Holomorphic functions and its integral representation

We want to prove that a holomorphic function $f:U\rightarrow\mathbb{C}$ is a infinitely smooth function, or $f$ has derivatives of any order. This is based on the integral representation (or Cauchy integral formula) of $f$ : $f(z)=\frac{1}{2\pi i}\int_{C}\frac{f(\xi)}{\xi-z}\mathrm d\xi$ (1). Here $C$ is a circle in the counterclockwise sense and $z$ is a point inside $C$ .

1. Prove the following Lemma (see lecture notes).

Lemma 1 Let $\varphi(\xi)$ be a continuous funtion on $C$ . Then, the function $F_n(z)=\int_C\frac{\varphi(\xi)}{(\xi-z)^n}\mathrm{d}\xi,n=1,2,3,\dots$ , is holomorphic for any $z$ inside $C$ , and its derivative is $F_n'(z)=nF_{n+1}(z)$ . (2)
You should first prove the statement for $n=1$ , and the by assuming the statement is true for $n-1$ , and prove the statement for $n$ .
Guideline: this can be done by first choosing a disk $D(z_0,r)$ which is inside $C$ for an arbitrary point $z_0$ . Let $\xi\in C$ and let $z$ be any point such that $|z-z_0|\leq\frac{r}{2}$ , then $|z-\xi|>\frac{r}{2},|z_0-\xi|>r$ . The above estimates is always useful in the proof. For any $n$ , assume that $F'_{n-1}(z)=(n-1)F_n(z)$ (3) is true. Then, you need to compute $F_n(z)-F_n(z_0)$ and also $\frac{F_n(z)-F_n(z_0)}{z-z_0}$ . You need to use the following identity $\frac{1}{(\xi-z)^n}=\frac{1}{(\xi-z)^{n-1}}(\frac{1}{\xi-z}-\frac{1}{\xi-z_0}+\frac{1}{\xi-z_0})=\frac{1}{(\xi-z)^{n-1}}(\frac{z-z_0}{(\xi-z)(\xi-z_0)}+\frac{1}{\xi-z_0})=\frac{1}{(\xi-z)^{n-1}}\frac{z-z_0}{(\xi-z)(\xi-z_0)}+\frac{1}{(\xi-z)^{n-1}}\frac{1}{\xi-z_0}$ , and also the assumption (3) to prove (2).

For all $z_0$ inside $C$ , let $\rho=\inf_{\xi\in C}|\xi-z_0|>0,r=\min(1,\frac{\rho}{2}),M=\sup_{\xi\in C}|\varphi(\xi)|$ . Therefore when $\xi\in C,z\in D(z_0,r)$ , we have $|z-z_0|\leq\frac{r}{2},|z-\xi|>\frac{r}{2},|z_0-\xi|>r$ . As $\frac{1}{\xi-z}=\frac{z-z_0}{(\xi-z)(\xi-z_0)}+\frac{1}{\xi-z_0}$ , we can get $|\frac{F_n(z)-F_n(z_0)}{z-z_0}-nF_{n+1}(z)|=|\int_C\{\frac{1}{z-z_0}[(\frac{z-z_0}{(\xi-z)(\xi-z_0)}+\frac{1}{\xi-z_0})^n-\frac{1}{(\xi-z_0)^n}]-\frac{n}{(\xi-z)^{n+1}}\}\varphi(\xi)\mathrm d\xi|=|\int_C(\frac{1}{z-z_0}\sum_{i=1}^{n}C_n^i\frac{(z-z_0)^i}{(\xi-z)^i(\xi-z_0)^n}-\frac{n}{(\xi-z)^{n+1}})\mathrm d\xi|=|\int_C(\frac{n}{(\xi-z)(\xi-z_0)^n}-\frac{n}{(\xi-z)^{n+1}}+\sum_{i=2}^nC_n^i\frac{(z-z_0)^{i-1}}{(\xi-z)^i(\xi-z_0)^n})\varphi(\xi)\mathrm d\xi|=|\int_C(\frac{n(z_0-z)\sum_{i=0}^{n-1}(\xi-z)^i(\xi-z_0)^{n-1-i}}{(\xi-z)^{n+1}(\xi-z_0)^n}+\sum_{i=2}^nC_n^i\frac{(z-z_0)^{i-1}}{(\xi-z)^i(\xi-z_0)^n})\varphi(\xi)\mathrm d\xi|\leq n|z-z_0|\sum_{i=0}^{n-1}|\int_C\frac{\varphi(\xi)}{(\xi-z)^{n+1-i}(\xi-z_0)^{i+1}}\mathrm d\xi|+\sum_{i=2}^nC_n^i|z-z_0|^{i-1}|\int_C\frac{\varphi(\xi)}{(\xi-z)^i(\xi-z_0)^n}\mathrm d\xi|\leq n|z-z_0|\sum_{i=0}^{n-1}\frac{2^{n+1-i}M|C|}{r^{n+2}}+\sum_{i=2}^nC_n^i|z-z_0|^{i-1}\frac{2^iM|C|}{r^{n+i}}$ . Let $z\rightarrow z_0$ , the equation turns into $F_n'(z_0)-nF_{n+1}(z_0)=0$ , which means $F_n'(z)=nF_{n+1}(z)$ and $F_n(z)$ is holomorphic. (If $n=1$ , consider the summation terms $\sum_{i=2}^n\dots$ as $0$ , the proof still works.)

2. Let $f$ be in its integral representation (1), give the expression of its $n$ th-order derivative.

As $F_n'(z)=nF_{n+1}(z),\forall n$ , we can get $F_1^{(n)}=1F_2^{(n-1)}=\cdots=n!F_{n+1}$ , in other words, $f^{(n)}(z)=\frac{1}{2\pi i}F_1^{(n)}(z)=\frac{n!}{2\pi i}\int_C\frac{f(\xi)}{(\xi-z)^{n+1}}\mathrm d\xi$ .

We are going to derive a finite Taylor expansion of $f(z)$ around a point $z_0$ that is $f(z)=a_0+(z-z_0)a_1+\dots+(z-z_0)^{n-1}a_{n-1}+(z-z_0)^nh(z)$ (4), where $h(z)$ is also a holomorphic function. To do so, let $F_1(z)=\frac{f(z)-f(z_0)}{z-z_0}$ .

3. Prove that $F_1(z)$ is a holomorphic function for any point $z$ inside $C$ ( $z_0$ is also inside $C$ ). In particular give the expression of $F_1(z)$ . In our language of singularity, $z_0$ is a removable singularity which roughly means that the singularity is fake.

Consider $F_1(z)=\frac{1}{2\pi i(z-z_0)}\int_C\frac{f(\xi)}{\xi-z}-\frac{f(\xi)}{\xi-z_0}\mathrm d\xi=\frac{1}{2\pi i}\int_C\frac{f(\xi)}{(\xi-z)(\xi-z_0)}\mathrm d\xi$ . As $\frac{f(\xi)}{\xi-z}$ is continuous on $C$ , $F_1(z)$ is holomorphic for all $z$ inside $C$ supported by Lemma 1.

4. The above expression allows us to express $f(z)=f(z_0)+(z-z_0)F_1(z)$ . Again let $F_2(z)=\frac{F_1(z)-F_1(z_0)}{z-z_0}$ , this allows to expand $f(z)$ to the second order. Prove that $F_2(z)$ is holomorphic and give its expression.

Consider $F_2(z)=\frac{1}{2\pi i(z-z_0)}\int_C\frac{f(\xi)}{(\xi-z)(\xi-z_0)}-\frac{f(\xi)}{(\xi-z_0)^2}\mathrm d\xi=\frac{1}{2\pi i}\int_C\frac{f(\xi)}{(\xi-z)(\xi-z_0)^2}\mathrm d\xi$ . As $\frac{f(\xi)}{\xi-z}$ is continuous on $C$ , $F_2(z)$ is holomorphic for all $z$ inside $C$ supported by Lemma 1.

5. Continue the above process by letting $F_n(z) =\frac{F_{n−1}(z)−F_{n−1}(z_0)}{z−z_0}$ . Marching the expression you obtained with (4) you should be able to expression all the coefficients $a_j,j=0,1,\dots,n−1$ . Give the expression of $h(z)$ , and prove that it is holomorphic.

Assume that when $n=k$ , $F_n(z)=\frac{1}{2\pi i}\int_C\frac{f(\xi)}{(\xi-z)(\xi-z_0)^n}\mathrm d\xi$ . Now let $n=k+1$ , we have $F_n(z)=\frac{F_{n-1}(z)-F_{n-1}(z_0)}{z-z_0}=\frac{1}{2\pi i(z-z_0)}\int_C\frac{f(\xi)}{(\xi-z)(\xi-z_0)^k}-\frac{f(\xi)}{(\xi-z_0)^{k+1}}\mathrm d\xi=\frac{1}{2\pi i}\int_C\frac{f(\xi)}{(\xi-z)(\xi-z_0)^n}\mathrm d\xi$ . The situation $n=1$ is proved before, so for all $n$ , $F_n(z)=\frac{1}{2\pi i}\int_C\frac{f(\xi)}{(\xi-z)(\xi-z_0)^n}\mathrm d\xi$ . Notice that $f(z)=(z-z_0)F_1(z)+f(z_0)=(z-z_0)^2F_2(z)+(z-z_0)F_1(z_0)+f(z_0)=\dots=f(z_0)+\sum_{j=1}^{n-1}F_j(z_0)(z-z_0)^j+F_n(z)(z-z_0)^n$ . Compare with (4), we have $a_j=F_j(z_0)=\frac{1}{2\pi i}\int_C\frac{f(\xi)}{(\xi-z_0)^{j+1}}\mathrm d\xi,h(z)=F_n(z)=\frac{1}{2\pi i}\int_C\frac{f(\xi)}{(\xi-z)(\xi-z_0)^n}\mathrm d\xi$ . As $\frac{f(\xi)}{\xi-z}$ is continuous on $C$ , $h(z)$ is holomorphic for all $z$ inside $C$ supported by Lemma 1.

Analytic functions

6. Let $f(z) = z^a$ . If $a = 1/2$ , prove that this function is not a single-valued function for $z\ne0$ (hint: let $z = re^{iθ}, θ = θ_0+2n\pi, n \in \mathbb Z$ ). Provide a domain of definition on which $f(z)$ is a single-valued function. If $a\ne0$ is an irrational real number, provide a domain of definition on which $f(z)$ is a single-valued function.

Consider $f(z)=z^{\frac{1}{2}}=\sqrt re^{i(\frac{\theta_0}{2}+n\pi)}=\pm\sqrt re^{i\frac{\theta_0}{2}}$ , which means $f$ is not single-valued for $z\neq0$ . Restrict the $\theta\in[0,2\pi)$ , $f(z)$ is single-valued. When $a$ is an irrational real number, $f(z)=r^ae^{ia(\theta_0+2n\pi)}$ . So restrict $\theta\in[0,2\pi)$ , we can also let $f$ single-valued.

7. Consider a first-order ODE in the form $w' +P(z)w = 0$ , where $z$ is the independent variable. Prove the following theorem.

Theorem: $w$ has a solution of the form $w = z^ah(z)$ where $h(z)$ is a holomorphic function having a power series expansion around $z = 0$ , $h(z) = 1+c_1z +c_2z^2+\cdots$ , with a non-zero radius of convergence $r$ , if and only if, $P(z)$ is holomorphic on $0 <|z| < r$ with a first-order pole at $z = 0$ .

$P\Rightarrow Q$ : Assume that $a\neq0$ , which is essential for proof. Consider around $z=0$ , let $c_0=1$ , the equation can be turned into $z^ah'(z)+az^{a-1}h(z)+P(z)z^ah(z)=0$ , which means $P(z)=-\frac{zh'(z)+ah(z)}{zh(z)}$ . Notice that $\frac{1}{P(0)}=-\frac{0}{a}=0,(\frac{1}{P(z)})'|_{z=0}=-\frac{P'(0)}{P^2(0)}=\infty$ , which means $P(z)$ has a first-order pole at $z=0$ . As $h(z)$ is holomorphic, $P(z)$ is also holomorphic on $0<|z|<r$ .
$Q\Rightarrow P$ : $P(z)$ is holomorphic on $0 <|z| < r$ with a first-order pole at $z = 0$ means $P(z)$ has Laurent series $\frac{c}{z}+Q(z)$ around $z=0$ , and $Q(z)$ is holomorphic on $z=0$ . Consider $w(z)=c\frac{h(z)}{z}$ , the equation turns into $\frac{ch'(z)z-ch(z)}{z^2}+(\frac{c}{z}+Q(z))\frac{h(z)}{z}=0\Rightarrow ch'(z)+Q(z)h(z)=0\Rightarrow h(z)=\exp(-\frac{\int Q(z)\mathrm dz}{c})$ . As $Q(z)$ is holomorphic, $h(z)$ is also holomorphic. Consider the integral constant, we can always get a $h(z)$ satisfied $h(0)=1$ , which means $h(z)$ have a power series expansion around $z = 0$ , $h(z) = 1+c_1z +c_2z^2+\cdots$ , with a non-zero radius of convergence $r$ .

Power series and Laurent series

8. Give the power series or Laurent series expansion around $0$ of the function $\frac{1}{z −a}$ , for $|z|<|a|$ and $|z| > |a|$ , where $a$ is a non-zero complex number.

When $|z|<|a|$ , $\frac{1}{z-a}=-\frac{1}{a}\frac{1}{1-\frac{z}{a}}=-\frac{1}{a}\sum_{n=0}^\infty(\frac{z}{a})^n=\sum_{n=0}^\infty\frac{-1}{a^{n+1}}z^n$ .
When $|z|>|a|$ , $\frac{1}{z-a}=\frac{1}{z}\frac{1}{1-\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^\infty(\frac{a}{z})^n=\sum_{n=1}^\infty a^{n-1}z^{-n}$ .

9. Give the power series or Laurent series expansion around $2$ of the function $\frac{1}{z^2 −4}$ , for $|z - 2| < 4$ and $|z -2| > 4$ (hint: make a substitution $w = z -2$ ).

Let $w=z-2$ , $\frac{1}{z^2-4}=\frac{1}{w(w+4)}=\frac{1}{4}\frac{1}{w}-\frac{1}{4}\frac{1}{w+4}$
When $|w|<4$ , $\frac{1}{z^2-4}=\frac{1}{4}\frac{1}{w}-\frac{1}{16}\frac{1}{1+\frac{w}{4}}=\frac{1}{4}\frac{1}{w}-\frac{1}{16}\sum_{n=0}^\infty(-\frac{w}{4})^n=\frac{1}{4}(z-2)^{-1}-\sum_{n=0}^\infty\frac{(-1)^n}{4^{n+2}}(z-2)^n$ .
When $|w|>4$ , $\frac{1}{z^2-4}=\frac{1}{4}\frac{1}{w}-\frac{1}{4w}\frac{1}{1+\frac{4}{w}}=\frac{1}{4}\frac{1}{w}-\frac{1}{4w}\sum_{n=0}^\infty(-\frac{4}{w})^n=\sum_{n=2}^\infty(-4)^{n-2}(z-2)^{-n}$ .

Integration using complex integration

10. Compute the integral $\int_{-\infty}^\infty\frac{1}{1+x^4}\mathrm dx$ .

Hint: this integral is understood as $\int_{-\infty}^\infty\frac{1}{1+x^4}\mathrm dx=\lim_{R\rightarrow\infty}\int_{-R}^R\frac{1}{1+x^4}\mathrm dx$ . To compute it, we use a closed path $\gamma$ in the complex plane by connecting the line from $-R$ to $R$ on the real axis with a semi-circle in the upper complex plane (that can be denoted as $C_R$ ). You should be able to prove $\lim_{R\rightarrow\infty}\int_{C_R}\frac{1}{1+z^4}\mathrm dz=0$ , and compute $\int_\gamma\frac{1}{1+z^4}\mathrm dz$ .

Let $M(R)=\max_{z\in C_R}|\frac{1}{1+z^4}|=\frac{1}{1+R^4}$ , we have $|\int_{C_R}\frac{1}{1+z^4}\mathrm dz|=|\int_0^\pi\frac{1}{1+R^4e^{4i\theta}}Rie^{i\theta}\mathrm d\theta|\leq \frac{R\pi}{1+R^4}$ . Consider $R\rightarrow\infty$ , we get $\lim_{R\rightarrow\infty}\int_{C_R}\frac{1}{1+z^4}\mathrm dz=0$ . Consider $R>1$ . According to the residue theorem, $\int_\gamma\frac{1}{1+z^4}\mathrm dz=\int_{C_R}\frac{1}{1+z^4}\mathrm dz+\int_{-R}^R\frac{1}{1+x^4}\mathrm dx=2\pi i\sum_{k=0}^1\operatorname{Res}(\frac{1}{1+z^4},e^{i\frac{1+2k}{4}\pi})$ . By calculation, $\operatorname{Res}(\frac{1}{1+z^4},e^{i\frac{1+2k}{4}\pi})=\lim_{z\rightarrow e^{i\frac{1+2k}{4}\pi}}\frac{(z-e^{i\frac{1+2k}{4}\pi})}{1+z^4}=\frac{1}{4}e^{i\frac{5+2k}{4}\pi}$ . Therefore $\int_\gamma\frac{1}{1+z^4}\mathrm dz=\frac{\pi}{2} i(e^{i\frac{5}{4}\pi}+e^{i\frac{5+2}{4}\pi})=\frac{\sqrt2\pi}{2}$ .
Combine all formulas, $\int_{-\infty}^\infty\frac{1}{1+x^4}\mathrm dx=\lim_{R\rightarrow\infty}\int_{\gamma}\frac{1}{1+z^4}\mathrm dz=\frac{\sqrt2}{2}\pi$ .