely called
limaçon of Pascal as I discovered looking for it in Wikipedia (fr).
Taking x,y as the real and imag. parts of w=f(z)+1, its equation is
[E] (x^2+y^2-2*x) ^ 2 = c^2 * (x^2+y^2)
or in polar coordinates r = c+2*cos(phi) where w=r*exp(i*phi), i^2=-1.

The definition I found in the old book is quite different from
mine, but one can easily verify that all definitions / formulas
encountered are "essentially" equivalent. A external link in Wikipedia
led me to the following (French) site explaining the curve in detail:
http://www.mathcurve.com/courbes2d/limacon/limacon.shtml
there the different possible shapes are shown (with pictures) and
there is a good discussion of its properties etc. there, so that
I drop many details which I intended to include in this post ...

In fact there is a little problem with [E] when c>2: this equation
is always satisfied by the origin x=y=0, but for c>2 this is an
isolated point of the (real) algebraic 'curve' defined by [E] which
the polar coord. version and my definition above don't reach.
This is why I say that the def. / formulas are (only) "essentially"
equivalent ...
-------------
Now the optical phenomenon behind this all. Consider the situation:
there is a small short cylinder (a normal one with circular sections
in planes perpendicular to the generatrices, i.e. obtainable by
the rotation of a straight line around an axis parallel to it), where
by 'short' is meant that it is limited by two near circular sections.
Length and diameter are both approx. 3 cm. (or 1 inch) , the axis
is horizontal. The room where it is has (among others) a wall
perpendicular to the cylinder axis, the wall being at about 85 cm. (or
a yard) of distance from the cylinder (which is at a hight of about
1m / 1 yard above the ground). A strong light source at 'infinity'
(the sun in a blue sky) shines through a window into the room,
obliquely (angles with the cylinder axis measured in horizontal
as well as vertical planes >0 but <30 deg. - estimated).

The cylinder is well shaped, metallic and smooth, so that it reflects
light after the classical law that a ray is reflected in a ray that is
symmetric to it with respect to the normal line to the reflecting
surface at the point where the ray meets the surface. This produces
on the wall a nearly complete annulus (= the area between two
concentric circles), rather narrow - so that it was just a circle for
me first - much brighter than the surrounding wall surface and
of large dimension (may-be with a diameter of more than 1m / yard).

Is the 'projection' of the cylinder to the wall an exact circle (when
the cylinder is very short, so that we may neglect its length) ?
It probably cannot be a complete circle because the external light
will not reach the whole cylinders and some reflected rays might
not be sent to the wall (I'm not sure about this 2d fact) - but my
analysis showed that already half of the cylinder surface might give
a complete circle / annulus. By my analysis of this thing by partly
geometrical, partly algebraic means, I found this: yes, if we can
neglect its diameter (and its center will even be at the intersection
of the axis and the wall). But if we don't want to do this, then
- after some rotation and resizing - the projection is "my" curve.
(From the result with a 'neglected diameter' it was easy to go to
the result without this neglecting. In this step I found it helpful
to use complex numbers to treat the plane geometry involved.)

BTW the precise result I got for the part where I neglect somehow
the diameter suggests that it might be possible to obtain it in a more
purely 'geometric' way. I wonder if someone can produce such
an argument: my result says that a 'naive simplification' that
'can only be wrong' a priori produces a correct result ... a vague
hint to what I mean here: although the plane spanned by a ray
from the source and the normal to the cylinder - containing also
the reflected ray - is in 'general position', what happens on the wall
results from the orthogonal projection to the wall of all 3 directions
as if the said plane was the wall ... up to a constant factor (for
the vectors involved) not depending on the point where reflection
takes place)

Another remark: the website mentioned above and others connected
with it also contain very similar considerations about reflecting
(surfaces or) curves, but I nevertheless find it difficult to see
a direct connection with the phenomenon I describe - which is
"irreducibly" 3D whereas those discuss practically only 2D things.
.



Relevant Pages

  • Re: Attaching Cylinder to surface
    ... Picture a wall with a cylinder attached to the ... If the "Right" plane is the wall... ... Draw a circle, where one edge ... and give the cylinder it's 3rd dimension. ...
    (comp.cad.solidworks)
  • Re: Understanding raytracing code
    ... each pixel of the screen, this code gets the direction towards each ... cylinder you're viewing, and that the cylinder is aligned along the z ... Now, to find out if you're inside a circle, all you need to do is ... Now, once this while loop is done, its either outside the cylinder ...
    (comp.graphics.algorithms)
  • Re: need help with homework
    ... Bryana says that the teacher was teaching about inverses of functions ... OK, but I presume the other items were taught earlier, such as what ... a "cylinder" of any type, not just a right-circular cylinder, etc. ... formulas (area of circle and volume of general cylinder). ...
    (sci.math)
  • Re: need help with homework
    ... Good intuition as a starter. ... base is a circle then we have a circular cylinder. ... If you have a graph of a function, you make a rule as follows: ...
    (sci.math)
  • Re: Determining the curve
    ... Where I work I use a roller to role a flat plate of metal into a curve. ... The diameter of each cylinder is 74 mm. ... the plate into a curve which in the end curves to form a circle. ...
    (sci.math)