AS CANA 1

Let ${\omega }_1,{\omega }_2\in \mathbb{C}$ be linearly independent over $\mathbb{R}$ and let $L=\{m{\omega }_1+n{\omega }_2:m,n\in \mathbb{Z}\}$ be the lattice they generate.

Problem 1

Exercise 428.1(Exercise).

Prove that if real and imaginary part of both $f$ and $zf(z)$ are harmonic, then $f$ is analytic.

Let $f(z)=u(x,y)+iv(x,y)$, and assume $u$ and $v$ are harmonic, i.e., ${\nabla }^{2}u=u_{xx}+u_{yy}=0$ and ${\nabla }^{2}v=v_{xx}+v_{yy}=0$.

Consider $zf(z)=(x+iy)(u+iv)=(xu-yv)+i(xv+yu)$. Let $p(x,y)=xu-yv$ and $q(x,y)=xv+yu$. Assume $p$ and $q$ are also harmonic, i.e., ${\nabla }^{2}p=0$ and ${\nabla }^{2}q=0$.

Compute the second partial derivatives of $p$:

$$p_x=u+x{u}_x-y{v}_x,p_{xx}=2u_x+x{u}_{xx}-y{v}_{xx},$$ $$p_y=x{u}_y-v-y{v}_y,p_{yy}=x{u}_{yy}-2v_y-y{v}_{yy}.$$

Thus,

$${\nabla }^{2}p=p_{xx}+p_{yy}=x(u_{xx}+u_{yy})-y(v_{xx}+v_{yy})+2u_x-2v_y=2u_x-2v_y=0,$$

This simplifies to $u_x=v_y$.

Now compute the second partial derivatives of $q$:

$$q_x=v+x{v}_x+y{u}_x,q_{xx}=2v_x+x{v}_{xx}+y{u}_{xx},$$ $$q_y=u+x{v}_y+y{u}_y,q_{yy}=2u_y+x{v}_{yy}+y{u}_{yy}.$$

Thus,

$${\nabla }^{2}q=q_{xx}+q_{yy}=x(v_{xx}+v_{yy})+y(u_{xx}+u_{yy})+2v_x+2u_y=2v_x+2u_y=0,$$

This simplifies to $v_x=-u_y$.

The equations $u_x=v_y$ and $u_y=-v_x$ are the Cauchy-Riemann equations. Since $u$ and $v$ are harmonic, they are $C^2$. Therefore, $f$ is analytic.

Problem 2

Exercise 428.2(Exercise).

Prove that $G_{2k}(L)$ converges absolutely for all $k>1$. Here $G_{2k}(L)$ is defined as

$$G_{2k}(L)=\sum _{\omega \in L\setminus \{0\}}{\omega }^{-2k}.$$

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For $r\in {\mathbb{Z}}_{\ge 0}$, define $L_r:=\{m{\omega }_1+n{\omega }_2:|m|,|n|⩽r\}$. For $r\in {\mathbb{Z}}_{>0}$, define $S_r:=L_r\setminus L_{r-1}$. Then, $L\setminus \{0\}={\bigcup }_{r>0}{S}_r$ and $|S_r|=(2r+1{)}^2-(2r-1{)}^2=8r$ for $r>0$. WLOG, suppose $|{\omega }_1|⩽|{\omega }_2|$. Then, for every $\omega \in S_r$, $|\omega |⩾r|{\omega }_1|$. Therefore, we can write

$$\sum _{\omega \in L\setminus \{0\}}\frac{1}{|\omega {|}^{2k}}=\sum _{r=1}^{\infty }\sum _{\omega \in S_r}\frac{1}{|\omega {|}^{2k}}⩽\sum _{r=1}^{\infty }\frac{8r}{|r{\omega }_1{|}^{2k}}=\frac{8}{|{\omega }_1{|}^{2k}}\sum _{r=1}^{\infty }\frac{1}{r^{2k-1}}.$$

Since $2k-1>1$, $\sum 1/r^{2k-1}$ converges. it follows from the comparison test that $G_{2k}(L)$ converges absolutely.

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Problem 3

Exercise 428.3(Exercise).

Find all limit points of the set $\{x_n\}$

$$x_n=\frac{1}{n}\sum _{k=1}^n{k}^{ib},n=1,2,\dots$$

where $b$ is a nonzero real number.

Claim 428.4(Claim).

All limit points of $\{x_n\}$ lie on the circle of radius $1/\sqrt{1+b^2}$.

It suffices to show that $|x_n|\to 1/\sqrt{1+b^2}$ as $n\to \infty$.

$$\begin{array}{rl}|x_n{|}^2& ={\left(\frac{1}{n}\sum _{k=1}^n\cos (b\ln k)\right)}^2+{\left(\frac{1}{n}\sum _{k=1}^n\sin (b\ln k)\right)}^2\\ & =\frac{1}{n^2}\sum _{k_1,k_2=1}^n\cos \left(b\ln k_2-b\ln k_1\right).\end{array}$$

When $n\to \infty$, this is equal to the integral

$${\int }_0^1{\int }_0^1\cos (b\ln x-b\ln y)dxdy.$$E1

We evaluate the inner integral first. Let $t=b\ln x-b\ln y$. We have

$$\begin{array}{r}{\int }_0^1\cos (b\ln x-b\ln y)dx=\frac{y}{b}{\int }_{-\infty }^{-b\ln y}\cos (t)e^{t/b}dt.\end{array}$$

Integrating by parts, we obtain

$$\begin{array}{rl}\frac{y}{b}{\left(\frac{b{e}^{t/b}\cos t+b^2{e}^{t/b}\sin t}{1+b^2}\right)}_{t=-\infty }^{t=-b\ln y}& =\frac{\cos (b\ln y)-b\sin (b\ln y)}{1+b^2}.\end{array}$$

Thus, (E1) becomes

$${\int }_0^1\frac{\cos (b\ln y)-b\sin (b\ln y)}{1+b^2}dy.$$

One substituting $t=b\ln y$ and integrating by parts again, we see that the integral is equal to $1/\sqrt{1+b^2}$.

Claim 428.5(Claim).

For any $\phi \in [0,2\pi )$, there exists a subsequence of $S_n:={\sum }_{k=1}^n{k}^{ib}$ whose amplitudes converge to $\phi$.

Let $f(x)=x^{ib}$. Write

$$\begin{array}{r}{S}_n-{\int }_0^n{x}^{ib}dx=\sum _{k=1}^n\left(k^{ib}-{\int }_{k-1}^k{x}^{ib}dx\right).\end{array}$$E2

For $x\in [k-1,k]$, the mean value theorem gives

$$|x^{ib}-k^{ib}|⩽|f^{\prime }(c)|(k-x)$$

for some $c\in (k-1,k)$. Since $|f^{\prime }(c)|=|b|/c$, it follows that

$$|x^{ib}-k^{ib}|⩽\frac{|b|}{k-1}(k-x).$$

We have for each term of (E2)

$$\begin{array}{rl}\left|k^{ib}-{\int }_{k-1}^k{x}^{ib}dx\right|& =\left|{\int }_{k-1}^k{k}^{ib}-x^{ib}dx\right|\\ & ⩽{\int }_{k-1}^k\frac{|b|}{k-1}(k-x)dx\\ & ⩽\frac{C}{k}.\end{array}$$

Therefore

$$\left|S_n-{\int }_1^n{x}^{ib}dx\right|⩽\sum _{k=1}^{n}C/k=O(\ln n).$$

Since

$${\int }_1^n{x}^{ib}dx=\frac{n^{1+ib}}{1+ib}+O(1),$$

we get

$$S_n=\frac{n^{1+ib}}{1+ib}+{\mathbf{ϵ}}_n,$$

where $|{\mathbf{ϵ}}_n|=O(\ln n)$. Since $(\ln n)/n\to 0$ as $n\to \infty$,

$$\text{arg}(S_n)=b\ln n-\text{arg}(1+ib)+{\phi }_n,$$

where ${\phi }_n\to 0$ as $n\to \infty$.

Next, since $\ln (n+1)-\ln (n)\to 0$ as $n\to \infty$ and $\ln n\to \infty$ as $n\to \infty$, we see that the set $\{b\ln n\mathrm{mod}2\pi :n\in \mathbb{N}\}$ is dense in $[0,2\pi ]$. The claim follows.

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Problem 4

Exercise 428.6(Exercise).

Prove that $\varphi (z)$ converges absolutely and uniformly on every compact subset of $\mathbb{C}\setminus L$. Here $\varphi (z)$ is defined as

$$\varphi (z)=\frac{1}{z^2}+\sum _{\omega \in L\setminus \{0\}}\left(\frac{1}{(z-\omega {)}^2}-\frac{1}{{\omega }^2}\right).$$

Let $C$ be a compact subset of $\mathbb{C}\setminus L$. Being a compact subset of $\mathbb{C}$, $C$ is bounded. Suppose $|z|⩽M$ for all $z\in C$.

$$\begin{array}{r}\left|\frac{1}{(z-\omega {)}^2}-\frac{1}{{\omega }^2}\right|=|z|\frac{|z-2\omega |}{|\omega {|}^2|z-\omega {|}^2}.\end{array}$$

For all $z\in C$, when $|\omega |>2M⩾2|z|$, we have $|z-2\omega |⩽3|\omega |$ and $\frac{1}{2}|\omega |⩽|z-\omega |$. Thus, for all $z\in C$ and for $|\omega |>2M$,

$$\begin{array}{r}\left|\frac{1}{(z-\omega {)}^2}-\frac{1}{{\omega }^2}\right|⩽c\frac{1}{|\omega {|}^3},\end{array}$$

where $c>0$ is a constant that only depends on $M$. Thus, for all $z\in C$,

$$\begin{array}{rl}\sum _{\omega \in L\setminus \{0\}}\left|\frac{1}{(z-\omega {)}^2}-\frac{1}{{\omega }^2}\right|& =\sum _{\begin{array}{c}\omega \in L\setminus \{0\}\\ |\omega |⩽2M\end{array}}(\ast )+\sum _{\begin{array}{c}\omega \in L\setminus \{0\}\\ |\omega |>2M\end{array}}(\ast )\\ & ⩽\sum _{\begin{array}{c}\omega \in L\setminus \{0\}\\ |\omega |⩽2M\end{array}}(\ast )+c\sum _{\begin{array}{c}\omega \in L\setminus \{0\}\end{array}}\frac{1}{|\omega {|}^3},\end{array}$$

which converges by Exr 2. Clearly, the bound is uniform.

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Exercise 428.7(Exercise).

Prove that $\varphi (z+w)=\varphi (z)$ for all $z\in \mathbb{C}$ and $w\in L$, and that it is meromorphic on $\mathbb{C}$ with a double pole at each point of $L$ and no other poles.

By Stein & Shakarchi (2003) 2.5.2, $\varphi$ is holomorphic on $\mathbb{C}\setminus L$. For any $\tilde{\omega }\in L$,

$$\begin{array}{rl}{\varphi }^{\prime }(z)& =\frac{-2}{z^3}+\sum _{\omega \in L\setminus \{0\}}\frac{-2}{(z-\omega {)}^3}.\\ & =\sum _{\omega \in L}\frac{-2}{(z-\omega {)}^3}\\ & =\frac{-2}{(z+\tilde{\omega }{)}^3}+\sum _{\omega \in L\setminus \{-\tilde{\omega }\}}\frac{-2}{(z-\omega {)}^3}\\ & =\frac{-2}{(z+\tilde{\omega }{)}^3}+\sum _{\omega \in L\setminus \{0\}}\frac{-2}{(z+\tilde{\omega }-\omega {)}^3}\\ & ={\varphi }^{\prime }(z+\tilde{\omega }).\end{array}$$

Thus for fixed $\tilde{\omega }\in L$, $\varphi (z+\tilde{\omega })-\varphi (z)$ is constant. Next, observe that $\varphi$ is an even function. Thus, for $\tilde{\omega }=w_1$ (and ${\omega }_2$), we have $\varphi (-{\omega }_1/2+{\omega }_1)-\varphi (-{\omega }_1/2)=0$, i.e, $\varphi (z+{\omega }_1)-\varphi (z)=0$.

We also have that $\varphi (z+{\tilde{\omega }}_1)-\varphi (z+{\tilde{\omega }}_2)$ is constant for any ${\tilde{\omega }}_1,{\tilde{\omega }}_2\in L$. Therefore,

$$\begin{array}{rl}& \varphi (z+(k+1){\omega }_1)-\varphi (z+k{\omega }_1)=0,\\ & \varphi (z+(k+1){\omega }_2)-\varphi (z+k{\omega }_2)=0\end{array}$$

by plugging $z=-(k+1/2){\omega }_1$ or $z=-(k+1/2){\omega }_2$. Finally, for arbitrary $\tilde{\omega }=m{\omega }_1+n{\omega }_2$, we can write

$$\begin{array}{rl}\varphi (z+\tilde{\omega })-\varphi (z)& =\sum _{i=m}^1\varphi (z+i{\omega }_1+n{\omega }_2)-\varphi (z+(i-1){\omega }_1+n{\omega }_2)\\ & +\sum _{i=n}^1\varphi (z+i{\omega }_2)-\varphi (z+(i-1){\omega }_2)\\ & =0.\end{array}$$

It is clear from the definition of $\varphi$ that it has a pole of order 2 at $0$. For $\tilde{\omega }\in L\setminus \{0\}$,

$$\varphi (z)=\frac{1}{(z-\tilde{\omega }{)}^2}-\underset{\text{homolorphic in a neighborhood of }\tilde{\omega }}{\underbrace{\frac{1}{{\tilde{\omega }}^2}+\left(\frac{1}{z^2}+\sum _{\omega \in L\setminus \{0,\tilde{\omega }\}}\left(\frac{1}{(z-\omega {)}^2}-\frac{1}{{\omega }^2}\right)\right)}}.$$

Thus, $\varphi$ has a double pole at each point of $L$. Since $\varphi$ is defined on all of $\mathbb{C}\setminus L$, it cannot have any other poles.

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References

Stein, E. M., & Shakarchi, R. (2003). Complex Analysis. Princeton University Press.