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Exercise [05.09]
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Joined: 02 May 2008, 20:47
Posts: 5
Location: Brazil
Exercise [05.09]
Logarithmic Spiral

In parametric form, the curve can be written as

with complex constants , , and a real variable .

is the radius vector´ from the origin to the curve.
is the velocity vector´, therefore it is tangent to the curve.
The angle between two complex numbers is the argument of the quotient of these numbers. Therefore, the angle between the tangent vector´, , and the radius vector´, is
,
that is constant in conformity with the spiral definition. So the value of the argument of control how tightly and in which direction the spiral spirals. In the particular case that is pure imaginary, the spiral becomes a circle of radius . When is real, the spiral becomes a straight line from origin to the direction of vector´ .

Complex Powers

We define by

Because of the ambiguity in , we can add any integer multiple of to . This means that we can multiply or divide any particular choice of by any number of times and still get an allowable . Let us call the particular value of chosen, then the sequence of all allowable can be written as:

for each integer .

See that all the allowable lies in the spiral given by the following equation:
(1)
at the points for which is an integral multiple of . So, we have just found one of the spirals we want, then all we have to do is to find another spiral that intersects spiral (1) exactly at these points.

We can show that this spiral can be given by the following equation.
(2)

Indeed, the points of intersection are found solving the following equation, in which and are the unknowns:

eliminating , we get:
.

Writing in terms of its real and imaginary parts, , the equation becomes:

this equation holds only if the two sides of the equation have the same modulus and the difference of their arguments be an integral multiple of , that is:

and
, where k is any integer.

Assuming that , the first of these equations holds only if , then substituting for in the second one, we get:
.
Therefore is an integral multiple of at the intersections of spirals (1) and (2). But these are exactly the points of all allowable , as we wanted to show.

We could use a similar reasoning to show that another spiral intersects spiral (1) at and only at these points:
(3)

There are no other spiral that intersects the first spiral at and only at all points of allowable .

The figure below illustrate the case where and . Then, we can choose . The red spiral corresponds to spiral (1), the green one to (2), and the blue one to (3). Look that the green intersects the red at the same points that the blue intersects the red, these points are marked with black spots and represents the allowable .

In the case is real, both spirals degenerate to a circle of radius centered at the origin, making it impossible to distinguish where the answers points are. In order to see where they are, we could break the degeneracy by adding a small imaginary part to z.

In the case is pure imaginary, the first spiral becomes a straight line from origin to the direction of vector´ .

Last edited by susumu on 06 May 2008, 12:36, edited 10 times in total.

02 May 2008, 20:50

Joined: 13 Mar 2008, 14:06
Posts: 42
Location: Ithaca NY
Re: Exercise [5.9]
I'm glad somebody tackled one of the more interesting ones!
You didn't quite get it, though - the problem is to show that of the answers to are in the intersection of two spirals. Not that a point in the intersection of 2 spirals is an answer to .
As I remember, when you get different log. spirals using for different integers k, there is a tightest one that spirals in the clockwise direction, and a tightest one that spirals counterclockwise, and you need to show that the intersection of spirals contains all answers to .
Also, a curve with constant arg() isn't necessarily a log spiral, because it could be parameterized in a different way. The parameter could move at different speeds along the curve at different points, and still be tracing out the same path in the complex plane. But it'd be easy to show that you can reparameterize a curve with constant arg() to be a log spiral.
Laura

02 May 2008, 21:53

Joined: 02 May 2008, 20:47
Posts: 5
Location: Brazil
Re: Exercise [5.9]
Laura wrote:
...the problem is to show that of the answers to are in the intersection of two spirals. Not that a point in the intersection of 2 spirals is an answer to .

I did show that all of the answers of are in the intersection of two spirals. Well, maybe my explanation was not clear enough, so I rephrased it above, see if you can convince yourself.

Laura wrote:
...a curve with constant arg() isn't necessarily a log spiral, because it could be parameterized in a different way.

Let be a new parameter such that the curve is now given by. where is a real monotonic function. Even with this re-parametrization, does not change its direction, but it just gets multiplied by real :

so its argument remains constant, or whether is crescent () or decrescent () respectively. As expected, since this is a property of the curve, the angle between the curve and the radial line, it must not depend on the particular parametrization.

05 May 2008, 17:25

Joined: 13 Mar 2008, 14:06
Posts: 42
Location: Ithaca NY
Re: Exercise [5.9]
susumu wrote:
Laura wrote:
...the problem is to show that of the answers to are in the intersection of two spirals. Not that a point in the intersection of 2 spirals is an answer to .

I did show that all of the answers of are in the intersection of two spirals. Well, maybe my explanation was not clear enough, so I rephrased it above, see if you can convince yourself.

OK. You're right that if you look at the expression, integer multiple of works. It'd be a good idea to say this explicitly though - as written it looks like you're saying "look, comes out to be a multiple of , so I'm done".

If you subtract an integer from so the real part is in the interval [0,1), then one spiral goes one direction around the circle and the other spiral goes in the opposite direction. Then it's obvious that they have to go around by a total angle of .
The real part of could be 1,000,005, say.
Quote:
Laura wrote:
...a curve with constant arg() isn't necessarily a log spiral, because it could be parameterized in a different way.

Let be a new parameter such that the curve is now given by. where is a real monotonic function. Even with this re-parametrization, does not change its direction, but it just gets multiplied by real :

so its argument remains constant, or whether is crescent () or decrescent () respectively. As expected, since this is a property of the curve, the angle between the curve and the radial line, it must not depend on the particular parametrization.

Well, that is a technical detail, I didn't bother with it myself. He's only asking you to show that two such log spirals exist, anyways, as I remember, it isn't necessary to solve the problem to reduce the general parameterization to the special one where it's a log spiral. So that is an unnecessary point.
Laura

06 May 2008, 19:03

Joined: 02 May 2008, 20:47
Posts: 5
Location: Brazil
The simplest spirals
Laura wrote:
If you subtract an integer from so the real part is in the interval [0,1), then one spiral goes one direction around the circle and the other spiral goes in the opposite direction. Then it's obvious that they have to go around by a total angle of .
The real part of could be 1,000,005, say.

The spirals in my first post are algebraically the most obvious, but geometrically are not the simplest ones.

The real part of equals the number of turns the spiral (1) goes between two consecutive intersections, the spiral (2) goes one turn less, and spiral (3) one turn more. When the real part of is large the spirals goes through dizzying many turns, one chasing the other, between two intersections.

But there is a solution, the simplest geometrically, in which both spirals goes only a fraction of turn between intersections, one clockwise and other counterclockwise. These spirals are given by the following equations:
(5)
and
(6)
where is a chosen value of ,
is the real part of and
is the floor function, the largest integer less than or equals to .

Then, it's easy to see that when the real part of is integer all the intersections are collinear, since the spiral (5) becomes a line from origin into the direction of .

07 May 2008, 18:55
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