# References/Analysis (cont.d)

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Math 448 Complex Variables, Wu (Fall 2017)

• Read proofs of Theorems 1, 2, and 3 in Section 2.1 [Fisher1999].

An alternative proof of the second half of Theorem 2.

Theorem 2: Let $f = u + iv$ be an analytic function on a domain $D$. If $u$ or $|f|$ is constant, then $f$ is constant on the domain $D$.

Sketch of Proof. We shall provide another proof for the case where $|f|$ is constant on $D$. Since $|f| = \sqrt{u^2 + v^2}$ is constant. Let $|f| = R$. If $R = 0$, we fall back to the first part of Theorem 2 since $R = 0$ implies $u = v = 0$.

Now assume $R > 0$. We have $u^2 + v^2 = R^2$. Take partials w.r.t. $x$ and $y$ respectively, then we have

\begin{align*} \tag{*} \label{eqn:eq1_thm2} 2 \frac{\pdv u}{\pdv x} u + 2 \frac{\pdv v}{\pdv x} v &= 0, \\ 2 \frac{\pdv u}{\pdv y} u + 2 \frac{\pdv v}{\pdv y} v &= 0. \end{align*}

By analyticity of $f$, apply Cauchy-Riemann equations, $u_x = v_y$ and $u_y = -v_x$ to \eqref{eqn:eq1_thm2}. Further rewrite the two equations of \eqref{eqn:eq1_thm2} in matrix form,

$\tag{**} \label{eqn:eq2_thm2} \left( \begin{array}{rr} u & -v \\ v & u \end{array} \right) \left( \begin{array}{rr} u_x \\ u_y \end{array} \right) = 0.$

However $\det \left( \begin{array}{rr} u & -v \\ v & u \end{array} \right) = R^2 \neq 0$. We must have a unique zero solution to \eqref{eqn:eq2_thm2}, i.e., $u_x = u_y = 0$. So $u$ is constant. Apply the first half of Theorem 2. Then we are done. Q.E.D.

• A remark on proof of Theorem 3.

Theorem 3: Let $f = u + iv$ be a complex valued function on a domain $D$. If $u$, $v$, $u_x$, $v_x$, $u_y$ and $v_y$ exist in a neighborhood of $z = z_0$ and they are all continuous at $z = z_0$, plus $f$ also satisfies Cauchy-Riemann conditions at $z = z_0$, then $f$ is differentiable at $z = z_0$.

Remark. The proof is based on Taylor series of two variables for $u(x, y)$ and $v(x, y)$ at point $z_0 = (x_0, y_0)$. There are two tricks involved in writing down the difference $u(x_0 + \Delta x, y_0 + \Delta y) - u(x_0, y_0)$.

The first trick, add and subtract $u(x_0 + \Delta x, y_0)$, i.e.,

\begin{align*} \tag{*} \label{eqn:eq1_thm3} u(x_0 + \Delta x, y_0 + \Delta y) - u(x_0, y_0) &= u(x_0 + \Delta x, y_0 + \Delta y) \\ & \hspace{4mm} - u(x_0 + \Delta x, y_0) \\ & \hspace{4mm} + u(x_0 + \Delta x, y_0) - u(x_0, y_0) \\ &= u_y(x_0 + \Delta x, \xi) \Delta y + u_x(\zeta, y_0) \Delta x \\ &= u_y(x_0, y_0) \Delta y + u_x(x_0, y_0) \Delta x \\ & \hspace{4mm} + \big(\underbrace{u_y(x_0 + \Delta x, \xi) - u_y(x_0,y_0)}_{:=~ \varepsilon_1 \to 0 \mbox{ as } z \to z_0}\big) \Delta y \\ & \hspace{4mm} + \big(\underbrace{u_x(\zeta, y_0) - u_x(x_0, y_0)}_{:=~\varepsilon_2 \to 0 \mbox{ as } z \to z_0}\big) \Delta x \end{align*}

where $\xi$ is between $y_0$ and $y_0 + \Delta y$, and $\zeta$ is between $x_0$ and $x_0 + \Delta x$. The second equality follows from mean value theorem of a single variable. The third equality follows from Taylor series in two variables and $\varepsilon_1 \to 0$ and $\varepsilon_2 \to 0$ as $z \to z_0$ follow from continuity of $u_y$ and $u_x$ at $z = z_0$ respectively.

By the same token, we can write down an equation similar to \eqref{eqn:eq1_thm3} using $v(x,y)$ involving $\varepsilon_3$ and $\varepsilon_4$.

The second trick was used when we estimated the term

$\frac{\varepsilon_1 \Delta y + \varepsilon_2 \Delta x + \varepsilon_3 \Delta y + \varepsilon_4 \Delta x}{\Delta z},$

we have it goes to $0$ as $z \to z_0$ since each $\varepsilon_i \to 0$ as $z \to z_0$ and $\left | \frac{\Delta y}{\Delta z} \right| \leq 1$ and $\left | \frac{\Delta x}{\Delta z} \right| \leq 1$ are bounded.

• Conversion between Cartesian coordinates and polar coordinates in multivariable calculus, i.e., $dx dy = rdr d\theta$ follows from $dx dy = J dr d\theta$, where

\begin{align*} J &= \left| \begin{array}{rr} \frac{\pdv x}{\pdv r} & \frac{\pdv x}{\pdv \theta} \\ \frac{\pdv y}{\pdv r} & \frac{\pdv y}{\pdv \theta} \end{array}\right| \\ &= \left| \begin{array}{ll} \cos\theta & r(-\sin\theta) \\ \sin\theta & r\cos\theta \end{array} \right| \\ \end{align*}

is the Jacobian between the sets of coordinates $(x,y)$ and $(r, \theta)$.

• Let $f(z)$ be an analytic function on a domain $D$ and $\gamma$ be a simple piecewise smooth closed contour in $D$, then Cauchy’s integral formula says we can recover the function value of any point inside $\gamma$ from investigating the integral over the points on the boundary $\gamma$, specifically,

$\tag{*} \label{eqn:cif} f(z_0) = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z - z_0} dz,$

where $z_0$ is a point inside $\gamma$. We also know that $f(z)$ is infinitely many times differentiable; this is due to the fact that, with some justification, we can interchange integral sign and differentiation on the right hand side of \eqref{eqn:cif} and the kernel $\frac{1}{z - z_0}$ (a function of any point $z_0$ inside $\gamma$) is a very nice function, arbitrarily many times differentiable w.r.t. $z_0$.

The interchange of a derivative and integral, i.e., differentiation under integral sign can be justified by Leibniz rule.

• The closed contour in Morera’s Theorem does not require it be simple. Morera’s Theorem is equivalent to $f$ has antiderivative in domain $D$.
• Rouché’s original theorem says for two analytic functions $f, g$ on a region $D$, if $|f| > |g|$ holds on the boundary $\partial D$, then $f$ and $f+g$ have the same number of roots inside D. Therefore if we consider an analytic function when it comes to its roots in some region, we can focus on the dominant part only. The one presented in [Palka1991] is the stronger version.
• Consider a polynomial $f(z) = a_0 z^n + a_1 z^{n - 1} + \cdots$ . If it has a zero of order $m > 1$, then $f(z) = z^m (b_0 + b_1 z + \cdots )$, where $b_0 \not = 0$. Near $z = 0$, what $f$ does is approximately mapping $z$ to $b_0 z^m$, i.e., rotating $z$ to an argument $m$ times larger following a scaling of modulus.
• For an analytic function $f(z)$ on a disc centered at $z = z_0$, it has Taylor series valid around a , i.e.,

$f(z) = \sum_{n = 0}^{\infty} a_n (z - z_0)^n.$

To prove the uniqueness of each $a_n$, integrate both sides along a closed contour $\gamma$ in the disc, the left hand side is a fixed number. After applying uniform continuity and switch the order of integration and summation, the only integral left on the right hand side is the one with $n = 1$, because

$\int_\gamma \frac{1}{(z - z_0)^n} dz = \left\{ \begin{array}{ll} 0 & \text{if n \not = 1,} \\ 2\pi i & \text{otherwise.} \end{array} \right.$

• Consider the mean value property of analytic function $f$ on a disc $\{z \in \CC : |z - z_0| \leq r\}$, we have

$f(z_0) = \frac{1}{2\pi}\int_0^{2 \pi} f(z_0 + re^{i\theta}) d\theta.$

It does not hold generally for differentiable functions on the real line $\RR^1$. It is a striking result when $\RR^1$ is extended to $\CC = \RR^2$.

• Question: Why when we compute improper integrals involving $\sin(x)$ and $\cos(x)$ functions, we want to consider $e^{iz}$ rather than $\sin(z)$ and $\cos(z)$ in complex variables?

Answer: Unlike in the case of real variables, $\sin(z) = \sin(x)\cosh(y) + i \cos(x)\sinh(y)$ is not bounded in the upper half plane. For example, along the commonly used semicircle in the upper half plane, $z = R e^{i \theta}$, we cannot use ML estimate to claim $| \int_\gamma \sin(z) dz | \leq \pi R \cdot 1$. More specifically, let $z = x + iy$, $x, y \in \RR$

\begin{align*} \sin(z) &= \frac{e^{iz} - e^{-iz}}{2i} \\ &= \frac{e^{-y} e^{ix} - e^{y} e^{-ix}}{2i}, \\ \cos(z) &= \frac{e^{iz} + e^{-iz}}{2} \\ &= \frac{e^{-y} e^{ix} + e^{y} e^{-ix}}{2}. \end{align*}

as $x \to \infty$ or $y \to \infty$, $e^x$ or $e^y$ blows up.

Another reason is, after all, complex sine and cosine functions are combinations of $e^{iz}$, $z \in \CC$.

• A poem was written on the side blackboard before today’s class (Nov 6, 2017):

Someday all of your memories will be lost,
like tears in the rain.
What is vain?

— Unknown

• A note on antiderivatives. Consider the example with two complex valued functions $\frac{1}z$, $\frac{1}{z^2}$ and closed contour $\gamma: |z| = 1$. It is known that $\int_\gamma \frac{1}z dz = 2 \pi i$ and $\int_\gamma \frac{1}{z^2} dz = 0$. The latter case can be explained by the fact that $\frac{1}{z^2}$ has antiderivative on $\CC \setminus \{ 0 \}$ and $\frac{1}{z^2}$ is analytic on the interior of $\gamma$ except the origin, i.e., on a region where a hole might exist, a function analytic on that region along any closed contour inside that region is $0$. This is not the case when it comes to $\frac{1}z$ since it only has antiderivative on $\CC \setminus (-\infty, 0]$, and the point $(-1, 0)$ on $\gamma$ does not belong to the region where its antiderivative exists. That is why the integral of $\frac{1}z$ along $\gamma$ is nonzero.
• (TODO) Theorems require $f$ be analytic
• on the closed contour itself
• Theorem ( ) dummy text
• inside a region that does not have holes (simply connected)
• Taylor expansion.
• inside a region that might have holes
• Theorem (Residue Theorem): Let $f : D\setminus\{z_1, \ldots, z_k\} \to \CC$ be an analytic function and $\gamma : [a, b] \to D\setminus\{z1, \ldots, z_k\}$ a closed piecewise smooth curve. Then

$\int_\gamma f = 2\pi i \sum_{j=1}^k {\rm Res}(f;z_j) \chi(\gamma; z_j).$

• Theorem (Argument Principle): Let $f: D \to \CC$ be an analytic function and $\gamma$ a closed curve whose image does not vanish. Then

$\frac{1}{2\pi i}\int_\gamma \frac{f'}f = \chi(f\circ\gamma; 0).$

Remark: The right hand side is the winding number of $f(\gamma)$ around $0$, in particular, an integer.

• Laurent expansion.
• on the closure of a region
• Find the image of strip $\{ z \in \CC : |\Re(z)|<1 \}$ under map $f(z) = \frac{1}z$.

First note that inversion $\frac{1}z$ is a special form of Möbius transform $\mathcal{M}(z) = \frac{az + b}{cz + d}$ which forms group structure between sets of circles and lines. More specifically, circles that do not touch origin will be mapped to circles (depending on the map, in this case, $1/z$ blows up at origin or $0$ is a pole of the map. See next example), otherwise images of circles collapse into lines, i.e., points within finite distances to origin have images that have finite distances to origin but $0$ will be mapped to infinity hence the collapse. Similarly, lines not passing origin will be mapped to circles. (Points on a line within finite distances have images that are within finite distances to origin but infinity on the line is mapped to $0$ hence all images cannot be arbitrarily far from origin.) Lines touching origin will be mapped to lines.

Figure 1: Strip $\{ z \in \CC : |\Re(z)|<1 \}$

Figure 2: Image of strip $\{ z \in \CC : |\Re(z)|<1 \}$ under $\frac{1}z$ is Region $3$

Since the strip divides the extended complex plane $\hat{\CC}$ into $3$ components, (one line divides the plane into $2$, and one more line further divides one of the two into two.) the image of the strip should be one of the Regions $1$, $2$ or $3$. They are interior of circle in the right half plane, interior of the circle in the left half plane and exterior of the union of those two circles respectively. It turns out to be Region $3$. Indeed, if one walks along the line $\Re(z) = -1$ from negative infinity to positive infinity, the strip stays on the right hand side, correspondingly one walks along its image—the circle in the left half plane anticlockwise, the image should stay on the right hand side.

See also the code for the figures in this example.

#!/usr/bin/env sage

import sys
from sage.all import *

'''
create illustration of map: image of a strip under inversion
-------------------------------------------------------------

by Y\"un Han

2017-11-28
'''
# preimage figure
size       = 4
percnt     = .995
strip_lbnd = line([(-1, -size),(-1, size)], thickness = 2)
strip_rbnd = line([( 1, -size),( 1, size)], thickness = 2)
shd        = polygon([[-1,-size*percnt], [-1,size*percnt],
[ 1, size*percnt], [ 1,-size*percnt]],
color='lightgray')
fig2       = strip_lbnd + strip_rbnd + shd
#fig2.show(xmin=-size, xmax=size, ticks=[1, 1], tick_formatter='latex', aspect_ratio=1)
fig2.save('map_strip_pre.svg',xmin=-size, xmax=size, ticks=[1, 1],
tick_formatter='latex', aspect_ratio=1)
# image figure
circ1      = circle((-.5,0),.5, thickness = 3)
circ2      = circle(( .5,0),.5, thickness = 3)
reg1       = text('$\\rm{Region\ } 1$', ( .5, .25), fontsize=21, color='black')
reg2       = text('$\\rm{Region\ } 2$', (-.5, .25), fontsize=21, color='black')
reg3       = text('$\\rm{Region\ } 3$', ( -1, .75), fontsize=21, color='black')
# upper half
shd_path   = [[1,0],[1.5,0],[1.5,1],[-1.5,1],[-1.5,0],[-1,0]]
shd_path  += [[-.5+.5*cos(pi - pi*_/200),.5*sin(pi - pi*_/200)] for _ in range(200)]
shd_path  += [ [.5+.5*cos(pi - pi*_/200),.5*sin(pi - pi*_/200)] for _ in range(200)]
# use 'sorted(colors)' to see all named colors
shd        = polygon2d(shd_path, color='lightgray')
# lower half
shd_path   = [[1,0],[1.5,0],[1.5,-1],[-1.5,-1],[-1.5,0],[-1,0]]
shd_path  += [[-.5+.5*cos(pi + pi*_/200),.5*sin(pi + pi*_/200)] for _ in range(200)]
shd_path  += [ [.5+.5*cos(pi + pi*_/200),.5*sin(pi + pi*_/200)] for _ in range(200)]
shd       += polygon2d(shd_path, color='lightgray')
fig3       = circ1 + circ2 + shd + reg1 + reg2 + reg3
# save figure
fig3.save('map_strip_img.svg',xmin=-1.5, xmax=1.5, ymin=-1, ymax=1,
ticks=[.5, .5], tick_formatter='latex', aspect_ratio=1)

• A follow-up example to the above. The map $f:= z \mapsto \frac{z+1}{-z+1}$ establishes an isomorphism of the upper semidisc and first quadrant.

The major difference here is that the circular arc is part of a unit circle that does not touch origin but it is not mapped to a circle. It is mapped to a ray in the image in Figure 4. Here in this case, circular arc touches $1$ (the would-be pole of the target transform) which blows up $f(z)$ hence the collapse of the circle into a line.

Figure 3: Upper semidisc

Figure 4: First quadrant as image of upper semidisc under $z \mapsto \frac{z+1}{-z+1}$

• It is known that any comformal map $f : \mathbb{H} \mapsto \mathbb{E}$, where $\mathbb{H}$ is the open upper half plane and $\mathbb{E}$ is open unit disc, is of the form

$z \mapsto e^{i \varphi} \frac{z - \lambda}{z - \bar{\lambda}},$

where $\lambda \in \mathbb{H}$, $\varphi \in \mathbb{R}$. The special case, map $z \mapsto \frac{z - i}{z + i}$ is called Cayley. The following two animated Figures 5 and 6 show how Cayley map carries horizontal lines and vertical lines in $z$-plane to their images in $w$-plane. Figure 5: Horizontal lines and vertical lines in $z$-plane Figure 6: Images of horizontal lines and vertical lines under Cayley map

#!/usr/bin/env sage

import sys
from sage.all import *

import numpy as np

'''
create animated images of horizontal/vertical lines under Cayley map
--------------------------------------------------------------------

by Y\"un Han

2017-12-02

'''
canvasSize =    2
shft       = .025
# in preimage z-plane
# draw background canvas first
shd_path   = [[-canvasSize-shft*2,0],[canvasSize+shft*2,0],\
[canvasSize+shft*2,canvasSize+shft*2],
[-canvasSize-shft*2,canvasSize+shft*2]]
canvas     = polygon2d(shd_path, color='lightgray')

# draw horizontal lines and overlay them
cnt = 0
for i in np.arange(canvasSize, -canvasSize-.25, -.25):
# horizontal lines first
# highlight current line with red
lni_r     = line([(-canvasSize-shft,i), (canvasSize+shft,i)],\
thickness=2.25, color='red')
canvas_r  = canvas + lni_r
seq       = str(cnt+1).zfill(3)
canvas_r.save('zplane' + seq + '.png',
xmin=-2.05,xmax=2.05,ymin=-2.05,ymax=2.05,
ticks=[1, 1], tick_formatter='latex', aspect_ratio=1)
# move forward one line, update canvas
lni       = line([(-canvasSize-shft,i), (canvasSize+shft,i)],\
thickness=1.25)
canvas   += lni
cnt      += 1

for i in np.arange(-canvasSize, canvasSize+.25,.25):
# vertical lines
# highlight current line with red
lni_r     = line([(i,-canvasSize-shft), (i,canvasSize+shft)],\
thickness=2.25, color='red')
canvas_r  = canvas + lni_r
seq       = str(cnt+1).zfill(3)
canvas_r.save('zplane'+ seq +'.png',
xmin=-2.05,xmax=2.05,ymin=-2.05,ymax=2.05,
ticks=[1, 1], tick_formatter='latex', aspect_ratio=1)
lni       = line([(i,-canvasSize-shft), (i,canvasSize+shft)],\
thickness=1.25)
# move forward one line, update canvas
canvas   += lni
cnt      += 1

# in image w-plane
# draw background canvas first
shd_path  = [[cos(2*pi*_/300),sin(2*pi*_/300)] for _ in range(300)]
canvas    = polygon2d(shd_path, color='lightgray')

# draw images of horizontal lines and overlay them
cnt = 0
for i in np.arange(canvasSize, -canvasSize-.25, -.25):
# image of horizontal lines first
if i != -1:
# highlight current image with red
ciri_r    = circle((i/(i+1),0), abs(1/(i+1)), thickness=2.25, color='red')
canvas_r  = canvas + ciri_r
seq       = str(cnt+1).zfill(3)
canvas_r.save('wplane'+ seq +'.png',
xmin=-2.05,xmax=2.05,ymin=-2.05,ymax=2.05,
ticks=[1, 1], tick_formatter='latex', aspect_ratio=1)
# move forward one circle, update canvas
ciri      = circle((i/(i+1),0), abs(1/(i+1)), thickness=1.25)
canvas   += ciri
cnt      += 1
else:
# image of horizontal line y = -1 is a line
lni_r     = line([(1,-canvasSize-shft*5), (1,canvasSize+shft*5)],\
thickness=2.25, color='red')
canvas_r  = canvas + lni_r
seq       = str(cnt+1).zfill(3)
canvas_r.save('wplane'+ seq +'.png',
xmin=-2.05,xmax=2.05,ymin=-2.05,ymax=2.05,
ticks=[1, 1], tick_formatter='latex', aspect_ratio=1)
# move forward one circle, update canvas
lni       = line([(1,-canvasSize-shft*5), (1,canvasSize+shft*5)],\
thickness=1.25)
canvas   += lni
cnt      += 1

y = var('y') # dummy var for parametric plot
for i in np.arange(-canvasSize, canvasSize+.25,.25):
# image of vertical lines
if i != 0:
# image of x = 0 (Im) in z-plane is a line y = 0 (Re) in w-plane
# highlight current image with red
ciri_r    = parametric_plot((real((i+y*I-I)/(i+y*I+I)),
imag((i+y*I-I)/(i+y*I+I))),
(y,-200,200),color='red',
plot_points=5000,thickness=2.25)
canvas_r  = canvas + ciri_r
seq       = str(cnt+1).zfill(3)
canvas_r.save('wplane'+ seq +'.png',
xmin=-2.05,xmax=2.05,ymin=-2.05,ymax=2.05,
ticks=[1, 1], tick_formatter='latex', aspect_ratio=1)
# move forward one circle, update canvas
ciri      = parametric_plot((real((i+y*I-I)/(i+y*I+I)),
imag((i+y*I-I)/(i+y*I+I))),
(y,-200,200),
plot_points=5000,thickness=1.25,
color='blue')
canvas   += ciri
cnt      += 1
else:
# image of x = 0 (Im) in z-plane is a line y = 0 (Re) in w-plane
lni_r     = line([(-canvasSize-shft*5,0), (canvasSize+shft*5,0)],
thickness=2.25, color='red')
canvas_r  = canvas + lni_r
seq       = str(cnt+1).zfill(3)
canvas_r.save('wplane'+ seq +'.png',
xmin=-2.05,xmax=2.05,ymin=-2.05,ymax=2.05,
ticks=[1, 1], tick_formatter='latex', aspect_ratio=1)
# move forward one circle, update canvas
lni       = line([(-canvasSize-shft*5,0), (canvasSize+shft*5,0)],
thickness=1.25,color='blue')
canvas   += lni
cnt      += 1

# generate GIF animated plot
# do this from terminal using imagemagick

• Computing integral involving fractional power. Consider computing

$\int_0^\infty \frac{x^{\frac{1}3}}{x^2 + 4x + 8}\, dx$

using contour integral $\int_{C_\varepsilon + C_R} f$ along a closed path (circular arcs plus slit of full circles) that does not pass singularities of $f(z) = \frac{z^{\frac{1}3}}{z^2 + 4z + 8}$, where $C_\varepsilon, C_R$ are two circles with radii $\varepsilon, R$ respectively. Note without this slit, Residue Theorem (general version or not) does not apply, since this slit contains all the singularities (non isolated) of $z^{\frac{1}3} = e^{\frac{1}3 \log(z)}$ when we choose a branch cut of $\log(z)$ right along this slit.

Next we need to break the contour into two pieces as shown in Figure 7. Note that along the branch cut of $\log(z)$ there is an argument jump from $2\pi$ to $0$, therefore integration on each piece along this slit does not cancel, which is in contrast to the situation of integration along $\ell_{C_\varepsilon}$ and $\ell_{C_R}$ where integrals cancel since there is no argument jumps in terms of curve parameterizations.

Figure 7: Contour used when integrand involves fractional powers

See sage code to draw the contour.

#!/usr/bin/env sage

import sys
from sage.all import *

'''
create a contour plot for integral involving fractional power
-------------------------------------------------------------

by Y\"un Han

2017-11-14

'''
R = 5
e = 1
# small angle
anR = 0.03
# ane will be calculated based on geometry
# R*sin(anR) = e*sin(ane)
ane = arcsin(R*sin(anR)/e)
# pure trial and error to disguise bezier curve along which arrow resides
ane2 = ane/4
# where cut is
ancut = pi*5/4

# gxxx stands for graphics
# outer circular arc
gCR  = arc((0,0), R, angle=0, sector=(anR, 2*pi - anR), thickness=3)
gCR += text('$C_R$' , (  1.5, 5.10), fontsize=16, color='black')
gCR += text('$\\Im$', (-0.25,  5.5), fontsize=16, color='black')
gCR += text('$\\Re$', (  5.5, 0.25), fontsize=16, color='black')
# add arrow along path using bezier: path=[[start, ctrl_pt1, ctrl_pt2, end]]
gCR += arrow2d(path=[[(R*cos(pi/4-2*ane2),R*sin(pi/4-2*ane2)),
(R/cos(ane2)*cos(pi/4-ane2),R/cos(ane2)*sin(pi/4-ane2)),
(R/cos(ane2)*cos(pi/4-ane2),R/cos(ane2)*sin(pi/4-ane2)),
(R*cos(pi/4),R*sin(pi/4))]],arrowsize=4.5)

# inner circular arc
gCe  = arc((0,0), e, angle=0, sector=(ane, 2*pi - ane), thickness=3)
gCe += text('$C_\\varepsilon$', (1,.95), fontsize=16, color='black')
gCe += text('$O$', (-.35,-.35), fontsize=16, color='black')
# add arrow along path using bezier
anbase = pi*3/8  # corrections since arrow head too big on a small circle
gCe += arrow2d(path=[[(e*cos(anbase-2*ane2),-e*sin(anbase-2*ane2)),
(e/cos(ane2)*cos(anbase-ane2),-e/cos(ane2)*sin(anbase-ane2)),
(e/cos(ane2)*cos(anbase-ane2),-e/cos(ane2)*sin(anbase-ane2)),
(e*cos(anbase),-e*sin(anbase))]],arrowsize=4.5)

# join inner and outer circular arcs with lines
gln1 = line([(e*cos(ane),  e*sin(ane)), (R*cos(anR),  R*sin(anR))], thickness=3)
gln2 = line([(e*cos(ane), -e*sin(ane)), (R*cos(anR), -R*sin(anR))], thickness=3)

# draw cut which divdes contour into two pieces
gcut  = line([(e*cos(ancut), e*sin(ancut)), (R*cos(ancut), R*sin(ancut))], linestyle='--')
gcut += text('$\\ell_{C_R}$', (-2.5, -1.5), fontsize=16, color='black')
gcut += text('$\\ell_{C_\\varepsilon}$', (-1.5, -2.25), fontsize=16, color='black')

# move arrow up/down by shft amount
r1    = 1.15
r2    = 4.85
shft  = 0.2
gcut += arrow((R*cos(ancut), R*sin(ancut)+shft), (r1*cos(ancut), r1*sin(ancut)+shft),
arrowsize=3, width=2.5)
gcut += arrow((e*cos(ancut), e*sin(ancut)-shft), (r2*cos(ancut), r2*sin(ancut)-shft),
arrowsize=3, width=2.5)

fig = gCR + gCe + gln1 + gln2 + gcut
# show figure
# show(gCR + gCe + gln1 + gln2, axes=False)
# selective ticks (none) and latex format
#fig.show(ticks=[[],[]], tick_formatter='latex')

# save figure as vector graphics
fig.save('path_int_frac_pow.svg',ticks=[[],[]])


Math 542 Complex Analysis (Fall 2017)

• Infinitesimally, given a differentiable function $f(z)$ differentiation keeps shape intact, i.e., a circle around $z_0$ in the domain of $f(z)$ is mapped to a circle around $f(z_0)$. Indeed, consider the principal part of the linearization of $f(z)$ at $z = z_0$, we have

$f(z) \approx f(z_0) + f'(z_0) (z - z_0).$

It says the value of $f(z)$ around $z = z_0$ can be obtained by three successive operations on vector $z - z_0$:

• rotation by $\Arg (f'(z_0))$;
• dilation further by $|f'(z_0)|$;
• finally translation by $f(z_0)$.
• Convergence tests: Dirichlet and Abel.
• Dirichlet’s test: Given a sequence $\{ a_n \}$ in $\RR$ and a sequence $\{ b_n \}$ in $\CC$ with

• $a_n \geq a_{n+1}$;
• $\lim_{n \to \infty} a_n = 0$;
• $\left| \sum_{n=1}^N b_n \right| \leq M$ for all $N \in \ZZ_{>0}$, where $M$ is a positive constant.

Then $\sum_{n=1}^\infty a_n b_n$ converges. (not necessarily absolutely converges.)

Remark. It reads, a non-increasing real sequence with limit $0$ and a complex sequence with all partial sums bounded, their “inner product” converges.

• Abel’s test:
• The real case. The real case looks very similar to the Dirichlet test above.
• The complex case. Test the convergence of series on the boundary of unit disc except one point $z = 1$.
• An interesting point about the integral $\int_0^\infty e^{-t^2} dt = \frac{\sqrt{\pi}}{2}$. To compute it, we square it first therefore go from one dimension to two dimensions. Then convert the squared version of the integral to double integral using polar coordinates

\begin{align*} \left( \int_0^\infty e^{-t^2} dt \right)^2 &= \int_0^\infty \int_0^\infty e^{-t^2-s^2} dtds \\ &= \int_0^{\frac{\pi}{2}} \int_0^\infty e^{-r^2} rdrd\theta. \end{align*}

For higher dimensions in $\RR^d$, $\int_{\RR^d} e^{-|x|^2} dx_1 dx_2 \ldots dx_n = (\sqrt{\pi})^d$. It is interesting (according to Prof Erdoğan) because we use 2D computation to get the 1D value first then generalize it to $d$D case.

• Cantor’s intersection theorem. For a decreasing sequence of nested nonempty compact subsets of a set, their intersection is a nonempty set. The Cantor’s theorem in the hint of [Palka1991] VII. 5.7 refers to this theorem, not the theorem which states the cardinality of a set is less than or equal to the cardinality of its power set.
• Lambert series $\sum_{n=1}^\infty d(n) z^n$ is the limit function of normally convergent series $\sum_{n=1}^\infty \frac{z^n}{1 - z^n}$ on unit disk $B(0,1)$, where $d(n)$ is the number of divisors function of $n$. The analyticity of Lambert series is guaranteed by theorem of doubly series of Weierstrass since one can write

$\sum_{n=1}^\infty \frac{z^n}{1 - z^n} = \sum_{n=1}^\infty z^n \sum_{k=0}^\infty z^{kn}$

when $|z^n|<1$ for $z \in B(0,1)$.

As for the coefficient $d(n)$ in Lambert series, its interpretation is as follows: for each $n \in \NN$, all the $z^k$ terms for $k < n$ can be collected with $z^n$ if $k \mid n$. Further for $k > n$, there will not be any $z^n$ terms in the rest of the series. Hence the number of $z^n$ terms is $\sum_{k\, \mid \, n} 1 = d(n)$.

• The set of zeros of an analytic function that is non-constant on a domain $D$ must not have any accumulation point in $D$. (Otherwise all coefficients of its power series that is valid on $D$ would vanish, i.e., the analytic function is identically $0$ on $D$.)
• The concept of homology is to associate algebraic objects with topological spaces. Originally it was used to identify holes.
• Classification of singularities of a complex function. Unlike the real case, they can be completely characterized. There could be three cases.
• A removable singularity: $z = z_0$ is a removable singularity of $f(z)$ on $U$ if there is analytic function $g(z)$ which coincides with $f(z)$ on $U \setminus \{z_0\}$. This is equivalent to
• $f(z)$ admits some neighborhood of $z_0$ where $f(z)$ is bounded, or
• $\lim_{z \to z_0} (z - z_0)f(z) = 0$.
• A pole: There is a positive integer $m \in \ZZ_{>0}$ such that $\lim_{z \to z_0} (z - z_0)^{m+1} f(z) = 0$. $m$ is called the order of the pole. When $m = 1$, $z = z_0$ is a simple pole. A removable singularit is a pole of order $0$. $f(z)$ blows up uniformly around $z_0$.
• An essential singularity: None of the above two. Picard’s theorem says that in this case, $f(z)$ maps every deleted neighborhood of $z = z_0$ to the entire complex plane with the possibility of missing at most one point.
• The image of a connected set under continuous functions is connected. The image of a compact set under continuous functions is compact. A continuous function on a compact set is uniformly continuous.
• Zhukovsky transform $J(z) = \frac{1}2\left(z + \frac{1}z\right)$. Originally, it was invented to study the design of airfoil. Further details can be found here.

For example, it is a map that maps unit circle to line segment $[-2, 2]$; and circles with radii $\not= 1$ to ellipses. See Example 1.6, page 386 of [Palka1991].

For a circle through $-1$ and with $+1$ as its interior point, Zhukovsky map will carry the circle into a profile of an airfoil. The original streamlines of the profile of a circle will be mapped to streamlines of the profile of an airfoil. See mathematica illustration on StackExchange here. Also see Example 8, page 275 of [Fisher1999].

Harvard Math 213a Complex Analysis I (Fall 2017)

• Quick introduction to differential forms.

Emacs 26.1 (Org mode 9.1.13)