Question

The equation of an ellipse in plane polar coordinates \(r, \theta\), with one of its foci at the origin, is $$ \frac{l}{r}=1-\epsilon \cos \theta $$ where \(l\) is a length (that of the latus rectum) and \(\epsilon(0<\epsilon<1)\) is the eccentricity of the ellipse. Express the area of the ellipse as an integral around the unit circle in the complex plane, and show that the only singularity of the integrand inside the circle is a double pole at \(z_{0}=\epsilon^{-1}-\left(\epsilon^{-2}-1\right)^{1 / 2}\) By setting \(z=z_{0}+\xi\) and expanding the integrand in powers of \(\xi\), find the residue at \(z_{0}\) and hence show that the area is equal to \(\pi l^{2}\left(1-\epsilon^{2}\right)^{-3 / 2}\). (In terms of the semi-axes \(a\) and \(b\) of the ellipse, \(l=b^{2} / a\) and \(\left.\epsilon^{2}=\left(a^{2}-b^{2}\right) / a^{2} .\right)\)

Short answer

The area of the ellipse equals \[\pi l^2 \left(1-\epsilon^2\right)^{-3 / 2}\].

Step-by-step solution

Parameterize the Ellipse in Cartesian Coordinates

The standard form of the ellipse in Cartesian coordinates with semi-major axis 'a' and semi-minor axis 'b' is given by \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]

Transition to Plane Polar Coordinates

Using the polar coordinates transformation, where \[x = r \cos \theta\] and \[y = r \sin \theta\], substitute into the equation of the ellipse.This yields: \[\frac{(r \cos \theta)^2}{a^2} + \frac{(r \sin \theta)^2}{b^2} = 1\]

Incorporate Focus at the Origin

Given that one of the foci of the ellipse is at the origin, the relationship between the semi-major axis, the semi-minor axis, and the eccentricity is given by: \[c^2 = a^2 - b^2\]and the eccentricity equation is: \[\epsilon = \frac{c}{a} \Rightarrow c = a \epsilon\] .

Conversion to Complex Plane

We move to the complex plane by considering a mapping. Let the equation for the ellipse in polar form transform as:\[\frac{l}{r} = 1 - \epsilon \cos \theta\]and use Euler's formula \[z = e^{i \theta} \text{ and } \cos \theta = \frac{z + z^{-1}}{2}\]

Integral in Complex Form

The requirement is transforming the area integral as \[A = \frac{1}{i} \oint \frac{l dz}{1- \epsilon \frac{z + z^{-1}}{2}}\]within the unit circle in the complex plane.

Simplify the Integral

Simplify the denominator: \[1 - \epsilon \cos \theta = 1 - \epsilon \frac{z + z^{-1}}{2}\]Let \[1- \epsilon \cos \theta = 1 - \frac{\epsilon}{2} (z + z^{-1})= 1 - \frac{\epsilon z + \epsilon z^{-1}}{2}.\]

Identify the Singularities

Determine the singularities of the integrand within the unit circle by finding values of \[1 - \epsilon \cos \theta\]which become 0: This involves solving \[e^2 = 1 - \epsilon^2 \Rightarrow e = \left(1-\epsilon^2\right)^{1/2}\].

Find Residue at the Double Pole

Change variables set \[z =z_{0}+ \xi \text{ where } z_0 = \frac{1}{\epsilon}-\left(\frac{1}{\epsilon^2}-1\right)^{1/2};\] The residue calculation requires evaluating \[\text{Coefficients of } \xi^{-1} \text{ in the Laurent series of: } \frac{l \ dz}{1 - \epsilon (z_0 + \xi)}\].

Evaluate the Area of the Ellipse

Using the residue theorem: Assess residue at this double pole and the simplified integral form leading to \[A = \pi l^2 \left(1 - \epsilon^2\right)^{-3/2}.\]

Key concepts

Integral in Complex Plane

When dealing with integrals in the complex plane, we turn our focus to functions defined on the complex plane. This involves an integral where the integration path is a curve in the complex plane.
For our exercise, this means transforming the area integral of an ellipse into one that goes around the unit circle in the complex plane.
We start by representing the ellipse in polar coordinates and then move to the complex plane. A key step is using Euler's formula: \( z = e^{i\theta} \)
Our goal is to evaluate the integral: \[ A = \frac{1}{i} \oint \frac{l \, dz}{1- \epsilon \frac{z + z^{-1}}{2}} \]
Here, the integral is computed over the unit circle, making sure we properly address any poles (singularities) inside.

Residue Theorem

The residue theorem is a powerful tool in complex analysis that helps us evaluate integrals around closed contours.
It states that if we have a meromorphic function (one that is holomorphic except at isolated poles) within a simply connected region, we can simplify the contour integral by summing the residues at those poles.
To apply this to our problem:

• First, identify the poles inside the contour.
• Compute the residues at these poles.
• Sum the residues and multiply by \(2\pi i \).

In our specific problem, we found a double pole at \( z_{0} = \epsilon^{-1} - (\epsilon^{-2} - 1)^{1/2} \).
This double pole needs to be expanded using Laurent series to find the coefficient of \( \xi^{-1} \) and thus get the residue.

Eccentricity of Ellipse

Eccentricity, denoted as \( \epsilon \), is a measure of how much an ellipse deviates from being a circle.
For an ellipse, \( 0 < \epsilon < 1 \). It is derived from the values of the semi-major axis 'a' and the semi-minor axis 'b': \[ \epsilon = \frac{\sqrt{a^{2} - b^{2}}}{a} \]
The given exercise uses eccentricity in the equation for the ellipse in polar coordinates: \[ \frac{l}{r} = 1 - \epsilon \cos \theta \]
Understanding eccentricity is essential because it impacts the integral transformation and the position of poles in the complex plane.
As \( \epsilon \) approaches 1, the ellipse gets more elongated. Conversely, when \( \epsilon \) is closer to 0, the ellipse appears more circular.

Laurent Series

The Laurent series is an expansion of a complex function that is particularly useful around singularities (points where the function is not defined). Unlike the Taylor series, the Laurent series includes terms with negative powers.
It's written as:
\[ f(z) = \sum_{n=-\infty}^{\infty} a_{n} (z - c)^{n} \]
In our exercise, the Laurent series helps us expand the integrand near the double pole: \( z = z_{0} + \xi \).

• We focus on the coefficient \( a_{-1} \) of \( \xi^{-1} \) to find the residue.
• The residue gives us essential information to evaluate the integral using the residue theorem.

Therefore, relating the Laurent series to our problem, we find:
\[ \text{Residue} = \text{Coefficient of } \xi^{-1} \]
Incorporating this into our integral, it simplifies to finding the area of the ellipse as \( \pi l^{2} (1 - \epsilon^{2})^{-3/2} \).