Homework 04

2. Winding number

Suppose $f(z)$ is holomorphic in the disc $|z| ≤ ϵ$ and has a zero at $z = 0$ but nowhere else in the disc $|z| ≤ ϵ$. Show by direct integration that

\[ \frac{1}{2πi} \int_{|z|=ϵ} \frac{f'(z)}{f(z)} dz \]

equals the winding number of the argument of f around the circle $|z| = ϵ$. Then use the residue theorem to show that this equals the degree of the zero, in agreement with the argument principle.

The integral

\[ \int_{C} \frac{f'(z)}{f(z)} dz = \int_{C} \frac{1}{f(z)} df(z) = \int_{f(C)} \frac{1}{w} dw \]

effectively measures the total change in the argument of $f(z)$ as $z$ traverses the circle, which equals the winding number of the argument of $f$.

Since $f(z)$ is holomorphic in the disc $|z| \leq \epsilon$ and has a zero at $z = 0$, $f(z)$ can be locally expressed as $z^n g(z)$, where $n$ is the degree of the zero.

Then, $f'(z) = n z^{n−1} g(z) + z^n g'(z)$, and so

\[ \frac{f'(z)}{f(z)} = \frac{n}{z} + \frac{g'(z)}{g(z)} \]

The residue at $z=0$ is the coefficient of $\frac{1}{z}$ in this expression, which is $n$, the degree of the zero.

4. Analytic continuation and Fourier coefficents

Give an analytic continuation of $\cos θ$ from the unit circle $z = e^{iθ}$ to the complex plane minus the origin.

Conclude that the Fourier coefficients $c_n$ of $e^{- \cos θ}$ decrease faster than any exponential, meaning $c_n = o(e^{−αn})$ for all $α$ as $n → ±∞$. Compare this to the Fourier series of $1/(cos θ −3/2)$, what is the decay of its Fourier coefficents?

Analytic continuation

On the unit circle $z=e ^ {iθ}$, $\cos \theta = \frac{1}{2} (e^{iθ} +e^{−iθ}) = \frac{1}{2} (z+ z^{-1})$. This expression provides an analytic continuation of $\cos \theta$ to the complex plane minus the origin, as it is well-defined for all $z \neq 0$.

Fourier coefficents of $e^{−\cos \theta}$

The function $e^{−\cos \theta}$ is smooth and periodic. The Fourier coefficients $c_n$ of a periodic function $f(θ)$ are given by:

\[ c_n= 1/2π ∫_{0}^{2π} f(θ)e^{inθ}dθ = 1/2π \int_{0}^{2π} e^{-\cos θ + i n θ}dθ \]

\[ 1/2π \int_{0}^{2π} e^{-\cos θ + i n θ} dθ = 1/2π \int_{|z|=1} e^{-z/2-1/2z} z^n \frac{dz}{i z} = Res_{z=0} { e^{-z/2-1/2z} z^{n-1} } \]

The coefficient $a_{-n}$ of $g(z)=e^{-z/2-1/2z}$ at the point of $z=0$ can de derived by expanding by separately $e^{-z/2}$ as Talyor Series and $e^{−1/(2z)}$ as a Laurent Series and then multiplying these series together

\[ a_{-n} = \sum_{k=0}^{\infty} (-\frac{1}{2})^k \frac{1}{k!} (-\frac{1}{2})^{(n+k)} \frac{1}{(n+k)!} \]

So fourier coefficients $c_n$

\[ c_n = a_{-n} \leq |a_{-n} | \leq \frac{1}{2^{n}}\frac{1}{n!} \]

decrease faster than any exponential.

Fourier coefficents of $\frac{1}{\cos \theta - 3/2}$

The Fourier coefficients $c_n$ of a periodic function $f(θ)$ are given by:

\[ \begin{aligned} c_n &= \frac{1}{2π} \int_{0}^{2π} f(θ)e^{inθ}dθ \\ &= \frac{1}{2π} \int_{0}^{2π} \frac{1}{\cos \theta - 3/2}e^{i n θ}dθ \\ &= \frac{1}{2πi} \int_{|z| = 1} \frac{2z}{z^2-3z+1} z^{n-1} dz \end{aligned} \]

Let $z^2 - 3z+1=(z-z_1)(z-z_2)$, where $z_1 = \frac{1}{2} \left(3-\sqrt{5}\right), z_2=\frac{1}{2} \left(3 + \sqrt{5}\right)$.

\[ \begin{aligned} c_n &= Res_{z=z_1} \frac{2z}{(z-z_1)(z-z_2)} z^{n-1} \\ &= \frac{2 z_1}{z_1-z_2} {z_1}^{n-1} \end{aligned} \]

5. Laurent series and singularity

Let’s consider a function $f$ that is holomorphic in a disc around $z_0$ except at $z_0$ itself.

Removable Singularity:

If $f$ has a removable singularity at $z_0$, it means that $f$ can be extended to a holomorphic function at $z_0$. In terms of the Laurent series, this implies that all the coefficients $a_n$ for $n < 0$ are zero because it reduces to its Taylor series.

Conversely, if all $a_n = 0$ for $n < 0$, the Laurent series reduces to a Taylor series, implying that $f$ is holomorphic at $z_0$ (since it can be expressed as a power series), and thus the singularity is removable.
Pole of Order $m$ :

If $f$ has a pole of order $m$ at $z_0$, it means that in the Laurent series, there is a term with $(z - z_0)^{-m}$ (where $a_{-m} \neq 0$) and no terms with higher negative powers.

Conversely, if there is some $m < 0$ such that $a_m \neq 0$ but $a_n = 0$ for all $n < m$, then the Laurent series has a term $a_m (z - z_0)^m$ as its term with the highest negative power, indicating a pole of order $m$ .
Essential Singularity:

If the singularity at $z_0$ is neither removable nor a pole, it must be an essential singularity. This is characterized by the fact that there are infinitely many negative powers of $z - z_0$ in the Laurent series with non-zero coefficients. In other words, if the Laurent series has non-zero $a_n$ for infinitely many $n < 0$, then $z_0$ is an essential singularity.

6. Euler proof of Basel problem

Using the result of the bonus problem, prove that

\[ \sin π z = \prod (1 − z/n)e^{z/n} = π z \prod (1 − z^2/n^2) \]

Then compare the Taylor series of sin πz to the first couple terms in the expansion of the infinite product to conclude

\[ \sum_{n=1}^{\infty} 1/n^2 = π^2/ 6 \]

According to the Weierstrass factorization theorem, an entire function can be represented as a product over its zeros. The function $\sin \pi z$ is entire and has zeros at all integers. The product representation for $\sin \pi z$ is given by:

\[ \sin \pi z = \pi z \prod_{n=1}^{\infty} \left(1 - \frac{z}{n}\right)\left(1 + \frac{z}{n}\right) = \sin \pi z = \pi z \prod_{n=1}^{\infty} \left(1 - \frac{z^2}{n^2}\right) \]

The Taylor series expansion of $\sin \pi z$ around $z = 0$ is:

\[ \sin \pi z = \pi z - \frac{\pi^3 z^3}{3!} + \frac{\pi^5 z^5}{5!} - \frac{\pi^7 z^7}{7!} + \cdots \]

Now, let’s expand the infinite product to the first couple of terms and keeping terms up to $z^3$, we get:

\[ \begin{aligned} \pi z \prod_{n=1}^{\infty} \left(1 - \frac{z^2}{n^2}\right) &= \pi z \left(1 - \frac{z^2}{1^2}\right)\left(1 - \frac{z^2}{2^2}\right)\left(1 - \frac{z^2}{3^2}\right) \cdots \\ &= \pi z \left(1 - z^2 \left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots\right) + \cdots \right) \end{aligned} \]

Comparing the coefficient of $z^3$ from the Taylor series and the product expansion, we have:

\[ -\frac{\pi^3}{6} = -\pi \left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots\right) \]

This is the result we want.

--- title: Homework 04 format: html: default typst: output-file: phys231_hw04_raw --- ::: {.content-visible when-format="html"} - [Complete PDF](./phys231_hw04_final.pdf) - [Mathematica Notebook](./phys231_hw04.nb) ::: ## 2. Winding number > Suppose $f(z)$ is holomorphic in the disc $|z| ≤ ϵ$ and has a zero at $z = 0$ but nowhere else in the disc $|z| ≤ ϵ$. Show by direct integration that > > $$ > \frac{1}{2πi} \int_{|z|=ϵ} \frac{f'(z)}{f(z)} dz > $$ > > equals the winding number of the argument of f around the circle $|z| = ϵ$. Then use the residue theorem to show that this equals the degree of the zero, in agreement with the argument principle. The integral $$ \int_{C} \frac{f'(z)}{f(z)} dz = \int_{C} \frac{1}{f(z)} df(z) = \int_{f(C)} \frac{1}{w} dw $$ effectively measures the total change in the argument of $f(z)$ as $z$ traverses the circle, which equals the winding number of the argument of $f$. Since $f(z)$ is holomorphic in the disc $|z| \leq \epsilon$ and has a zero at $z = 0$, $f(z)$ can be locally expressed as $z^n g(z)$, where $n$ is the degree of the zero. Then, $f'(z) = n z^{n−1} g(z) + z^n g'(z)$, and so $$ \frac{f'(z)}{f(z)} = \frac{n}{z} + \frac{g'(z)}{g(z)} $$ The residue at $z=0$ is the coefficient of $\frac{1}{z}$ in this expression, which is $n$, the degree of the zero. ## 4. Analytic continuation and Fourier coefficents > Give an analytic continuation of $\cos θ$ from the unit circle $z = e^{iθ}$ to the complex plane minus the origin. > > Conclude that the Fourier coefficients $c_n$ of $e^{- \cos θ}$ decrease faster than any exponential, meaning $c_n = o(e^{−αn})$ for all $α$ as $n → ±∞$. Compare this to the Fourier series of $1/(cos θ −3/2)$, what is the decay of its Fourier coefficents? ### Analytic continuation On the unit circle $z=e ^ {iθ}$, $\cos \theta = \frac{1}{2} (e^{iθ} +e^{−iθ}) = \frac{1}{2} (z+ z^{-1})$. This expression provides an analytic continuation of $\cos \theta$ to the complex plane minus the origin, as it is well-defined for all $z \neq 0$. ### Fourier coefficents of $e^{−\cos \theta}$ The function $e^{−\cos \theta}$ is smooth and periodic. The Fourier coefficients $c_n$ of a periodic function $f(θ)$ are given by: $$ c_n= 1/2π ∫_{0}^{2π} f(θ)e^{inθ}dθ = 1/2π \int_{0}^{2π} e^{-\cos θ + i n θ}dθ $$ $$ 1/2π \int_{0}^{2π} e^{-\cos θ + i n θ} dθ = 1/2π \int_{|z|=1} e^{-z/2-1/2z} z^n \frac{dz}{i z} = Res_{z=0} { e^{-z/2-1/2z} z^{n-1} } $$ The coefficient $a_{-n}$ of $g(z)=e^{-z/2-1/2z}$ at the point of $z=0$ can de derived by expanding by separately $e^{-z/2}$ as Talyor Series and $e^{−1/(2z)}$ as a Laurent Series and then multiplying these series together $$ a_{-n} = \sum_{k=0}^{\infty} (-\frac{1}{2})^k \frac{1}{k!} (-\frac{1}{2})^{(n+k)} \frac{1}{(n+k)!} $$ So fourier coefficients $c_n$ $$ c_n = a_{-n} \leq |a_{-n} | \leq \frac{1}{2^{n}}\frac{1}{n!} $$ decrease faster than any exponential. ### Fourier coefficents of $\frac{1}{\cos \theta - 3/2}$ The Fourier coefficients $c_n$ of a periodic function $f(θ)$ are given by: $$ \begin{aligned} c_n &= \frac{1}{2π} \int_{0}^{2π} f(θ)e^{inθ}dθ \\ &= \frac{1}{2π} \int_{0}^{2π} \frac{1}{\cos \theta - 3/2}e^{i n θ}dθ \\ &= \frac{1}{2πi} \int_{|z| = 1} \frac{2z}{z^2-3z+1} z^{n-1} dz \end{aligned} $$ Let $z^2 - 3z+1=(z-z_1)(z-z_2)$, where $z_1 = \frac{1}{2} \left(3-\sqrt{5}\right), z_2=\frac{1}{2} \left(3 + \sqrt{5}\right)$. $$ \begin{aligned} c_n &= Res_{z=z_1} \frac{2z}{(z-z_1)(z-z_2)} z^{n-1} \\ &= \frac{2 z_1}{z_1-z_2} {z_1}^{n-1} \end{aligned} $$ ## 5. Laurent series and singularity Let's consider a function $f$ that is holomorphic in a disc around $z_0$ except at $z_0$ itself. 1. **Removable Singularity:** If $f$ has a removable singularity at $z_0$, it means that $f$ can be extended to a holomorphic function at $z_0$. In terms of the Laurent series, this implies that all the coefficients $a_n$ for $n < 0$ are zero because it reduces to its Taylor series. Conversely, if all $a_n = 0$ for $n < 0$, the Laurent series reduces to a Taylor series, implying that $f$ is holomorphic at $z_0$ (since it can be expressed as a power series), and thus the singularity is removable. 2. **Pole of Order $m$ :** If $f$ has a pole of order $m$ at $z_0$, it means that in the Laurent series, there is a term with $(z - z_0)^{-m}$ (where $a_{-m} \neq 0$) and no terms with higher negative powers. Conversely, if there is some $m < 0$ such that $a_m \neq 0$ but $a_n = 0$ for all $n < m$, then the Laurent series has a term $a_m (z - z_0)^m$ as its term with the highest negative power, indicating a pole of order $m$ . 3. **Essential Singularity:** If the singularity at $z_0$ is neither removable nor a pole, it must be an essential singularity. This is characterized by the fact that there are infinitely many negative powers of $z - z_0$ in the Laurent series with non-zero coefficients. In other words, if the Laurent series has non-zero $a_n$ for infinitely many $n < 0$, then $z_0$ is an essential singularity. ## 6. Euler proof of Basel problem > Using the result of the bonus problem, prove that > > $$ > \sin π z = \prod (1 − z/n)e^{z/n} = π z \prod (1 − z^2/n^2) > $$ > > Then compare the Taylor series of sin πz to the first couple terms in the expansion of the infinite product to conclude > > $$ > \sum_{n=1}^{\infty} 1/n^2 = π^2/ 6 > $$ According to the Weierstrass factorization theorem, an entire function can be represented as a product over its zeros. The function $\sin \pi z$ is entire and has zeros at all integers. The product representation for $\sin \pi z$ is given by: $$ \sin \pi z = \pi z \prod_{n=1}^{\infty} \left(1 - \frac{z}{n}\right)\left(1 + \frac{z}{n}\right) = \sin \pi z = \pi z \prod_{n=1}^{\infty} \left(1 - \frac{z^2}{n^2}\right) $$ The Taylor series expansion of $\sin \pi z$ around $z = 0$ is: $$ \sin \pi z = \pi z - \frac{\pi^3 z^3}{3!} + \frac{\pi^5 z^5}{5!} - \frac{\pi^7 z^7}{7!} + \cdots $$ Now, let's expand the infinite product to the first couple of terms and keeping terms up to $z^3$, we get: $$ \begin{aligned} \pi z \prod_{n=1}^{\infty} \left(1 - \frac{z^2}{n^2}\right) &= \pi z \left(1 - \frac{z^2}{1^2}\right)\left(1 - \frac{z^2}{2^2}\right)\left(1 - \frac{z^2}{3^2}\right) \cdots \\ &= \pi z \left(1 - z^2 \left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots\right) + \cdots \right) \end{aligned} $$ Comparing the coefficient of $z^3$ from the Taylor series and the product expansion, we have: $$ -\frac{\pi^3}{6} = -\pi \left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots\right) $$ This is the result we want.