1. 1. What does it mean for a metric space to be sequentially compact? What does it mean for it to be totally bounded?
      Prove that a sequentially compact metric space is totally bounded.
   2. Let $X=𝒞[0,1]$ be the space of continuous real-valued functions on the closed interval $[0,1]$ with the supremum metric $$ d(f, g)=\sup _{x ∈[0,1]}{|f(x)-g(x)|},   \text { for } f, g ∈ X $$ [You do not need to show that this is a metric.]
      1. Let $Y$ be a totally bounded subset of $X$. Show that given $ε>0$ there exists $δ>0$ such that ${|f(x)-f(y)|}<ε$ whenever ${|x-y|}<δ$ and $f ∈ Y$. equicontinuous
      2. Assume instead now that $Y$ is merely a bounded subset of $X$. Does the same conclusion hold? Justify your answer.
   3. Let $Z$ be the space of binary sequences, that is, sequences $\mathbf{x}=\left(x_n\right)_{n=1}^∞$ with $x_n ∈\{0,1\}$ for $n ∈ ℕ$, together with the metric $$ d(\mathbf{x}, \mathbf{y})=\sum_{n=1}^∞ \frac{\left|x_n-y_n\right|}{2^n},   \text { for } \mathbf{x}=\left(x_n\right)_{n=1}^∞, \mathbf{y}=\left(y_n\right)_{n=1}^∞ ∈ Z $$ [You do not need to show that this is a metric.] Prove that $Z$ is sequentially compact.
2. 1. Let $X$ be a metric space. What does it mean for a map $f: X → X$ to be a contraction?
      Prove that a contraction $f: X → X$ can have at most one fixed point. Give an example of a metric space $X$ and contraction $f$ which has no fixed points.
   2. Now let $X$ be $ℝ^n$ endowed with the standard Euclidean metric $d$, and let $$ B=\left\{x ∈ ℝ^n: d(x, 0) ⩽ 1\right\} $$ be the usual closed unit ball in $ℝ^n$.
      1. Let $T: B → B$ be a 1-Lipschitz map, that is to say $$ d(T(x), T(y)) ⩽ d(x, y)   \text { for all } x, y ∈ B $$ Prove that $T$ has a fixed point.
         [Hint: for each positive integer $n$ consider the map $(1-1 / n) T$.]
      2. Give an example to show that this fixed point may be unique, and another to show that there may be many fixed points.
      3. Need the conclusion in (i) hold for $B$ an arbitrary compact set in $ℝ^n$ and $T: B → B$ a 1-Lipschitz map? Justify your answer.
      4. Suppose now that we replace the standard Euclidean metric $d$ throughout by the $d_∞$ metric on $ℝ^n$. Need the conclusion in (i) then hold? And likewise for when we replace $d$ by the discrete metric on $ℝ^n$ ? Briefly justify your answers.
   3. Let $ℕ$ denote the positive integers. Let $d$ be the Euclidean metric on $ℕ$ and define the metric $$ δ(x, y):=\frac{d(x, y)}{x y}   \text { for } x, y ∈ ℕ $$ [You do not need to show that this is a metric.]
      1. Show that every subset of $ℕ$ is open with respect to both metrics $d$ and $δ$.
      2. Prove that $ℕ$ is complete with respect to $d$, but not complete with respect to $δ$.
3. 1. 1. Describe the Taylor expansion at $a ∈ ℂ$ of a holomorphic function $f$ defined on the disk $$ D(a, R)=\{z ∈ ℂ:{|z-a|}<R\} $$ giving the coefficients explicitly in terms of an integral.
      2. Describe the Laurent expansion of a holomorphic function $f$ defined on the annulus $$ A(a, r, R)=\{z ∈ ℂ: r<{|z-a|}<R\} $$
   2. Let $f(z)$ be a holomorphic function on $D(0,1)$ and let $0<r<1$.
      1. Show that there is a sequence $p_n(z)$ of polynomials which converges uniformly to $f(z)$ on $D(0, r)$.
      2. Need there be a sequence of polynomials which converges uniformly to $f(z)$ on $D(0,1)$? Justify your answer.
   3. Consider now holomorphic functions $f(z)$ on $ℂ∖\{0\}$.
      1. Give an example to show that there need not be a sequence $p_n(z)$ of polynomials which converges uniformly to $f(z)$ on $A(0,1,2)$.
      2. Show that there exists a sequence $q_n(z)$ of rational functions, whose only possible singularity is the origin, which converges uniformly to $f(z)$ on $A(0,1,2)$.
      3. Need there exist a sequence $q_n(z)$ of rational functions, whose only possible singularity is the origin, which converges uniformly to $f(z)$ on $D(0,1)∖\{0\}$ ? Justify your answer. [Recall that a rational function is one of the form $u(z) / v(z)$ where $u(z)$ and $v(z)$ are polynomials and $v(z)$ is not identically zero.]
4. 1. 1. Define the holomorphic branch $L(z)$ of $\log z$ on the cut plane $ℂ∖(-∞, 0]$ such that $L(1)=0$.
      2. Determine the real and imaginary parts of $L(z)$ and show directly that they satisfy the Cauchy-Riemann equations.
   2. For $z ∈ ℂ$, the hyperbolic tangent $\tanh z$ is defined by $$ \tanh z=\frac{\sinh z}{\cosh z}=\frac{\exp (2 z)-1}{\exp (2 z)+1} $$
      1. Write down, in terms of $L(z)$, the holomorphic branch of the inverse hyperbolic tangent $\tanh^{-1}z$ in the cut plane $$ X=ℂ∖((-∞,-1] ∪[1, ∞)) $$ and such that $\tanh^{-1}0=0$.
      2. What is the image of $X$ under $\tanh^{-1}z$ ? Justify your answer.
      3. Determine the Taylor expansion of $\tanh^{-1}z$ centred at 0. What is its radius of convergence?
   3. 1. Determine the Laurent expansion of $1 /(\tanh z)$, centred at 0, up to the $z$ term.
      2. Let $C_r$ denote the circle $|z|=r$, positively oriented. Evaluate $$ ∫_{C_3} \frac{\mathrm{d} z}{\tanh z}   \text { and }   ∫_{C_4} \frac{\mathrm{d} z}{\tanh z} $$
5. 1. By using a keyhole contour, or otherwise, calculate $$ ∫_0^∞ \frac{\sqrt[3]{x}}{(1+x)^2} d x $$ Carefully justify all the steps in your calculation.
   2. Let $-1<α<1$. Evaluate $$ ∫_0^{2 π} \frac{\mathrm{d} θ}{1-α \cos θ} $$ Deduce that for non-negative integers $n$ we have $$ ∫_0^{2 π} \cos ^{2 n} θ \mathrm{d} θ=\frac{(2 n) ! π}{2^{2 n-1}(n !)^2} $$ Carefully justify all the steps in your calculation.
6. 1. 1. Prove that every Möbius transformation is a composition of mappings of the form $$ T: z ↦ z+1,   I: z ↦ \frac{1}{z},   D_r: z ↦ r z,   R_θ: z ↦ e^{i θ} z $$ where $r ∈ ℝ_{>0}$ and $θ ∈[0,2 π)$.
      2. Let $$ L:=\{z ∈ ℂ:|z-i| ⩽ \sqrt{2} \text { and }|z+i| ⩽ \sqrt{2}\} . $$ Find a Möbius transformation which maps $L$ bijectively to the first quadrant $$ Q:=\{z ∈ ℂ: \operatorname{Re}(z) ⩾ 0, \operatorname{Im}(z) ⩾ 0\} ∪\{∞\} $$ of the extended complex plane $ℂ_∞=ℂ ∪\{∞\}$, and in addition maps 0 to $e^{i π / 4}$. Write this transformation as a composition of the mappings in part (i).
      3. Determine all Möbius transformations which map $L$ bijectively to $Q$, justifying your answer.
   2. Let $ϕ: ℂ_∞ → ℂ_∞$ be given by $$ ϕ(z):=z+\frac{1}{z} $$ and define $$ U:=\{z ∈ ℂ: 0<|z|<1\},   S^1:=\{z ∈ ℂ:|z|=1\} $$
      1. Prove that $ϕ$ restricts to a two-to-one map from $S^1$ onto the real interval $[-2,2]$. Further show that $ϕ$ maps $U$ bijectively to $ℂ∖[-2,2]$.
      2. By considering the complex function $ψ(z)$ defined implicitly by the equation $$ \frac{ψ(z)-1}{ψ(z)+1}=\left(\frac{z-1}{z+1}\right)^2, $$ or otherwise, write $ϕ(z)$ as a composition of Möbius transformations and the squaring map.
   3. State Riemann's mapping theorem. Let $W$ be an open connected and simply connected proper subset of $ℂ$ and $a ∈ W$. Let $D=\{z:|z|<1\}$ and take $b ∈ D$ and $α ∈ ℝ$. Show that there is a unique bijective conformal map $f$ from $W$ to $D$ such that $f(a)=b$ and $\arg f'(a)=α$.

Solution

1. 1. $X$ is sequentially compact if every sequence in $X$ has a convergent sub-sequence.
      $X$ is totally bounded if $∀ε>0$, $X$ can be covered by finitely many open balls of radius $ε$.
      Suppose $X$ is not totally bounded, so $∃ε$, finitely many balls of radius $ε$ can't cover $X$. Take $x_1∈X$. Suppose ∃ $(x_k)_{k=1}^n$ such that $∀i,j:{|x_i-x_j|}>ε$, take $x_{n+1}∈X∖⋃_{k=1}^nB(x_k,ε)$, then $(x_k)_{k=1}^{n+1}$ satisfy $∀i,j:{|x_i-x_j|}>ε$. In this way we construct a sequence $(x_k)_{k=1}^∞$ in $X$ that has no Cauchy sub-sequence, so $X$ is not sequentially compact.
   2. 1. For $Y$ finite, since $[0,1]$ is compact each $f_i∈Y$ is uniformly continuous, so $∃δ_i>0$ such that $∀x,y∈[0,1],{|x-y|}<δ_i:{|f_i(x)-f_i(y)|}<ε$, then set $δ=\min_iδ_i$.
         In general for $Y$ totally bounded, cover $Y$ by finitely many open balls $B(f_i,ε)$, we proved $∃δ>0$ such that $∀x,y∈[0,1],{|x-y|}<δ:{|f_i(x)-f_i(y)|}<ε$.
         $∀f∈Y,∃i:f∈B(f_i,ε)⇒∀x:{|f(x)-f_i(x)|}<ε$. By triangle inequality $∀x,y∈[0,1],{|x-y|}<δ:{|f(x)-f(y)|}≤{|f(x)-f_i(x)|}+{|f_i(x)-f_i(y)|}+{|f(y)-f_i(y)|}≤3ε$
      2. Counterexample: $f_n=\begin{cases}1&x≥1/n\\nx&x≤1/n\end{cases}$ are continuous and bounded by 1.
         Take $ε=1$. For any $δ>0$, we have $f_n(\fracδ2)-f_n(0)=1$ for $n>\frac2δ$
   3. $Z_0=\{0,1\},Z_1=\left\{0,\frac12\right\},Z_2=\left\{0,\frac14\right\},⋯$ with Euclidean metric are all sequentially compact.
      By Tychonoff’s theorem $Z=Z_0×Z_1×Z_2×⋯$ with product metric\[d((x_1,x_2,…),(y_1,y_2,…))=\left\|\left(d_{Z_1}(x_1,y_{1}),d_{Z_2}(x_2,y_2),…\right)\right\|_1\]is sequentially compact.
2. 1. $∃K∈(0,1):∀x_1,x_2∈X:d_X(f(x_1),f(x_2))≤K⋅d_X(x_1,x_2)$
      Let $x_1≠x_2$ be fixed points\begin{align*}\text{By contraction }d_X(x_1,x_2)&=d_X(f(x_1),f(x_2))\\&≤K⋅d_X(x_1,x_2)\\d_X(x_1,x_2)>0&⇒1≤K\text{, contradiction.}\end{align*}$X=ℝ^+,f(x)=\frac x2$ is a contraction that has no fixed points
   2. 1. For $n∈ℕ$ the map $T_n≔(1-\frac1n)T$ is a contraction and $T_n(B)=B(0,1-\frac1n)⊂B$
         So $T_n:B→B$ has a fixed point $a_n$. Since $B$ is sequentially compact, $(a_n)$ has a convergent subsequence $a_{n_k}→a∈B$\[(1-\frac1{n_k})T(a_{n_k})=a_{n_k}⇒\lim_k T(a_{n_k})=\lim_k a_{n_k}=a\]Since $T$ is continuous, $T(a)=T(\lim_k a_{n_k})=\lim_k T(a_{n_k})=a$
      2. If $T$ is constant, fixed point is unique. If $T=\text{id}_B$ any point of $B$ is a fixed point.
      3. For $[-2,-1]∪[1,2]$, $x↦-x$ has no fixed point.
         If the subset need to be connected, take an annulus and rotate about center.
      4. For $d_∞$ the closed unit ball $B=[-1,1]^n$ is still compact, $T_n(B)⊂B$.
         For discrete metric $B=ℝ^n$ is not compact, the conclusion does not hold for $x↦x+1$
   3. 1. Union of open sets is open, it suffices to prove $\{x\}$ open: $\{x\}=B_d(x,\frac12)=B_δ(x,\frac1{x^2+x})$
         To prove $B_δ(x,\frac1{x^2+x})=\{x\}$, for any $y≠x$\[δ(x,y)=\left|\frac1x-\frac1y\right|≥\left|\frac1x-\frac1{x+1}\right|=\frac1{x^2+x}⇒y∉B_δ\left(x,\frac1{x^2+x}\right)\]
      2. Let $a_n$ be a Cauchy sequence in $(ℕ,d)$.$$∃N:∀n,m≥N:d(a_n,a_m)<1⇒d(a_n,a_N)<1⇒a_n=a_N$$so $a_n→a_N$, so $(ℕ,d)$ is complete. To prove $(ℕ,δ)$ is not complete:
         Consider the sequence $a_n=n$ in $(ℕ,δ)$. We show $a_n$ is Cauchy. $∀ε>0$, let $N>\frac1ε$,$$∀x>y≥N:δ(x,y)=\left|\frac1x-\frac1y\right|<\frac1y≤\frac1N<\frac1ε$$Suppose $a_n→a∈ℕ$, for $n>a,δ(a,n)=\frac1a-\frac1n$ is strictly increasing, contraction.
3. 1. 1. There exist unique $c_n∈ℂ$ such that $f(z)=\sum_{k=0}^∞c_k(z-a)^k$ for all $z∈D(a,R)$\[c_k=\frac1{2πi}\int_{γ(a,ρ)}\frac{f(w)}{(w-a)^{k+1}}\mathrm{d}w 0<ρ<R\]
      2. There exist unique $c_n∈ℂ$ such that $f(z)=\sum_{k=-∞}^∞c_k(z-a)^k$ for all $z∈A(a,r,R)$\[c_k=\frac1{2πi}\int_{γ(a,ρ)}\frac{f(w)}{(w-a)^{k+1}}\mathrm{d}w r<ρ<R\]
   2. 1. By the estimation lemma, take $ρ∈(r,1)$, let $M=\max_{z∈γ(0,ρ)}{|f(z)|}$\[{|c_k|}≤\frac1{2π}\int_{γ(0,ρ)}\left|\frac{f(w)}{w^{k+1}}\right|\mathrm{d}w≤\frac1{2π}⋅2πρ⋅\frac{M}{ρ^{k+1}}=\frac{M}{ρ^k}\]Let $p_n(z)$ be the partial sum $\sum_{k=0}^{n-1}c_kz^k$. For $z∈D(0,r)$\[\left|p_n(z)-f(z)\right|≤\sum_{k=n}^∞{|c_k|}r^k≤\sum_{k=n}^∞M\left(r\overρ\right)^k\text{ converges}\]By Weierstrass M-test, $p_n(z)→f(z)$ uniformly on $D(0,r)$.
      2. $f(z)=\log(1-z)$ is holomorphic on the open disk $D(0,1)$. The idea is that $f(z)$ is unbounded near 1 but $p(z)$ is bounded.
         Suppose $p_n(z)→f$ uniformly on $D(0,1)$, then $∃N∀n>N:\sup_{D(0,1)}{|f-p_n|}≤1⇒f$ is bounded, contradiction.
   3. 1. $f(z)=z^{-1}$ is holomorphic on $ℂ∖\{0\}$. If a sequence of polynomials $p_n(z)→f(z)$ uniformly on $A(0,1,2)$, take $r∈(1,2)$, then $γ(0,r)⊂A(0,1,2)$,\[0=∫_{γ(0,r)}p_n→∫_{γ(0,r)}f=2πi\text{ contradiction.}\]
      2. To prove $q_n(z)=\sum_{k=-n}^nc_kz^k→f(z)$ uniformly on $A(0,1,2)$.
         Like b(i) take $ρ∈(2,∞)$, let $M=\max_{z∈γ(0,ρ)}{|f(z)|}$, \[{|c_k|}≤\frac1{2π}\int_{γ(0,ρ)}\left|\frac{f(w)}{w^{k+1}}\right|\mathrm{d}w≤\frac1{2π}⋅2πρ⋅\frac{M}{ρ^{k+1}}={M\overρ^k}\] \[⇒ \left|\sum_{k=n}^∞c_kz^k\right|≤\sum_{k=n}^∞\frac{M}{ρ^k}{|z|}^k≤\sum_{k=n}^∞M\left(2\over ρ\right)^k\text{ converges} \] By Weierstrass M-test, $\sum_{k=n}^∞c_kz^k→0$ uniformly on ${|z|}≤2$.
         Take $ρ∈(0,1)$, let $M=\max_{z∈γ(0,ρ)}{|f(z)|}$, \[{|c_k|}≤\frac1{2π}\int_{γ(0,ρ)}\left|\frac{f(w)}{w^{k+1}}\right|\mathrm{d}w≤\frac1{2π}⋅2πρ⋅\frac{M}{ρ^{k+1}}={M\overρ^k}\] \[⇒ \left|\sum_{k=-∞}^{-n}c_kz^k\right|≤\sum_{k=-∞}^{-n}{M\overρ^k}{|z|}^k≤\sum_{k=-∞}^{-n}{M\overρ^k}\text{ converges} \]By Weierstrass M-test, $\sum_{k=-∞}^{-n}c_kz^k→0$ uniformly on ${|z|}≥1$.
      3. Suppose $q_n(z)$ converges to $f(z)=\sin\left(\frac1z\right)$ uniformly on $D(0,1)∖\{0\}$.
         Take $0<ε<\frac12$, then $∃N:∀n>N:\sup_{0<{|z|}≤1}{|q_n(z)-f(z)|}<ε$.
         For $z_k=\frac1{kπ},f(z_k)=0$, so $∀n>N:{|q_n(z_k)|}<ε$, let $k→∞$, ${|q_n(0)|}≤ε$
         For $z_k=\frac1{\fracπ2+2kπ},f(z_k)=1$, so $∀n>N:{|q_n(z_k)-1|}<ε$, let $k→∞$, ${|q_n(0)-1|}≤ε$
         But $ε<\frac12$, contradiction.
4. 1. 1. $L(z)=\log{|z|}+i\arg z,\arg z∈(-π,π)$
      2. $u(x,y)=\log{|x+iy|}=\frac12\log(x^2+y^2),v(x,y)=\arg(x+iy)$ \begin{align*} {∂u\over∂x}&=\frac{2x}{2(x^2+y^2)}=\frac x{x^2+y^2}&{∂u\over∂y}&=\frac{2y}{2(x^2+y^2)}=\frac y{x^2+y^2}\\ {∂v\over∂x}&=\frac{-y/x^2}{1+y^2/x^2}=-\frac y{x^2+y^2}&{∂v\over∂y}&=\frac{1/x}{1+y^2/x^2}=\frac x{x^2+y^2} \end{align*}They satisfy ${∂u\over∂x}={∂v\over∂y},{∂u\over∂y}=-{∂v\over∂x}$
   2. 1. $w=\tanh^{-1} z⇒z=\tanh w=\frac{\exp(2w)-1}{\exp(2w)+1}⇒\exp(2w)=\frac{z+1}{1-z}⇒w=\frac{L(z+1)-L(1-z)}2$
      2. $\frac{z+1}{1-z}$ is a bijection on $ℂ∪\{∞\}$ so the image of $X$ is $ℂ∖(-∞,0]$
         The image of $|z|$ for $z∈ℂ∖(-∞,0]$ is $(0,∞)$
         The image of $\log{|z|}$ for $z∈ℂ∖(-∞,0]$ is ℝ
         The image of $\arg z$ for $z∈ℂ∖(-∞,0]$ is $(-π,π)$
         So the image of $X$ under the composition $\tanh^{-1}z=\frac12L\left(\frac{z+1}{1-z}\right)$ is $ℝ×\left(-\fracπ2,\fracπ2\right)$
      3. $\tanh^{-1}z=\frac{L(z+1)-L(1-z)}2=\frac12\left(\sum_{k=1}^∞-\frac{(-1)^kz^k}k+\frac{z^k}k\right)=\sum_{k=1}^∞\frac{z^{2k-1}}{2k-1}$
         $\tanh^{-1}z$ has branch points $±1$, so the radius of convergence is 1
         Also $\frac1{\lim_{k→∞}\sqrt[2k-1]{2k-1}}=1$
   3. 1. $\frac1{\tanh z}=\frac{\cosh z}{\sinh z}=\frac{1+\frac{z^2}2+O(z^4)}{z\left(1+\frac{z^2}6+O(z^4)\right)}=\frac1z\left(1+\frac{z^2}2+O(z^4)\right)\left(1-\frac{z^2}6+O(z^4)\right)=\frac1z+\frac z3+O(z^3)$
         Or use $\frac1{\tanh z}=\tanh(z-\fracπ2i)$
      2. The poles of $\frac1{\tanh z}$ is $iπn,n∈ℤ$ and $\frac1{\tanh z}$ is $iπ$-periodic. By (i) $\DeclareMathOperator*{\res}{Res}$the residue at 0 is 1, so the residue at $iπn,n∈ℤ$ are all 1. \begin{align*} ∫_{C_3} \frac{\mathrm{d} z}{\tanh z}&=2πi\res_0\frac1{\tanh z}=2πi \\∫_{C_4} \frac{\mathrm{d} z}{\tanh z}&=2πi\res_{0,±iπ}\frac1{\tanh z}=6πi \end{align*}
5. 1. Using the keyhole contour, outer arc $γ_R$ of radius $R$, inner arc $γ_ϵ$ of radius $ϵ$, \[2πi\res_{-1}f(z)=\int_{γ_R}f(z)dz-\int_{γ_ϵ}f(z)dz+\int_{ϵ\exp(δi)}^{R\exp(δi)}f(z)dz-\int_{ϵ\exp(-δi)}^{R\exp(-δi)}f(z)dz\]As $ϵ,δ→0$, \begin{split} \int_{ϵ\exp(δi)}^{R\exp(δi)}f(z)dz&→\int_0^∞f(x)dx \\\int_{ϵ\exp(-δi)}^{R\exp(-δi)}f(z)dz&→e^{2πi/3}\int_0^∞f(x)dx \end{split} Inside the contour the integrand has a double pole $-1$ with residue$$\res_{-1}f(z)=\left.\frac d{dz}[(z-1)^2f(z)]\right|_{-1}=\frac1{3e^{2πi/3}}$$By estimation lemma \begin{split} \left|\int_{γ_R}f(z)dz\right|&≤2πR\sup_{γ_R}\left|\frac{\sqrt[3]z}{(z+1)^2}\right|≤2πR\frac{\sqrt[3]R}{(R-1)^2}→0 \text{as }R→∞\\ \left|\int_{γ_ϵ}f(z)dz\right|&≤2πϵ\sup_{γ_ϵ}\left|\frac{\sqrt[3]z}{(z+1)^2}\right|≤2πϵ\frac{\sqrt[3]ϵ}{(1-ϵ)^2}→0 \text{as }ϵ→0 \end{split}Substituting in\[∫_0^∞ \frac{\sqrt[3]{x}}{(1+x)^2} d x=\frac{2πi}{3e^{2πi/3}(1-e^{2πi/3})}=\frac{2π}{3\sqrt3}\]
   2. Let γ be the unit circle.\[∫_0^{2π}\frac{dθ}{1-α\cosθ}=\int_γ\frac1{1-α\frac{z+z^{-1}}2}{dz\over iz}=\frac{2i}α\int_γ\frac{dz}{z^2-\frac2αz+1}\]The integrand has 2 simple poles $α^{-1}±\sqrt{α^{-2}-1}$
      $z_1=α^{-1}+\sqrt{α^{-2}-1}>α^{-1}>1$ is outside γ
      $z_2=α^{-1}-\sqrt{α^{-2}-1}=\frac1{z_1}<1$ is inside γ with residue $\frac1{(z^2-\frac2αz+1)'|_{z_1}}=-\frac1{2\sqrt{α^{-2}-1}}$ \begin{align*}∫_0^{2π}\frac{dθ}{1-α\cosθ}&=\frac{2i}α⋅2πi⋅\res_{z_2}f(z)\\&=\frac{2π}{\sqrt{1-α^2}}\\&=2π\sum_{n=0}^∞\binom{-\frac12}n(-1)^nα^{2n}\end{align*} By the Weierstrass M-Test, $\sum_{n=0}^∞(α\cosθ)^n≤\sum_{n=0}^∞α^n<∞$ converges uniformly on θ∈[0,2π] and so can swap the order of integration and summation \begin{align*}∫_0^{2π}\frac{dθ}{1-α\cosθ}&=\sum_{n=0}^∞∫_0^{2π}(α\cosθ)^ndθ\\&=\sum_{n=0}^∞α^n∫_0^{2π}\cos^nθdθ \end{align*}Comparing coefficient of $α^{2n}$ \[∫_0^{2π}\cos^{2n}θdθ=2π\binom{-\frac12}n(-1)^n=2π\frac{1×3×⋯×(2n-1)}{2^nn!}=2π\frac{(2n)!}{2^{2n}(n!)^2}\]
6. 1. 1. $\frac{az+b}{cz+d}=\frac ac\left(\frac{\frac{bc}{ad}-1}{\frac cdz+1}+1\right)⇒\frac{az+b}{cz+d}=D_{\frac ac}∘T∘D_{\frac{bc}{ad}-1}∘I∘T∘D_{\frac cd}$
         Then use $D_{re^{iθ}}=D_rR_θ$
      2. A Möbius transformation $f$ such that $f(-1)=0,f(1)=∞$ has the form $f(z)=C\frac{z+1}{1-z}$
         Let $f(0)=e^{iπ/4}$, we find $C=e^{iπ/4}$. To verify $f(L)=Q$\[z∈L⇔\arg{z-1\over z+1}∈\left[-\fracπ4,\fracπ4\right]⇔\arg f(z)∈\left[0,\fracπ2\right]⇔f(z)∈Q\]Write as a composition $f=(-e^{iπ/4}z)∘(z+1)∘(-2z)∘\frac1z∘(z+1)∘(-z)$
      3. It suffices to determine all Möbius bijection $T:L→L$. Then $T$ maps points ±1 to ±1. Assume $T(1)=1$ [if $T(1)=-1$ precompose with the map $z↦-z$]. Take a point $p=(\sqrt2-1)i$ on the upper boundary of $L$. Then $T(p)$ is any point other than ±1 on the upper boundary because $T$ preserves orientation of three points.
         $T$ is determined by $T(p),T(1),T(-1)$. Then $T$ maps the upper boundary to itself as circline containing $±1,T(p)$. Also $T$ is conformal, the lower boundary must be mapped to the lower boundary, since it is the unique circline intersecting upper boundary at ±1 with $±π/2$.
   2. 1. For $w∈[-2,2]$, let $θ=\arccos\frac w2$, $w=ϕ(z)⇒z=e^{±iθ}$. So ${ϕ|}_{S^1}$ is a two-to-one map.
         For $w∈ℂ∖[-2,2]$, $ϕ(z)=w⇔z^2-wz+1=0$ has two solutions $z_1,z_2({|z_1|}≤{|z_2|})$.
         ${|z_1|}≠1,{|z_1|}⋅{|z_2|}=1⇒z_1∈U$. So $w↦z_1$ is inverse of ${ϕ|}_U^{-1}$. So ${ϕ|}_U$ is a bijection.
      2. Rearranging $\frac{ψ(z)-1}{ψ(z)+1}=\left(\frac{z-1}{z+1}\right)^2$ we find\[ψ(z)={(z+1)^2+(z-1)^2\over(z+1)^2-(z-1)^2}=\frac12\left(z+\frac1z\right)=\frac12ϕ(z)\]So $ϕ=2ψ=\frac{2(z+1)}{1-z}∘z^2∘\frac{z-1}{z+1}$
   3. (Riemann's mapping theorem) Let $W$ be an open connected and simply-connected proper subset of ℂ. Then for any $z_0∈W$ there is a unique bijective conformal map $g:W→D$ such that $g(z_0)=0,g'(z_0)>0$
      So there is a bijective conformal map $g:W→D,g(a)=0,g'(a)>0$
      Let $h(z)=\frac{b+z}{\overline b z+1}$. Then $h'(0)=1-b\bar b>0$.
      For $z∈D$, $\left|\bar bz+1\right|^2-\left|b+z\right|^2=(1-{|b|}^2)(1-{|z|}^2)>0⇒h(z)∈D$, so $h(D)⊂D$.
      Also $h^{-1}(z)=\frac{-b+z}{-\bar b z+1}=-h(-z)⇒h^{-1}(D)⊂D$, so $h$ is a bijection $D→D$.
      So $f=h∘R_α∘g$ satisfies $f(a)=h(R_α(0))=h(0)=b$ and by chain rule \[\arg f'(a)=\arg h'(0)+\arg R_α'(0)+\arg g'(a)=0+α+0=α\]

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      In general a Möbius transformation is an involution when it is the identity or traceless