It is currently 29 Jul 2014, 06:38




Post new topic Reply to topic  [ 5 posts ] 
 Exercise [05.09] 
Author Message
User avatar

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

In parametric form, the curve can be written as
z=a e^{b t}
with complex constants a, b, and a real variable t.

z is the `radius vector´ from the origin to the curve.
dz/dt 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´, dz/dt, and the `radius vector´, z is
\phi=\arg(\frac{dz/dt}{z})=\arg b,
that is constant in conformity with the spiral definition. So the value of the argument of b control how tightly and in which direction the spiral spirals. In the particular case that b is pure imaginary, the spiral becomes a circle of radius |a|. When b is real, the spiral becomes a straight line from origin to the direction of `vector´ a.

Complex Powers

We define w^z by
w^z=e^{z \log w}
Because of the ambiguity in \log w, we can add any integer multiple of 2\pi\imath to \log w. This means that we can multiply or divide any particular choice of w^z by e^{z\cdot 2\pi\imath} any number of times and still get an allowable w^z. Let us call p_0 the particular value of w^z chosen, then the sequence of all allowable w^z can be written as:
w^z=p_0 \left(e^{z\imath 2\pi}\right)^k=p_0 e^{z\imath 2k\pi}
for each integer k.

See that all the allowable w^z lies in the spiral given by the following equation:
z_1 = p_0 e^{z\imath t_1} (1)
at the points for which t_1 is an integral multiple of 2\pi. 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.
z_2 = p_0 e^{(z-1)\imath t_2} (2)

Indeed, the points of intersection are found solving the following equation, in which t_1 and t_2 are the unknowns:
p_0 e^{z\imath t_1} = p_0 e^{(z-1)\imath t_2}
eliminating p_0, we get:
e^{z\imath t_1} = e^{(z-1)\imath t_2}.

Writing z in terms of its real and imaginary parts, z=z_x+z_y\imath, the equation becomes:
e^{(-z_y+z_x\imath)t_1}=e^{(-z_y+(z_x-1)\imath)t_2}
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 2\pi, that is:
e^{-z_y t_1}=e^{-z_y t_2}
and
z_x t_1 - (z_x - 1)t_2 = 2k\pi, where k is any integer.

Assuming that z_y\not=0, the first of these equations holds only if t_1=t_2, then substituting t_1 for t_2 in the second one, we get:
t_1 = 2k\pi.
Therefore t_1 is an integral multiple of 2\pi at the intersections of spirals (1) and (2). But these are exactly the points of all allowable w^z, 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:
z_2 = p_0 e^{(z+1) \imath t_2} (3)

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

The figure below illustrate the case where w=1 and z=\frac{4-\imath}{5}. Then, we can choose p_0=1. 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 w^z.
Image

In the case z is real, both spirals degenerate to a circle of radius |p_0| 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 \varepsilon\imath to z.

In the case z is pure imaginary, the first spiral becomes a straight line from origin to the direction of `vector´ p_0.


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

02 May 2008, 20:50
Profile

Joined: 13 Mar 2008, 14:06
Posts: 42
Location: Ithaca NY
Post 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 all of the answers to w^z are in the intersection of two spirals. Not that a point in the intersection of 2 spirals is an answer to w^z.
As I remember, when you get different log. spirals using e^{it(z-k)}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 those spirals contains all answers to w^z.
Also, a curve f(t) with constant arg(f'(t)/f(t)) isn't necessarily a log spiral, because it could be parameterized in a different way. The parameter t 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(f'(t)/f(t)) to be a log spiral.
Laura


02 May 2008, 21:53
Profile
User avatar

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


I did show that all of the answers of w^z 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 f(t) with constant arg(f'(t)/f(t)) isn't necessarily a log spiral, because it could be parameterized in a different way.


Let s be a new parameter such that the curve is now given byz=f(g(s)). where g(s) is a real monotonic function. Even with this re-parametrization, {dz/ds} \over z does not change its direction, but it just gets multiplied by real g'(s):
\frac {dz/ds} z = g'(s) \frac{f'(g(s))} {f(g(s))} = g'(s) b
so its argument remains constant, \arg b or \arg b-\pi whether g(s) is crescent (g'(s)>0) or decrescent (g'(s)<0) 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
Profile

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


I did show that all of the answers of w^z 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, t_1=any integer multiple of 2\pi works. It'd be a good idea to say this explicitly though - as written it looks like you're saying "look, t_1 comes out to be a multiple of 2\pi, so I'm done".

If you subtract an integer from z 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 2\pi.
The real part of z could be 1,000,005, say.
Quote:
Laura wrote:
...a curve f(t) with constant arg(f'(t)/f(t)) isn't necessarily a log spiral, because it could be parameterized in a different way.


Let s be a new parameter such that the curve is now given byz=f(g(s)). where g(s) is a real monotonic function. Even with this re-parametrization, {dz/ds} \over z does not change its direction, but it just gets multiplied by real g'(s):
\frac {dz/ds} z = g'(s) \frac{f'(g(s))} {f(g(s))} = g'(s) b
so its argument remains constant, \arg b or \arg b-\pi whether g(s) is crescent (g'(s)>0) or decrescent (g'(s)<0) 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
Profile
User avatar

Joined: 02 May 2008, 20:47
Posts: 5
Location: Brazil
Post The simplest spirals
Laura wrote:
If you subtract an integer from z 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 2\pi.
The real part of z 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 z 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 z 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:
z_1=p_0 e^{(z-\lfloor z_x\rfloor)\imath t_1} (5)
and
z_2 = p_0 e^{(z-\lfloor z_x\rfloor-1)\imath t_2} (6)
where p_0 is a chosen value of w^z,
z_x is the real part of z and
\lfloor z_x\rfloor is the floor function, the largest integer less than or equals to z_x.

Then, it's easy to see that when the real part of z is integer all the intersections are collinear, since the spiral (5) becomes a line from origin into the direction of p_0.


07 May 2008, 18:55
Profile
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 5 posts ] 


Who is online

Users browsing this forum: No registered users and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
Powered by phpBB © phpBB Group.
Designed by Vjacheslav Trushkin for Free Forums/DivisionCore.