$\newcommand{\abs}[1]{\left|#1\right|}
\newcommand{\d}{\mathrm{~d}}
\DeclareMathOperator{\Re}{Re}
\DeclareMathOperator{\Im}{Im}
\newcommand{\res}[1]{\operatorname*{Res}_{#1}}$
- Evaluate, using a keyhole contour cut along the positive real axis, or otherwise,
\[
\int_0^∞\frac{x^{1/2}\log x}{(1+x)^2}\d x
\]
Solution.
$z^{1/2}=e^{\log z/2}$, so $z^{1/2}$ has same branch cut as $\log z$, normally is the negative real axis, $\arg z∈(-π,π]$, now it is the positive real axis, $\arg z ∈[0,2π)$.
Consider the keyhole contour $Γ_{R,ϵ}$. The function has a double pole at $z=-1$ with Laurent series \[(1+z)^{-2}(-1+(1+z))^{1/2}\log z=(1+z)^{-2}\left(i-\frac i2(1+z)+⋯\right)(πi-(1+z)+⋯)\] So\[\res{z=-1}\frac{z^{1/2}\log z}{(1+z)^2}=-\frac i2⋅πi-i=\fracπ2-i\] For $R>1>ϵ$, $-1$ is inside $Γ_{R,ϵ}$. By residue theorem,$$\int_{Γ_{R,ϵ}}\frac{z^{1/2}\log z}{(1+z)^2}\d z=2πi\left(\fracπ2-i\right)=2π+π^2i\tag1$$For $\abs z=R$, $\abs{\frac{z^{1/2}\log z}{(1+z)^2}}=\abs{\frac{z^{1/2}(\log\abs z+i\arg z)}{(1+z)^2}}≤R^{1/2}\frac{\log R+2π}{(R-1)^2}$. By estimation lemma, $$\abs{\int_{Γ_R}\frac{z^{1/2}\log z}{(1+z)^2}\d z}≤2πR⋅R^{1/2}\frac{\log R+2π}{(R-1)^2}→0\text{ as }R→∞$$Similarly,$$\abs{\int_{Γ_ϵ}\frac{z^{1/2}\log z}{(1+z)^2}\d z}≤2πϵ⋅ϵ^{1/2}\frac{\logϵ+2π}{(1-ϵ)^2}→0\text{ as }ϵ→0$$The integral of $f$ over the line segment above real axis tends to$$\int_0^∞\frac{x^{1/2}\log x}{(1+x)^2}\d x\tag2$$The integral of $f$ over the line segment below real axis tends to \[\int_∞^0\frac{(-x^{1/2})(\log(x)+2\pi i)}{(1+x)^2}\d x=\int_0^∞\frac{x^{1/2}(\log(x)+2\pi i)}{(1+x)^2}\d x\tag3\] (1) is the sum of (2) and (3). Taking the real part, we get $$\int_0^∞x^{1/2}\frac{\log(x)}{(1+x)^2}\d x=π$$ - By considering the integral
\[
\int_{Γ_n}\frac{π\d w}{w^2\sinπw}
\]
where $Γ_n$ is the square in ℂ with vertices $±(n+1/2)(1±i)$ show that
\[
\frac{π^2}{12}=1-\frac14+\frac19-\frac1{16}+⋯
\]
Proof.
We may write $\sin(z)=z(1-zh(z))$ where $h(z)=z/3!-z^3/5!+O(z^5)$ is holomorphic at $z=0$. Then $$\frac1{\sin z}=\frac1z(1-zh(z))^{-1}=\frac1z\left(1+\sum_{n≥1}z^nh(z)^n\right)=\frac1z+\frac z6+O(z^3)$$ Substitute $πw$ for $z$ and multiply $π/w^2$, $$f(w)=\fracπ{w^2\sinπw}=\frac1{w^3}+\frac{π^2}{6w}+O(w^3)$$ So $f(w)$ has a pole of order 3 at $w=0$, with residue $π^2/6$.
Similarly, we find the Laurent series for $f(w)$ at non-zero integer $n$, \begin{align*} f(w)&=\frac{(-1)^nπ}{(n+(w-n))^2\sin(π(w-n))}\\ &=(-1)^nπ\left(\frac1{n^2}+O(w-n)\right)\left(\frac1{π(w-n)}+\frac{π(w-n)}6+O((w-n)^3)\right) \end{align*} So $f(w)$ has a simple pole at each non-zero integer $n$, with residue $(-1)^n/n^2$. By residue theorem, as $n→∞$, the integral of $f$ over $Γ_n$ tends to $$\frac{π^2}6+2\sum_{n=1}^∞\frac{(-1)^n}{n^2}$$It remains to prove the integral tends to 0.
On $Γ_n$, we have $\abs{w}≥\left(n+\frac12\right)π$. As shown on the right, $\abs{\cscπw}≤C$.
By estimation lemma, $$\abs{\int_{Γ_n}\frac{π\d w}{w^2\sinπw}}≤4(2n+1)⋅\frac{πC}{\left(n+\frac12\right)^2π^2}=\frac{8C}{\left(n+\frac12\right)π}$$ tends to 0 as $n→∞$.For all $z=x+iy$, $$\abs{\sin z}=\sqrt{\sin^2 x\cosh^2 y+\cos^2x\sinh^2y}=\sqrt{\sin^2 x+\sinh^2y}$$ Therefore $\abs{\sin z}≥\abs{\sin x}$, $\abs{\sin z}≥\abs{\sinh y}$.$∃C,∀w∈Γ_n,∀n≥1:\abs{\cscπw}≤C$
Proof.
On the vertical sides, $\abs{\sinπw}≥\abs{\sin(\Re(πw))}=\abs{\sin\left(π\left(n+\frac12\right)\right)}=1$.
On the horizontal sides, $\abs{\sinπw}≥\abs{\sinh(\Im(πw))}=\sinh\left(\left(n+\frac12\right)π\right)≥\sinh\frac{3π}2$. - Let $n≥2$. By using the contour comprising $[0,R]$, the circular arc from $R$ to $R e^{2πi/n}$, and $\left[0,Re^{2πi/n}\right]$, show that
\[
\int_0^∞\frac{\mathrm{d}x}{1+x^n}=\fracπ{n} \csc \left(\fracπ{n}\right)
\]
Proof.
$f(z)=\frac1{1+z^n}$ has only one pole inside $Γ_R\,(R>1)$ at $z=e^{πi/n}$, with residue $$\res{z=e^{πi/n}}\frac1{1+z^n}=\lim_{z→e^{πi/n}}\frac {z-e^{πi/n}}{z^n+1}=-\frac1ne^{πi/n}$$So the integral is $-\frac {2πi}ne^{πi/n}$.
Since $n>1$, by estimation lemma, as $R→∞$, the integral on arc vanishes. Leaving us with $$-\frac {2πi}ne^{πi/n}=\int_0^∞f(x)\d x-e^{2\pi i/n}\int_0^∞f(e^{2πi/n}x)\d x$$ Using $f(e^{2πi/n}x)=f(x)$ with $\sin\left(π\over n\right)=\frac{e^{πi/n}-e^{-πi/n}}{2i}$,$$\int_0^∞\frac{\mathrm{d}x}{1+x^n}=\fracπn\csc\left(π\over n\right)$$ - In each of the following cases find a conformal mapping from the given region $G$ onto the open unit disc $𝔻=B(0,1)$.
- $G=\{z∈ℂ:\Im z>0\}$,
- $G=\{z∈ℂ:z≠0$ and $-π/4<\arg z< π/4\}$,
- $G=\{z∈ℂ:\abs{z-i}<\sqrt2$ and $\abs{z+i}>\sqrt2\}$.
- Möbius transform $κ(z)=\frac{z-i}{z+i}$ maps $∞,1,-1$ to $1,-i,i$. Since κ maps circlines to circlines, κ maps the real line $L$ to the unit circle $S$.
By continuity, κ maps connected components of $ℂ∖L$ to connected components of $ℂ∖S$, but $κ(i)=0∈𝔻$, so $κ(G)=𝔻$. - First take $f(z)=iz^2$ to map $G$ onto the upper half-plane, then compose with $κ(z)=\frac{z-i}{z+i}$ to get $κ(f(z))=\frac{z^2-1}{z^2+1}$.
- $\arg\frac{z-1}{z+1}$ = the angle formed by $-1,z,1$
$\arg\frac{z-1}{z+1}∈\left(-\frac{3π}4,-\fracπ4\right)$ iff $z∈G$.
Let $f(z)=\frac{z-1}{z+1}$. Then $f(G)=\{z∈ℂ:z≠0$ and $\arg z∈\left(-\frac{3π}4,-\fracπ4\right)\}$. Similar to (ii), we find $g(z)=\frac{z^2+1}{z^2-1}$ maps $f(G)$ to 𝔻. Therefore $g(f(z))=-\frac{1+z^2}{2z}$ maps $G$ to 𝔻.
- Let
\[
A=\{z∈ℂ:\abs{z-2}< 2\text { and }\abs{z-1}>1\} ; \quad B=\{z∈ℂ:0<\Re z< π\} .
\]
- Find the image of $A$ under the map $z↦1/z$ and the image of $B$ under the map $z↦\exp(iz)$.
- Let $ℍ=\{z∈ℂ:\Im z>0\}$. Given $a, b∈ℍ$ find a conformal bijection $ℍ→ℍ$ of the form $z↦λz+μ$ which maps $a$ to $b$.
- Deduce that for any two points $a,b∈A$ there is a conformal bijection $f:A→A$ such that $f(a)=b$.
-
$A$ can be parametrized as $z=r(1+\cosθ+i\sinθ),\,θ∈(-π,π],\,r∈(1,2)$.
$$\Re\frac1z=\frac1r\Re\frac1{1+\cosθ+i\sinθ}=\frac1r\Re\frac{1+\cosθ-i\sinθ}{2+2\cosθ}=\frac1{2r}$$
So the image of $A$ under the map $z↦\frac1z$ is $\left\{\frac1{2r}+iy:r∈(1,2),y∈ℝ\right\}=\left\{z∈ℂ:\Re z∈\left(\frac14,\frac12\right)\right\}$.
$B$ can be parametrized as $z=x+iy,\,x∈(0,π),\,y∈ℝ$. $$\exp\big(i(x+iy)\big)=\mathrm{e}^{-y}(\cos x+i\sin x)$$ The range of $\mathrm{e}^{-y}$ is $ℝ^+$, so the image of $B$ under the map $z↦\exp(iz)$ is $\{z∈ℂ:\Im z>0\}$. -
Since $z↦λz+μ$ maps boundary of ℍ to itself, we have $λ∈ℝ$. If $λ=0$ the map is constant so it is not bijective.
If $λ<0$ it maps $ai,a>\frac{\Im μ}{-λ}$ to $aiλ+μ$, but $\Im(aiλ+μ)=aλ+\Im μ<0$, so the image of $ai$ is not in ℍ. Hence $λ>0$.
The image of ℍ is $\{z:\Im z>\Im μ\}$, as the map is surjective, $\Im μ=0$, so $μ∈ℝ$
$λa+μ=b⇒\Im(λa+μ)=\Im b⇒λ=\frac{\Im b}{\Im a},μ=b-λa=b-\frac{\Im b}{\Im a}a$ - $A$ is mapped to $\left\{x+iy:x∈\left(\frac14,\frac12\right),y∈ℝ\right\}$ by $z↦\frac1z$, then mapped to $B$ by $z↦π(4x-1)$, then mapped to ℍ by $z↦\exp(iz)$. Denote the composition map by $h$, then $h(a)∈ℍ,h(b)∈ℍ$, by (ii), there is a conformal bijection $f_1:ℍ→ℍ$ which maps $h(a)$ to $h(b)$. Therefore $f=h^{-1}∘f_1∘h$ is a conformal bijection $A→A$ which maps $a$ to $b$.
- Let $ℝ_∞=ℝ∪\{∞\}⊂ℂ_∞$.
- Show that group $Γ$ of Möbius transformations $T$ for which $T(ℝ_∞)=ℝ_∞$ are exactly those of the form $T(z)=(az+b)/(cz+d)$ where $a,b,c,d$ can be chosen to be real.
- Calculate the group $Γ_1$ of Möbius tranformations which preserve $ℍ=\{z∈ℂ:\Im(z)>0\}$.
[Hint: Why must $Γ_1$ be a subgroup of $Γ$ ?]
- If $c=0$, by definition of Möbius transformations, we have $ad-bc≠0$, so $d≠0$. Now $T(0)=\frac bd∈ℝ,T(1)-T(0)=\frac ad∈ℝ$, we may write $T(z)=\left(\frac adz+\frac bd\right)/\left(0z+1\right)$.
If $c≠0$, we may write $T(z)=\left(\frac acz+\frac bc\right)/\left(z+\frac dc\right)$, it remains to prove $\frac ac,\frac bc,\frac dc$ are real.
$T(z)=\frac{az+b}{cz+d},T^{-1}(z)=\frac{dz-b}{-cz+a}⟹T(∞)={a\over c}∈ℝ,T(0)={b\over d}∈ℝ_∞,T^{-1}(∞)=-\frac dc∈ℝ,T^{-1}(0)=-\frac ba∈ℝ_∞$. So we are done.
Alternate method:
Take 3 distinct points $a_1,a_2,a_3∈ℝ_∞$ that are mapped to $b_1,b_2,b_3∈ℝ_∞$. Then by invariance of cross-ratio we have for all $z∈ℂ$, $$\frac{z - a_1}{a_3 - a_1}:\frac{z - a_2}{a_3 - a_1}=\frac{T(z) - b_1}{b_3 - b_1}:\frac{T(z) - b_2}{b_3 - b_2}$$ Solving for $T(z)$, we can express $a,b,c,d$ in terms of $a_1,a_2,a_3,b_1,b_2,b_3$ with the appropriate conventions on $∞$. - Suppose $z∈ℍ$ and $T(z)∈ℍ$, then $\Im z>0$ and $\Im(T(z))>0$. $$\Im(T(z))=\Im\frac{(az+b)(c\bar z+d)}{\abs{cz+d}^2}=\frac{ad-bc}{\abs{cz+d}^2}⋅\Im(z)$$This implies $ad-bc>0$. We can normalize $ad-bc$ by dividing all of $a,b,c,d$ by $\sqrt{ad-bc}$. So $Γ_1=\left\{z↦\frac{az+b}{cz+d}:a,b,c,d∈ℝ,\,ad-bc=1\right\}$.
- Write down a bounded solution $u(x, y)$ to the Dirichlet problem for the following region and boundary conditions:
\[
U=\{x+iy:0≤y≤1\}, \quad u(x, 0)=0, \quad u(x, 1)=1 .
\]
Hence, using appropriate conformal maps, solve the Dirichlet problem for the following regions and boundary conditions.
- $U_1=\left\{z:r_1≤\abs z≤r_2\right\},\quad u(z)=0$ when $\abs{z}=r_1,\quad u(z)=1$ when $\abs{z}=r_2$.
- $U_2=\{z:\Im z≥0\},\quad u(x, 0)=0$ when $x>0,\quad u(x,0)=1$ when $x< 0$.
- $U_3=\{z:\abs z≤1\},\quad u(z)=0$ when $\abs z=1$ and $\Im z< 0,\quad u(z)=1$ when $\abs z=1$ and $\Im z>0$.
- $U_4=\{z:\Im z≥0\},\quad u(x, 0)=0$ when $\abs{x}>1, \quad u(x, 0)=1$ when $\abs{x}< 1$.
For $u(z)=\Im z$, we have $u(U)=[0,1],u(x,0)=0,u(x,1)=1$, so it is a bounded solution.- Let $f(z)=r_1\left(r_1\over r_2\right)^{iz}$. For $y∈[0,1]$, we have $\abs{f(x+iy)}=r_1\left(r_1\over r_2\right)^{-y}∈[r_1,r_2]$. As $\arg f(x+iy)=x\log\left(r_1\over r_2\right)$ ranges over ℝ, $f$ maps $U$ to $U_1$.
The inverse function is multivalued $$f^{-1}(w)= i\frac{ \log(w/r_1)}{\log(r_2/r_1)} $$ Starting with $\Im:U→ℝ$, since $\Re(\log(w/r_1))=\log(\abs w/r_1)$, we find the solution $\Im∘f^{-1}:U_1→ℝ$ $$\Im∘f^{-1}(w)=\frac{\log(\abs w/r_1)}{\log (r_2/r_1)}.$$ Note that even though $f^{-1}$ is multivalued, the solution $u$ is uniquely defined, since we only use $\log:ℝ^+→ℝ$. Alternative method: Let $u(w)=A\log\abs w+B$, constant $A,B$ to be determined \[\begin{cases}A\log r_1+B=0\\A\log r_2+B=1\end{cases}⇒\begin{cases}A=\frac1{\log(r_2/r_1)}\\B=-{\log r_1\over\log(r_2/r_1)}\end{cases}\] - $f_1(z)=e^{πz}$ maps $U$ to $U_2$, mapping the boundary $y=0$ to the positive real axis, and $y=1$ to the negative real axis.
Starting with $\Im:U→ℝ$, we find the solution $\Im∘f_1^{-1}:U_2→ℝ$ \[\Im∘f_1^{-1}(z)=\frac1π\arg z\] - $f_2(z)=\frac{z-i}{z+i}$ maps $U_3$ to $U_2$, mapping the upper unit semicircle to the positive real axis, the lower unit semicircle to the negative real axis.
$f_1^{-1}∘f_2^{-1}(z)=\frac1π\log\left(i\frac{1+z}{1-z}\right)=\frac12+\frac1π\log\left(\frac{1+z}{1-z}\right)$ maps $U_3$ to $U$, mapping the lower unit semicircle to $y=0$, and the upper unit semicircle to $y=1$.
Starting with $\Im:U→ℝ$, we find the solution $\Im∘f_1^{-1}∘f_2^{-1}:U_3→ℝ$ \[\Im∘f_1^{-1}∘f_2^{-1}(z)=\frac1π\arg\left(\frac{1+z}{1-z}\right)\] - By Q6(ii), any Möbius transform of the upper half plane has the form $T(z)=\frac{az+b}{cz+d}$ with $a,b,c,d∈ℝ$, $ad-bc>0$. We need to map $(-\infty, 0)$ to $(-1,1)$, where $u(x,0)=1$, and since conformal maps preserve the orientation, $T(∞)=-1,T(0)=1$.
Let $T(z) = \frac{1+z}{1-z}$, then we find the solution $\Im∘f_1^{-1}∘T^{-1}:U_4→ℝ$ \[\Im∘f_1^{-1}∘T^{-1}(z)=\frac1π\arg\left(\frac{z-1}{z+1}\right)\]
- Let $f:𝔻→𝔻$ be a holomorphic function, where $𝔻=B(0,1)$.
- Show that if $f(0)=0$ then $\abs{f(z)}≤\abs{z}$ for all $z∈𝔻$. (hint: show that $\abs{f(z)/z} ≤1/r$ if $\abs z≤r$, for any $r< 1$).
Show that if, moreover, $\abs{f(z_0)}=\abs{z_0}$ for some $z_0∈𝔻, z_0≠0$, then $f$ is a rotation. - Show that if $a∈𝔻$ the function \[ g_a(z)=\frac{a-z}{1-\bar{a} z} \] maps $𝔻$ to $𝔻$.
- Show that if $f:𝔻→𝔻$ is holomorphic and bijective then there is some $θ∈ℝ$ and $a∈𝔻$ such that $f(z)=e^{iθ}g_a(z)$.
- Since $f$ is analytic on 𝔻 and $f(0)=0$,$$f(z)=f'(0)z+\frac{f''(0)}2z^2+⋯$$
Then the function
$$g(z)=f'(0)+\frac{f''(0)}2z+⋯$$
is analytic on 𝔻.
By the maximum modulus principle, for $r< 1$, given any $z∈\bar B(0,r)$, there exists $z_r$ on the boundary of $\bar B(0,r)$ such that $$\abs{g(z)}≤\abs{g(z_r)}=\frac{|f(z_r)|}{|z_r|}=\frac{|f(z_r)|}r≤\frac1r$$ As $r→1$ we get $\abs{g(z)}≤1$. So $\abs{f(z)}≤\abs{z}$.
Moreover, suppose $\abs{f'(0)}=1$ or $∃z∈𝔻 ∖\{0\}:\abs{f(z)}=\abs{z}$. Then $∃z∈𝔻:\abs{g(z)}=1$. By the maximum modulus principle, $g(z)=a$ constant, then $\abs{a}=1$. Therefore, $f(z)=az$. - Since $\abs a< 1$, for $\bar z≤1$, $1-\bar az≠0$, the map $g_a$ is holomorphic. If $\abs z=1$ then $z=e^{iθ}$ and$$g_a\left(e^{iθ}\right)=\frac{a-e^{iθ}}{e^{iθ}\left(e^{-iθ}-\bar{a}\right)}=e^{-iθ}\frac{w}{-\bar{w}}$$where $w=a-e^{iθ}$, therefore $\abs{g_a(z)}=1$. By the maximum modulus principle, $\abs{g_a(z)}≤1$ for all $z∈𝔻$.
- Let $a=f^{-1}(0), g=f∘g_a$ and observe that $g(0)=f\left(g_a(0)\right)=f(a)=0$, by (i), $\abs{g(z)}≤\abs z$. Moreover, $g^{-1}(0)=0$, applying (i) to $g^{-1}$ yields $\abs{g^{-1}(z)}≤\abs z$, so $\abs z=\abs{g^{-1}(g(z))}≤\abs{g(z)}$, so $\abs{g(z)}=\abs z$. Hence, $g$ is a rotation. Recall that we had $f∘g_a = g$. As $g_a = g_a^{-1}$, $f = g∘g_a$ as desired.
- Show that if $f(0)=0$ then $\abs{f(z)}≤\abs{z}$ for all $z∈𝔻$. (hint: show that $\abs{f(z)/z} ≤1/r$ if $\abs z≤r$, for any $r< 1$).