Exercise [05.09]

susumu · 02 May 2008, 20:50

Logarithmic Spiral

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

,

, and a real variable

.

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´,

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

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 |a|

. When

is real, the spiral becomes a straight line from origin to the direction of `vector´

.

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

.

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

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

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

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

.

In the case

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

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

.

Exercise [05.09]

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