Homework 03

1. Conformal map

Consider the half-infinite strip

\[ S = \{z \,|\, \text{Re}\, z > 0, 2i < \text{Im}\, z < 5i \} \]

Find an invertible conformal map sending $S$ to the upper half plane

\[ H = \{z \,|\, \text{Im}\, z > 0 \}, \]

we can proceed in steps using standard conformal mappings.

Translate the Strip: First, we translate the strip downwards $T(z) = z - 2i$ so that its imaginary boundaries are on the real axis and at $3i$.
Scale the Strip: Next, we scale the strip so that its width becomes $\pi$. Define the scaling map $D(z) = \frac{\pi}{3}z$.
Apply the Exponential Function: The exponential function $E(z)=e^z$ maps horizontal strips to $\{z \,|\,\text{Im}\, z > 0, |z| > 1 \}$
Map to the Upper Half-Plane: $R(z)=\frac{1}{2} (z + 1/z)$ will map to the upper half-plane.

So, the complete conformal map $F$ from $S$ to $H$ is the composition of these maps:

\[ F(z) = R(E(D(T(z)))) = \sqrt{e^{\frac{\pi}{3}(z - 3.5i)}}. \]

This map is invertible and conformal.

Note that the inverse map of $R(z)$

\[ z = w+\sqrt{w^2 − 1} \]

has a branch cut at $w ∈ (−1, 1)$. However, for $|z| > 1$, we have $\text{Im}\, w > 0$, so the maps are inevitable.

2. Saddle point

Prove that if $f = u + iv$ is holomorphic at $z = 0$ and $f'(z)$ has a zero of degree 1 at $z = 0$, that both $u$ and $v$ have saddle points at $z = 0$.

At $z=0$, $f^{\prime}=u_x+i v_x=0$ \[ \Rightarrow u_x(0)=0 \text { and } v_x(0)=0 \]

$f$ is holomorphic

\[ \begin{array}{l} v_y=u_x=0 \\ u_y=-v_x=0 \end{array} \]

So the Hessian determinant is

\[ \begin{aligned} D_u=u_{x x} u_{y y}-\left(u_{x y}\right)^2 & =-v_{y x} v_{x y}-u_{x y}^2 \\ & =-v_{x y}^2-u_{x y}^2 \\ D_v=v_{x x} v_{y y}-\left(v_{x y}\right)^2= & -v_{x y}^2-u_{x y}^2 \end{aligned} \]

Since $f^{\prime}$ has a zero of degree $1 \Rightarrow$ second derivative of $u$ and $v$ are nonzero at $z=0$, which gives $D_u<0$ and $D_\nu<0$. Given first derivative is zero, $u$ and $v$ have saddle points at $z=0$

3. Holomorphic functions agree

Show that if two holomorphic functions agree on an interval of the real line, they agree everywhere.

Let’s say two holomorphic function $f$ and $g$ agree on Interval $I$. we show $h=f-g$ if $h \equiv 0$ on $I$, then $h$ is 0 everywhere.

\[ h(z)=\sum_{n=0}^{\infty} \frac{(z-c)^n}{n !} h^{(n)}(c) \]

for $c \in I$. On the real line, $h(x)$ and all its derivatives with respect to $x$ vanish. So $h^{(n)}(c) =0$ for all $n\ge0$. And because $h$ is an entire function the radius of convergence should be infinite. Therefore, for any $z ∈ C$ lies within the circle of convergence, we have $h(z) = 0$.

5. Mobius transformations

Show that Mobius transformations send circles and lines to circles and lines.

Mobius transformations

\[ f(z) = \frac{a z+b}{c z+d} = \frac{a}{c}+\frac{e}{z+\frac{d}{c}}, \]

can be decomposed into four simple transformation of translation, dilation, and inversion.

\[ f=f_4 \circ f_3 \circ f_2 \circ f_1 . \]

Since translation, dilation perserve geometrical lines and circles, we only need to show that inversion $I(z)=1/z$ sends circles and lines to circles and lines.

\[ I(z) = I(x+i y)=\frac{1}{x+i y}=\frac{x}{x^2+y^2}-\frac{y i}{x^2+y^2} \]

So $I$ maps $(x, y)$ into a $(u, v)$ with

\[ u=\frac{x}{x^2+y^2} \text { and } v=\frac{-y}{x^2+y^2} \]

For line of general form $Ax+By=C$, we have

\[ \begin{aligned} Au - B v= (u^2+v^2)C \end{aligned} \]

Thus $I$ maps a line to a circle ($C \neq 0$) or a line ($C = 0$).

For cicrcle of general form $D x+E y+F\left(x^2+y^2\right)=R$, we have

\[ D u -E v + F = R\left(u^2+v^2\right) \]

Thus $I$ maps a circle to a circle ($R \neq 0$) or a line ($R = 0$). Note here, $R$ is not the radius of the original circle.

6. cross-ratio under simultaneous Mobius transformations

Show that for any three points $z_1, z_2, z_3$, there is precisely one Mobius transformation sending $z_1$ to 0, $z_2$ to 1, and $z_3$ to infinity. The image of a fourth point $z_4$ under this map defines the “cross-ratio” of $\left(z_1, z_2, z_3, z_4\right)$. Show that the cross ratio is preserved under simultaneous Mobius transformations of these four points.

Let Möbius transformation $f(z) = (a z+b) /(c z+d)$, satisfying

\[ f\left(z_1\right)=0, f\left(z_2\right)=1, f\left(z_3\right)=\infty \]

Then the Möbius transformation is determined by

\[ \begin{aligned} & f\left(z_1\right)=0 \Rightarrow a z_1 + b=0 \\ & f\left(z_2\right)=1 \Rightarrow a z_2 +b - c z_2 - d =0 \\ & f\left(z_3\right)=\infty \Rightarrow c z_3 + d =0 \end{aligned} \]

The three linear equations can be solved in the sense of their relative ratio.

And the Möbius transformation can be written as

\[ f(z)=\frac{z_2-z_3}{z_2-z_1} \frac{z-z_1}{z-z_3} \]

So the cross ratio is

\[ \frac{z_2-z_3}{z_2-z_1} \frac{z_4-z_1}{z_4-z_3} \]

Then the cross ratio of the image under the transformation of any $f$ is

\[ \frac{f(z_2)-f(z_3)}{f(z_2)-f(z_1)} \frac{f(z_4)-f(z_1)}{f(z_4)-f(z_3)} \]

Note that

\[ f(x)-f(y)=\frac{a x+b}{c x+d}-\frac{a y+b}{c y+d}=\frac{(a d-b c)(x-y)}{(c x+d)(c y+d)} \]

and

\[ \frac{f(x)-f(y)}{f(x)-f(z)} = \frac{(x-y)(cz+d)}{(x-z)(cy+d)} \]

\[ \frac{f(z_2)-f(z_3)}{f(z_2)-f(z_1)} \frac{f(z_4)-f(z_1)}{f(z_4)-f(z_3)} = \frac{(z_2-z_3)(c z_1 +d)}{(z_2-z_1)(c z_3 + d)} \frac{(z_4-z_1)(c z_3 +d)}{(z_4-z_3)(c z_1 + d)} = \frac{z_2-z_3}{z_2-z_1} \frac{z_4-z_1}{z_4-z_3} \]

--- title: Homework 03 format: html: default typst: output-file: phys231_hw03_raw --- ::: {.content-visible when-format="html"} - [Complete PDF](./phys231_hw03_final.pdf) - [Mathematica Notebook](./phys231_hw03.nb) ::: ## 1. Conformal map > Consider the half-infinite strip > > $$ > S = \{z \,|\, \text{Re}\, z > 0, 2i < \text{Im}\, z < 5i \} > $$ > > Find an invertible conformal map sending $S$ to the upper half plane > > $$ > H = \{z \,|\, \text{Im}\, z > 0 \}, > $$ we can proceed in steps using standard conformal mappings. 1. **Translate the Strip**: First, we translate the strip downwards $T(z) = z - 2i$ so that its imaginary boundaries are on the real axis and at $3i$. 2. **Scale the Strip**: Next, we scale the strip so that its width becomes $\pi$. Define the scaling map $D(z) = \frac{\pi}{3}z$. 3. **Apply the Exponential Function**: The exponential function $E(z)=e^z$ maps horizontal strips to $\{z \,|\,\text{Im}\, z > 0, |z| > 1 \}$ 4. **Map to the Upper Half-Plane**: $R(z)=\frac{1}{2} (z + 1/z)$ will map to the upper half-plane. So, the complete conformal map $F$ from $S$ to $H$ is the composition of these maps: $$ F(z) = R(E(D(T(z)))) = \sqrt{e^{\frac{\pi}{3}(z - 3.5i)}}. $$ This map is invertible and conformal. Note that the inverse map of $R(z)$ $$ z = w+\sqrt{w^2 − 1} $$ has a branch cut at $w ∈ (−1, 1)$. However, for $|z| > 1$, we have $\text{Im}\, w > 0$, so the maps are inevitable. ## 2. Saddle point > Prove that if $f = u + iv$ is holomorphic at $z = 0$ and $f'(z)$ has a zero of degree 1 at $z = 0$, that both $u$ and $v$ have saddle points at $z = 0$. At $z=0$, $f^{\prime}=u_x+i v_x=0$ $$ \Rightarrow u_x(0)=0 \text { and } v_x(0)=0 $$ $f$ is holomorphic $$ \begin{array}{l} v_y=u_x=0 \\ u_y=-v_x=0 \end{array} $$ So the Hessian determinant is $$ \begin{aligned} D_u=u_{x x} u_{y y}-\left(u_{x y}\right)^2 & =-v_{y x} v_{x y}-u_{x y}^2 \\ & =-v_{x y}^2-u_{x y}^2 \\ D_v=v_{x x} v_{y y}-\left(v_{x y}\right)^2= & -v_{x y}^2-u_{x y}^2 \end{aligned} $$ Since $f^{\prime}$ has a zero of degree $1 \Rightarrow$ second derivative of $u$ and $v$ are nonzero at $z=0$, which gives $D_u<0$ and $D_\nu<0$. Given first derivative is zero, $u$ and $v$ have saddle points at $z=0$ ## 3. Holomorphic functions agree > Show that if two holomorphic functions agree on an interval of the real line, they agree everywhere. Let's say two holomorphic function $f$ and $g$ agree on Interval $I$. we show $h=f-g$ if $h \equiv 0$ on $I$, then $h$ is 0 everywhere. $$ h(z)=\sum_{n=0}^{\infty} \frac{(z-c)^n}{n !} h^{(n)}(c) $$ for $c \in I$. On the real line, $h(x)$ and all its derivatives with respect to $x$ vanish. So $h^{(n)}(c) =0$ for all $n\ge0$. And because $h$ is an entire function the radius of convergence should be infinite. Therefore, for any $z ∈ C$ lies within the circle of convergence, we have $h(z) = 0$. ## 5. Mobius transformations > Show that Mobius transformations send circles and lines to circles and lines. Mobius transformations $$ f(z) = \frac{a z+b}{c z+d} = \frac{a}{c}+\frac{e}{z+\frac{d}{c}}, $$ can be decomposed into four simple transformation of translation, dilation, and inversion. $$ f=f_4 \circ f_3 \circ f_2 \circ f_1 . $$ Since translation, dilation perserve geometrical lines and circles, we only need to show that inversion $I(z)=1/z$ sends circles and lines to circles and lines. $$ I(z) = I(x+i y)=\frac{1}{x+i y}=\frac{x}{x^2+y^2}-\frac{y i}{x^2+y^2} $$ So $I$ maps $(x, y)$ into a $(u, v)$ with $$ u=\frac{x}{x^2+y^2} \text { and } v=\frac{-y}{x^2+y^2} $$ For line of general form $Ax+By=C$, we have $$ \begin{aligned} Au - B v= (u^2+v^2)C \end{aligned} $$ Thus $I$ maps a line to a circle ($C \neq 0$) or a line ($C = 0$). For cicrcle of general form $D x+E y+F\left(x^2+y^2\right)=R$, we have $$ D u -E v + F = R\left(u^2+v^2\right) $$ Thus $I$ maps a circle to a circle ($R \neq 0$) or a line ($R = 0$). Note here, $R$ is not the radius of the original circle. ## 6. cross-ratio under simultaneous Mobius transformations > Show that for any three points $z_1, z_2, z_3$, there is precisely one Mobius transformation sending $z_1$ to 0, $z_2$ to 1, and $z_3$ to infinity. The image of a fourth point $z_4$ under this map defines the "cross-ratio" of $\left(z_1, z_2, z_3, z_4\right)$. Show that the cross ratio is preserved under simultaneous Mobius transformations of these four points. Let Möbius transformation $f(z) = (a z+b) /(c z+d)$, satisfying $$ f\left(z_1\right)=0, f\left(z_2\right)=1, f\left(z_3\right)=\infty $$ Then the Möbius transformation is determined by $$ \begin{aligned} & f\left(z_1\right)=0 \Rightarrow a z_1 + b=0 \\ & f\left(z_2\right)=1 \Rightarrow a z_2 +b - c z_2 - d =0 \\ & f\left(z_3\right)=\infty \Rightarrow c z_3 + d =0 \end{aligned} $$ The three linear equations can be solved in the sense of their relative ratio. And the Möbius transformation can be written as $$ f(z)=\frac{z_2-z_3}{z_2-z_1} \frac{z-z_1}{z-z_3} $$ So the cross ratio is $$ \frac{z_2-z_3}{z_2-z_1} \frac{z_4-z_1}{z_4-z_3} $$ Then the cross ratio of the image under the transformation of any $f$ is $$ \frac{f(z_2)-f(z_3)}{f(z_2)-f(z_1)} \frac{f(z_4)-f(z_1)}{f(z_4)-f(z_3)} $$ Note that $$ f(x)-f(y)=\frac{a x+b}{c x+d}-\frac{a y+b}{c y+d}=\frac{(a d-b c)(x-y)}{(c x+d)(c y+d)} $$ and $$ \frac{f(x)-f(y)}{f(x)-f(z)} = \frac{(x-y)(cz+d)}{(x-z)(cy+d)} $$ So $$ \frac{f(z_2)-f(z_3)}{f(z_2)-f(z_1)} \frac{f(z_4)-f(z_1)}{f(z_4)-f(z_3)} = \frac{(z_2-z_3)(c z_1 +d)}{(z_2-z_1)(c z_3 + d)} \frac{(z_4-z_1)(c z_3 +d)}{(z_4-z_3)(c z_1 + d)} = \frac{z_2-z_3}{z_2-z_1} \frac{z_4-z_1}{z_4-z_3} $$