Logarithmic SpiralIn 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 PowersWe 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´

.