Complex Analysis

12. Complex Analysis#

12.1. Complex functions, $f: \mathbb{C} \rightarrow \mathbb{C}$#

A complex function $f$ of complex variable $z = x + i y$, $f: \mathbb{C} \rightarrow \mathbb{C}$, can be written as

\[f(z) = \tilde{u}(z) + i \tilde{v}(z) = u(x,y) + i v(x,y) \ ,\]

as the sum of its real part $u(z)$ and $i$ times its imaginary part $v(x,y)$. Here $x,y \in \mathbb{R}$, while $\tilde{u}(z), \tilde{v}(z): \mathbb{C} \rightarrow \mathbb{R}$ and $u(x,y), v(x,y): \mathbb{R}^2 \rightarrow \mathbb{R}$. With some abuse of notation, tilde won’t be always explicitly written when arguments of real and imaginary parts of $f$ functions won’t be written.

12.1.1. Limit#

\[\lim_{z \rightarrow z_0} f(z) = f(z_0) \qquad , \qquad \forall \varepsilon > 0 \ \exists \delta > 0 \ \text{ s.t. } \ |f(z) - f(z_0)| < \delta \ \forall z \text{ s.t. } |z - z_0| < \varepsilon, \ z \ne z_0 \ .\]

12.1.2. Derivative#

Using the definition of limit of complex functions, the derivative of a function $f: \mathbb{C} \rightarrow \mathbb{C}$, if it exists, is the limit of incremental ratio,

\[f'(z) = \lim_{\Delta z \rightarrow 0} \frac{f(z + \Delta z) - f(z)}{\Delta z} \ .\]

12.1.3. Line Integrals#

Given a line $\gamma \in \mathbb{C}$, whose parametric form is $z(s)$, with regular parametrization with parameter $s \in [s_0, s_1]$,

\[\int_{\gamma} f(z) \, dz = \int_{s=s_0}^{s_1} f(z(s)) \, z'(s) \, ds \ .\]

12.2. Holomorphic Functions - Analytic Functions#

Definition 12.1

A holomorphic function is a function whose derivative exists.

Examples of analytic functions. todo…

12.2.1. Cauchy-Riemann conditions#

For a holomorphic function $f(z) = u(x,y) + i v(x,y)$, Cauchy-Riemann conditions

\[\begin{split}\begin{cases} u_{/x} = v_{/y} \\ u_{/y} = - v_{/x} \end{cases}\end{split}\]

hold. The evaluation of the derivative once with $\Delta z = \Delta x$ and once with $\Delta z = i \Delta y$

\[\begin{split}\begin{aligned} & f'(z) = \lim_{\Delta z \rightarrow 0} \frac{f(z+\Delta z) - f(z)}{\Delta z} = \\ & = \left\{ \begin{aligned} \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x,y) - f(x,y)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{u(x+\Delta x,y) + i v(x+\Delta x,y) - u(x,y) - i v(x,y)}{\Delta x} = u_{/x} + i v_{/x} \\ \lim_{\Delta y \rightarrow 0} \frac{f(x,y+\Delta y) - f(x,y)}{i \Delta y} = \lim_{\Delta y \rightarrow 0} \frac{u(x,y+\Delta y) + i v(x,y+\Delta y) - u(x,y) - i v(x,y)}{i \Delta y} = -i u_{/y} + v_{/y} \\ \end{aligned} \right. \end{aligned}\end{split}\]

provides the proof.

12.2.2. Cauchy Theorem#

For a holomorphic function $f$, $f: \Omega \subseteq \mathbb{C} \rightarrow \mathbb{C}$

\[\oint_{\gamma} f(z) \, dz = 0 \ ,\]

for $\forall \gamma \subset \Omega$. Proof follows from Green’s lemma, and Cauchy-Riemann conditions

\[\begin{split}\begin{aligned} \oint_{\gamma} f(z) dz & = \oint_{\gamma} \left( u(x,y) + i v(x,y) \right) \left( dx + i dy \right) = \\ & = \oint_{\gamma} \left( u dx - v dy \right) + i \oint_{\gamma} \left( u dy + v dx \right) = \\ & = - \int_{S} \left( \underbrace{u_{/y} + v_{/x}}_{=0} \right) \, dx \, dy + i \int_{S} \left( \underbrace{u_{/x} - v_{/y}}_{=0} \right) \, dx \, dy = 0 \ . \end{aligned}\end{split}\]

12.3. Useful integrals#

12.3.1. Independence of line integral for holomorphic functions#

For a function $f(z)$ analytic in $D$, the line integral on paths $\ell_{ab,i}$ with the same extreme points $a$, $b$ contained in $D$ is independent on the path, but only depends on the extreme points $a$, $b$,

\[\int_{\ell_{ab,1}} f(z) \, dz = \int_{\ell_{ab,2}} f(z) \, dz\]

The proof readily follows, using Cauchy theorem applied to a function $f(z): D \subseteq \mathbb{C} \rightarrow \mathbb{C}$, analytic in $D$, and splitting the closed path $\gamma$ into two paths $\ell_1$, $\ell_2$ with the same extreme points, $\gamma = \ell_1 \cup (- \ell_2)$

\[0 = \oint_{\gamma} f(z) \, dz = \int_{\ell_1} f(z) \, dz + \int_{-\ell_2} f(z) \, dz = \int_{\ell_1} f(z) \, dz - \int_{\ell_2} f(z) \, dz \ .\]

12.3.2. Sum and difference of line integrals#

12.3.3. Integral of $z^n$#

Given a path $\gamma$ embracing $z=0$ only once in counter-clockwise direction, and $n \in \mathbb{Z}$

\[\begin{split}\oint_{\gamma} z^n \, dz = \left\{ \begin{aligned} 2 \pi i & \qquad \text{if $n = -1$} \\ 0 & \qquad \text{otherwise} \end{aligned} \right.\end{split}\]

Since $z^n$ is analytic everywhere (todo prove it! Add a section with proofs for common functions) except for $z=0$, it’s possible to evaluate the integral on a circle with center $z=0$ and radius $R$. Using polar expression of the complex numbers on the circle, $z = R e^{i \theta}$, $\theta \in [0, 2 \pi]$, $R$ const, the differential becomes $dz = i R e^{i \theta} d \theta$ and the integral

\[\begin{split}\begin{aligned} \oint_{\gamma} z^n \, dz & = \int_{\theta=0}^{2 \pi} \left( R e^{i\theta}\right)^n i R e^{i \theta} d \theta = \\ & = i \int_{\theta=0}^{2 \pi} R^{n+1} e^{i (n+1) \theta} d \theta = \\ & = \left\{ \begin{aligned} & \text{if $n=-1$} & : & \quad i 2 \pi \\ & \text{otherwise} & : & \quad i R^{n+1} \frac{1}{i(n+1)} \left.e^{i(n+1)\theta}\right|_{\theta=0}^{2\pi} = \frac{R^{n+1}}{n+1} ( 1 - 1 ) = 0 \\ \end{aligned} \right.\\ \end{aligned}\end{split}\]

12.4. Meromorphic functions#

Definition 12.2

A meromorphic function in a domain is a function holomorphic everywhere except for a (finite?) number of poles. check

12.4.1. Singularities#

Definition 12.3 (Pole)

A pole of order $n$ of a function $f(z)$ is a complex number $a$ so that

\[f(z) = \frac{\phi(z)}{(z-a)^n} \ ,\]

with $\phi(z)$ holomorphic in $\phi(a) \ne 0$

Examples. …

Definition 12.4 (Branch)

Examples. $f(z) = z^{\frac{1}{2}}$

Definition 12.5 (Removable singularities)

Example. $f(z) = \frac{\sin z}{z}$

Other irregularities.

12.4.2. Laurent Series#

Given a function $f(z)$, in a disk $D_{a,\varepsilon}: 0 < |z-a| < \varepsilon$, its Laurent series centered in $a$ is the convergent (to $f(z)$, todo which type of convergnence?) series

(12.1)#\[f(z) \sim \sum_{n=-\infty}^{+\infty} a_n (z-a)^n \ ,\]

with

(12.2)#\[a_n = \frac{1}{2 \pi i}\int_{\gamma} f(z) \, (z-a)^{-(n+1)} \, dz\]

and $\gamma$ embracing $z = a$ once counter-clockwise. Proof follows immediately inserting the expressions of the coefficients $a_n$ and using the integral of $z^n$. Evaluating the integral (12.2) of the coefficients of the Laurent series, using (12.1) to replace $f(z)$ with its series

\[\begin{split}\begin{aligned} a_n & = \frac{1}{2 \pi i}\oint_{\gamma} \sum_{m=-\infty}^{+\infty} a_m (z-a)^m (z-a)^{-(n+1)} = \\ & = \frac{1}{2 \pi i} \oint_{\gamma} \sum_{m=-\infty}^{+\infty} a_m (z-a)^{m - n - 1} \, dz = \\ & = \frac{1}{2 \pi i} \oint_{\gamma} a_n \, z^{-1} \, dz = \\ & = a_n \ . \end{aligned}\end{split}\]

todo Some freestyle with function and its convergent series…add some detail, and the meaning of convergence

12.4.3. Cauchy formula#

For an analytic function $f(z)$,

\[f(a) = \frac{1}{2 \pi i} \oint_{\gamma} \frac{f(z)}{z-a} \, dz\]

Proof readily follows using the integral of $z^n$ on the Taylor series of $\frac{f(z)}{z-a}$ whose $0^{th}$ order term reads $f(a)$,

\[\frac{1}{2\pi i} \oint_{\gamma} \frac{f(a)+\sum_{m=1}^{+\infty} f'(a) (z-a)^m}{z-a} \, dz = \frac{1}{2\pi i} \oint_{\gamma} \frac{f(a)}{z-a} \, dz = f(a) \frac{2 \pi i}{2 \pi i} = f(a) \ .\]

12.4.4. Residues#

Definition 12.6 (Residue)

The residue of function $f$ in $a$, $\text{Res}(f,a)$ is a complex number $R$ so that $f(z) - \frac{R}{(z-a)}$ has analytic antiderivative in a disk $D_{a,\varepsilon}: \ 0 < |z-a| < \varepsilon$.

todo Explain this definition. Couldn’t be possible to use $\text{Res}(f,a) = \frac{1}{2 \pi i} \oint_{\gamma} f(z) \, dz = a_{-1}$ instead?

Properties.

If $f(z)$ is analytic in $D_{a,\varepsilon}$ and has a pole of order $n$ in $z = a$, its Laurent series has $a_m=0$ for $m < n$ and reads

(12.3)#\[f(z) = \sum_{m=-n}^{+\infty} a_m (z-a)^m \ ,\]

with $a_{-n} \ne 0$. Since $f(z)$ has a pole of order $n$ in $z = a$, it can be written as

\[f(z) = \frac{\phi(z)}{(z-a)^n} \ ,\]

with $\phi(z)$ analytic in $D_{a,\varepsilon}$ and $\phi(a) \ne 0$. Since $\phi(z)$ is analytic, it has a Taylor series (or a Laurent series with non-negative powers),

\[\phi(z) \sim \sum_{m=0}^{+\infty} b_m (z-a)^m \ ,\]

(todo prove it! Extension of the real case. Add a link to the proof) and thus

\[f(z) \sim \sum_{m=0}^{+\infty} b_m (z-a)^{m-n} = \sum_{m=-n}^{+\infty} b_{m+n} (z-a)^{m} = \sum_{m=-n}^{+\infty} a_{m} (z-a)^m \ , \]

with $a_m = b_{m+n}$.
For simple closed path $\gamma$ (embracing $a$ only once counter-clokwise) in $D_{a, \varepsilon}$,

(12.4)#\[\oint_{\gamma} f(z) \, dz = 2 \pi i a_{-1} = 2 \pi i \text{Res}(f,a)\]

The proof readily follows, using the integral of $z^n$ and Laurent series (12.1) of $f(z)$,

\[\oint_{\gamma} f(z) \, dz = \oint_{\gamma} \sum_{m=-\infty}^{+\infty} a_m (z-a)^m \, dz = 2 \pi i a_{-1} \ .\]
For a pole $a$ of order $n$, the following holds

\[a_{-1} = \frac{1}{(n+1)!} \lim_{z \rightarrow a} \frac{d^{n-1}}{dz^{n-1}} \left[ (z-a)^n \, f(z) \right]\]

The proof follows using Laurent series {eq}`eq:laurent:pole-n} for a function with pole of order $n$, and evaluating the $(n-1)^{th}$ order derivative

\[\begin{split}\begin{aligned} \frac{d^{n-1}}{dz^{n-1}} \left[ (z-a)^n f(z) \right] & = \frac{d^{n-1}}{dz^{n-1}} \left[ (z-a)^n \sum_{m=-n}^{+\infty} a_n (z-a)^m \right] = \\ & = \dfrac{d^{n-1}}{dz^{n-1}} \left[ \sum_{m=-n}^{+\infty} a_n (z-a)^{m+n} \right] = \\ & = \dfrac{d^{n-1}}{dz^{n-1}} \left[ \sum_{m=0}^{+\infty} a_{m-n} (z-a)^{m} \right] = \\ & = \dfrac{d^{n-2}}{dz^{n-2}} \left[ \sum_{m=0}^{+\infty} m a_{m-n} (z-a)^{m-1} \right] = \\ & = \dfrac{d^{n-3}}{dz^{n-3}} \left[ \sum_{m=0}^{+\infty} m(m-1) a_{m-n} (z-a)^{m-2} \right] = \\ & = \dots = \\ & = \left[ \sum_{m=0}^{+\infty} m! \, a_{m-n} (z-a)^{m-n+1} \right] \\ \end{aligned}\end{split}\]

and then letting $z \rightarrow a$, so that only the term with $m-n+1 = 0$ survives

\[\lim_{z \rightarrow a} \frac{d^{n-1}}{dz^{n-1}} \left[ (z-a)^n \sum_{m=-n}^{+\infty} a_n (z-a)^m \right] = (n-1)! \, a_{-1} \ .\]

12.4.5. Residue Theorem#

Theorem 12.1 (Residue Theorem)

Given $f(z)$ with a finite number of poles $p_n \in D$, then

\[\int_{\gamma} f(z) \, dz = 2 \pi i \ \sum_{n} I(\gamma, p_n) \text{Res}(f,p_n) \ ,\]

being $\gamma$ a path in $D$, and $I(\gamma, p_n)$ the winding index of the path $\gamma$ around pole $p_n$ (+1 for each counter-clockwise loop, -1 for each clockwise loop).

The proof readily follows extending the result for a single pole (12.4) to general number of poles and general paths $\gamma$ embracing (with sign) each pole $p_n$ $I(\gamma,p_n)$ times, with the same techinques shown in section Sum and difference of line integrals.

12.4.6. Evaluation of integrals#

12.4.7. Inverse Laplace Transform#

Given Laplace transform

\[F(s) := \mathscr{L}\{f(t)\}(s) := \int_{t=0^-}^{+\infty} f(t) e^{-st} \, dt \ ,\]

the inverse transform can be evaluated as

\[f(t) = \mathscr{L}^{-1}\{F(s)\}(t) := \lim_{T \rightarrow +\infty} \frac{1}{2 \pi i} \int_{s = a-iT}^{a+iT} e^{st} F(s) \, ds \ ,\]

with $a > \text{Re}\{p_n\}$ (todo why?) for each pole of the function $F(s)$, evaluated on the vertical line $s = a+iy$, $y \in [-T,T]$, $ds = i d y$,

\[\begin{split}\begin{aligned} \lim_{T \rightarrow +\infty} \frac{1}{2 \pi i} \int_{s = a-iT}^{a+iT} e^{st} F(s) \, ds & = \lim_{T \rightarrow +\infty} \frac{1}{2 \pi i} \int_{s = a-iT}^{a+iT} e^{st} \int_{\tau=0^-}^{+\infty} f(\tau) e^{-s\tau} \, d \tau \, ds = \\ & = \lim_{T \rightarrow +\infty} \frac{1}{2 \pi i} \int_{y = -T}^{T} e^{(a+iy)t} \int_{\tau=0^-}^{+\infty} f(\tau) e^{-(a+iy)\tau} \, d \tau \, i dy = \\ & = \lim_{T \rightarrow +\infty} \frac{1}{2 \pi} \int_{y = -T}^{T} \int_{\tau=0^-}^{+\infty} e^{iy(t-\tau)} e^{a(t-\tau)} f(\tau) \, d \tau \, dy = \\ & = \dots \\ & = \int_{\tau=0^-}^{+\infty} \delta(t-\tau) e^{a(t-\tau)} f(\tau) d \tau = f(t) \ . \end{aligned}\end{split}\]

having used the transform of Dirac’s delta $\delta(t) = \frac{1}{2\pi} \int_{\omega=-\infty}^{+\infty} e^{-j \omega t} \, d\omega$.

todo Ohter approach: if $a > \text{Re}\{p_n\}$, the contour built with the vertical line with real part $a$ and the arc of circumference on its…