-
(i) Suppose that $l(z)$ is holomorphic on $ℂ∖(-∞, 0]$ and satisfies $\exp l(z)=z$. Show that
$$
l(z)=\operatorname{Log}(z)+2nπi
$$
for some $n∈ℤ$ where $\operatorname{Log}(z)$ is the holomorphic branch of [log] defined in lectures.
(ii) Show that there is no holomorphic function $\lambda(z)$ on $ℂ ∖\{0\}$ such that $\exp λ(z)=z$.
(iii) There are unique holomorphic branches of $\log z, \sqrt{z}$ and $\sqrt[3]z$ on the cut plane $ℂ∖\{z∣z=-i{|z|}\}$ such that $\log 1=0 ; \sqrt1=1 ; \sqrt[3]1=1$. For these branches determine $$ \log (1+i), \quad \sqrt{-1-i}, \quad \sqrt[3]{-2}, \quad \sqrt{1-i} $$ (iv) Let $γ:[0,1]→ℂ$ be a $C^1$ curve with $γ(0)=i, γ(1)=-i$ such that $γ$ does not intersect $(-∞, 0]$. Calculate the integral $$ \int_γ\operatorname{Log}^2(z) \mathrm{d} z . $$ Solution.- Taking derivative on both sides of $\exp l(z)=z$, we find $l'(z)\exp l(z)=1$, so $l'(z)=\frac1z$, so $l(z)-\operatorname{Log}(z)$ is constant on $ℂ∖(-∞, 0]$.
From $\exp l(1)=1$ we have $l(1)=2nπi$ for some $n∈ℤ$, so $l(z)-\operatorname{Log}(z)=2nπi$. - Proof 1: By (i), $∀z∈ℂ∖(-∞,0],λ(z)=\operatorname{Log}z+2nπi$ for some $n∈ℤ$. But $\lim_{z→-1}\operatorname{Log}(z)$ does not exist, so $\lim_{z→-1}\log(z)$ does not exist.
Proof 2: If we had a holomorphic function $f(z)$ such that $\exp f(z) = z$ then differentiating gives $f'(z) = 1/z$ for all $z ≠ 0$.
Integrating on ${|z|}=1$, the left gives $0$ by Cauchy's Theorem but the right is $2πi$, a contradiction. - $\arg z$ jumps 2π across the branch cut, so $\arg z∈\left[-\fracπ2+2πn,\frac{3π}2+2πn\right]$
$\log 1=0⇒\arg 1=0⇒\arg z∈\left[-\fracπ2,\frac{3π}2\right]⇒\arg(1+i)=\fracπ4,\arg(-1-i)=\frac{5π}4,\arg(1-i)=-\fracπ4,\arg(-2)=π$
$\log(1+i)=\frac12\log2+i\arg(1+i)=\frac12\log2+i\fracπ4$
$\sqrt{-1-i}=\root4\of2\exp\left(i\frac12\arg(-1-i)\right)=\root4\of2\exp\left(i\frac{5π}8\right)$
$\sqrt{1-i}=\root4\of2\exp\left(i\frac12\arg(1-i)\right)=\root4\of2\exp\left(-i\fracπ8\right)$
$\root3\of{-2}=\root3\of2\exp\left(i\frac13\arg(-2)\right)=\root3\of2\exp\left(i\fracπ3\right)$ - $$\int_γ\log^2(z)\mathrm{~d}z=\int_γ\frac{\mathrm d}{\mathrm dz}(2 z+z\log ^2z-2z\log z)\mathrm{~d}z=[2 z+z\log ^2z-2z\log z]_i^{-i}=\frac{iπ^2}2-4i$$
- Taking derivative on both sides of $\exp l(z)=z$, we find $l'(z)\exp l(z)=1$, so $l'(z)=\frac1z$, so $l(z)-\operatorname{Log}(z)$ is constant on $ℂ∖(-∞, 0]$.
-
Calculate the integral$$\int_0^{2 π} e^{e^{it}} \mathrm{~d} t$$
Solution.
$e^z$ is holomorphic on the interior of $γ(t)=e^{it},t∈[0,2π]$. Thus \begin{aligned} &\int_0^{2 π} e^{e^{it}} \mathrm{~d} t\\ &=\int_γe^z⋅\frac1{iz}\mathrm{~d}z\hspace1em\text{Definition of path integral, }γ'(t)=ie^{it}\\ &=\frac1i\int_γ\frac{e^z}z\mathrm{~d}z\\ &=\frac1i⋅2πie^0\hspace1em\text{Cauchy’s Integral Formula}\\ &=2π \end{aligned} -
Green's Theorem states, for a region $D$ in the plane, bounded by (an) oriented closed curve(s) $C$ in $ℝ^2$ and for real-valued $L$ and $M$ with continuous partial derivatives on $D$, then
$$
\int_C(L \mathrm{~d} x+M \mathrm{~d} y)=\iint_D\left(M_x-L_y\right) \mathrm{~d} x \mathrm{~d} y .
$$
If we assume, for a holomorphic function $f=u+iv$, that $u_x,u_y,v_x,v_y$ are continuous, show that Cauchy's Theorem follows from Green's Theorem, that is, show that for a function $f$ which is holomorphic on the interior $D$ of a closed curve $C$, we have $\int_C f(z) \mathrm{~d}z=0$.
Solution. Applying Green's theorem to the closed curve $C$ with parameterization $𝐫(t)=(x(t),y(t))$ and the vector field $𝐅(x,y)=(u(x + iy), -v(x + iy))$. \begin{align*} \int_L f(z)\,dz &= \int_C (u+iv)(dx+i\,dy) \\ &=\int_C (u,-v)⋅(dx,dy) + i\int_C (u,-v)\cdot (dy,-dx) \\ &=\int_C 𝐅(𝐫)⋅d𝐫+ i\int_C 𝐅(𝐫)⋅d𝐫^⟂\\ &=\iint_D\operatorname{rot}𝐅+i\operatorname{div}𝐅\,dxdy\\ &=\iint_D(v_x+u_y)+i(u_x-v_y)\,dxdy\\ \small\text{by Cauchy-Riemann}&=0 \end{align*} -
Let
$$
f(z)=\frac{5 z^2-8}{z^3-2z^2}.
$$
Determine $\int_{γ(0,1)} f(z) \mathrm{~d} z$.
Describe different closed paths $γ$ in $ℂ$ such that $\int_γf(z) \mathrm{~d} z$ equals $14 π i, \; 18 π i, \;{-2 π i}$.
Solution.
Since $I(γ,0)=1,I(γ,2)=0,$ \begin{aligned} \int_{γ(0,1)}f(z)\mathrm{~d}z&=\int_{γ(0,1)}\frac2z+\frac3{z-2}+\frac4{z^2}\mathrm{~d}z\qquad\text{Partial fractions}\\ &=2⋅2πi+3⋅0+\int_{γ(0,1)}\frac4{z^2}\mathrm{~d}z \frac4{z^2}\text{ has primitive so integral is 0}\\ &=4πi \end{aligned}Integral on the path $z=\begin{cases}e^{8πit}&t∈[0,1/2]\\2+e^{(4t-1)πi}&t∈[1/2,1]\end{cases}$
equals $2⋅4πi+3⋅2πi=14πi$.
Integral on the path $z=2+e^{6πit},t∈[0,1]$
equals $2⋅0+3⋅6πi=18πi$.
Integral on the path $z=\begin{cases}e^{4πit}&t∈[0,1/2]\\2+e^{(1-4t)πi}&t∈[1/2,1]\end{cases}$
equals $2⋅2πi+3⋅-2πi=-2πi$.
-
Let
$$
I=\int_{γ(0,1)} \frac{\Re z}{2 z-i} \mathrm{~d} z \quad \text { and } \quad J=\int_0^{2π} \frac{\cos^2 θ}{5-4 \sin θ} \mathrm{~d} θ \text {.}
$$
Using only Cauchy's Integral Formula, evaluate $I$. (Take note that $\Re z$ is not holomorphic.)
By setting $z=e^{iθ}$ in the integral for $I$, determine $J$.
Solution. Note that $\Re z\,,z∈γ$ can be extended to $\frac{z+z^{-1}}2,z∈ℂ^×$
\begin{aligned}I&=\frac12\int_{γ(0,1)} \frac{z+z^{-1}}{2 z-i}\mathrm{~d} z\\ &=\frac12\int_{γ(0,1)} \frac {\frac z2-i}{z-\frac i2}+\frac iz\mathrm{~d} z\\ &=\frac12⋅2πi\left(\frac i4-i+i\right)\\ &=-\fracπ4 \end{aligned} Setting $z=e^{iθ},\mathrm{\ d}θ=-iz^{-1}\mathrm{\ d}z$, \begin{aligned}J&=\int_0^{2π}\frac{\cos^2θ}{5-4\sinθ}\mathrm{\ d}θ\\&=\int_{γ(0,1)}\frac{\left(z+z^{-1}\over2\right)^2}{5-4\frac{z-z^{-1}}{2i}}⋅-iz^{-1}\mathrm{\ d}z\\ &=\int_{γ(0,1)}-\frac12⋅\frac{\frac{z+z^{-1}}2}{2z-i}-\frac{i}{16z}+\frac1{8 z^2}\underbrace{-\frac{3 i}{16 (z-2 i)}}_\text{holomorphic}\mathrm{~d} z\\ &=-\frac I2+\fracπ8\\ &=\fracπ4 \end{aligned} -
(i) By making the substitution $z=re^{iθ}$, and making clear any special cases, for each integer $k$ determine $\int_{γ(0,r)}z^k\mathrm{~d}z$.
(ii) By writing $\sinθ=\frac{e^{iθ}-e^{-iθ}}{2i}$ rewrite the integral on the left as a path integral around $γ(0,1)$ and deduce that $$ \int_0^{2π}\sin^{2n}θ\mathrm{\ d}θ=\frac{2π}{4^n}\binom{2n}n $$ Solution.
(i) Substitute $z=re^{iθ},\mathrm{\ d}z=ire^{iθ}\mathrm{\ d}θ$
(ii)\begin{align*}\int_0^{2π}\sin^{2n}θ\mathrm{\ d}θ&=\int_0^{2π}\left(\frac{e^{iθ}-e^{-iθ}}{2i}\right)^{2n}\mathrm{\ d}θ\\ &=\frac1{4^n}\int_0^{2π}\sum_{k=0}^{2n}\binom{2n}k(-1)^ke^{2iθ(n-k)}\mathrm{\ d}θ\\ &=\frac{2π}{4^n}\binom{2n}n\qquad\text{all }k≠n\text{ terms vanishes}\end{align*}For $k=-1$, \begin{align*} \int_{γ(0,r)}z^{-1}\mathrm{~d}z&=\int_0^{2π}(re^{iθ})^{-1}rie^{iθ}\mathrm{~d}θ\\ &=\int_0^{2π}i\mathrm{~d}θ\\ &=2πi \end{align*} For $k≠-1$, \begin{align*} \int_{γ(0,r)}z^k\mathrm{~d}z&=ir^{k+1}\int_0^{2π}e^{i(k+1)θ}\mathrm{~d}θ\\ &=ir^{k+1}\left.\frac{e^{i(k+1)θ}}{i(k+1)}\right|_{θ=0}^{2π}\\&=0 \end{align*} -
(i) Use the estimation lemma to show that, if $γ:[0,1]→ℂ$ is a piecewise $C^1$ closed path, then the winding number $I(γ,z)$ is constant on the connected components of $ℂ∖γ^*$
(ii) Let $γ=γ(0,1)$ and let $a, b∈ℂ$ with ${|a|}≠{|b|}$. Calculate the integral $$ \int_γ \frac1{a z+b} \mathrm{~d} z $$ Proof.- Since $I(γ, z)$ is integer-valued, it suffices to show it is a continuous function of $z$ on $ℂ∖γ^*$.
Since $ℂ∖γ^*$ is open, there exists $r>0$ such that $B(a,2r)⊂ℂ∖γ^*$. For any $z∈γ^*,w∈B(a,r)$, we have ${|z-w|}>r⇒\left|\frac1{(z-w)^2}\right|<\frac1{r^2}$
For any $b∈B(a,r)$, \begin{aligned} {|I(γ,a)-I(γ,b)|}&=\left|\frac1{2πi} \int_γ \frac1{z-a} -\frac{1}{z-b} \mathrm{~d}z\right|\\ &=\frac1{2π}\left|\int_γ\int_a^b\frac1{(z-w)^2}\mathrm{~d}w\mathrm{~d}z\right|\\ \small\text{by Estimation Lemma}&≤\frac{L(γ)}{2π}\sup_{z∈γ}\left|\int_a^b\frac1{(z-w)^2}\mathrm{~d}w\right|\\ \small\text{by Estimation Lemma}&≤\frac{L(γ)}{2π}⋅{|a-b|}⋅\sup_{\substack{z∈γ\\w∈B(a,r)}}\left|\frac1{(z-w)^2}\right|\\ &<\frac{L(γ)}{2π}⋅r ⋅\frac1{r^2}→0 \text{as }r→0 \end{aligned} - If $a=0$, then function $\frac1b$ is constant, so it is holomorphic, so the integral is 0.
If ${|a|}<{|b|}$, then $\left|-\frac ba\right|>1$, so $I\left(γ,-\frac ba\right)=0$, so the integral is 0.
If ${|a|}>{|b|}$, then $\left|-\frac ba\right|< 1$, so $I\left(γ,-\frac ba\right)=1$, so the integral is $\int_γ \frac1{a z+b} \mathrm{~d} z=\frac1a\int_γ \frac1{z+b/a} \mathrm{~d} z=\frac{2πi}a$.
- Since $I(γ, z)$ is integer-valued, it suffices to show it is a continuous function of $z$ on $ℂ∖γ^*$.
-
Suppose that $f: U → ℂ$ is holomorphic on $U ∖\{p\}$ for some $p ∈ U$, and that $f$ is bounded near $p$ (thus there are constants $r, K \in ℝ_{>0}$ such that ${|f(z)|}< K$ for all $z ∈ B(p, r)$ ). Show that if $T$ is any triangle whose interior is entirely contained in $U$ then Cauchy's theorem for a triangle still holds, that is
$$
\int_T f(z)\mathrm{\ d}z=0 .
$$
Proof 1.
By Corollary 7.17 (Riemann’s removable singularity theorem) $f$ extends to a holomorphic function on all of $U$, then Cauchy's theorem for a triangle applies.
Proof 2.
Case 1. If $p$ is a vertex of $T$, let $T=T_{pbc}$. For any $ϵ>0$, let $δ=\min\left(r,\frac ϵ{4K}\right)$. Let $B(p,δ)∩T=\{u,v\}$.
The circumference of $T_{puv}={|p-u|}+{|p-v|}+{|u-v|}≤δ+δ+2δ=4δ$.
By Cauchy's theorem for triangle, $\int_{T_{vub}}f\mathrm{~d}z=\int_{T_{vbc}}f\mathrm{~d}z=0$,
so $\int_{T_{pbc}}f\mathrm{~d}z=\int_{T_{puv}}f\mathrm{~d}z+\int_{T_{vub}}f\mathrm{~d}z+\int_{T_{vbc}}f\mathrm{~d}z=\int_{T_{puv}}f\mathrm{~d}z$
\begin{align*}\left|\int_{T_{pbc}}f\mathrm{~d}z\right|&=\left|\int_{T_{puv}}f\mathrm{~d}z\right|\\&≤K\left|\int_{T_{puv}}\mathrm{~d}z\right|&&\text{by estimation lemma and }{|f(z)|}< K\text{ for all }z∈B(p,r)\\&≤4Kδ&&\text{The circumference of }T_{puv}≤4δ\\&≤ϵ\end{align*}
Case 2. If $p∉$interior or boundary of $T$, Cauchy's theorem for a triangle applies. If $p∈$interior or boundary of $T$, let $a,b,c$ be the vertices of $T$, the integral along $T$ is the sum of integral along triangles $pab,pbc,pca$, and each of them is zero by Case 1.
- Use Cauchy's Integral Formula and the holomorphic function \(f(z)=z^n(z-a)^{-1}\), where \(a \in ℝ\) with \(a>1\), to calculate the integral: \[ \int_0^{2π}\frac{\cos nθ}{1-2a\cos θ+a^2}\mathrm{\ d}θ \] Solution.\begin{align*}\int_0^{2π}\frac{\cos nθ}{1-2a\cos θ+a^2}\mathrm{\ d}θ&=\int_0^{2π}\frac{ℜ(\cosθ+i\sinθ)^n}{(\cosθ+i\sinθ-a)(\cosθ-i\sinθ-a)}\mathrm{\ d}θ\\ &=ℜ\int_{γ(0,1)}\frac{z^n}{(z-a)(z^{-1}-a)}⋅\frac1{iz}\mathrm{\ d}z\qquad\text{Substitute }z=e^{iθ},\mathrm{\ d}θ=\frac1{iz}\mathrm{\ d}z\\ &=ℑ\int_{γ(0,1)}\frac{z^n}{(z-a)(1-az)}\mathrm{\ d}z\\ &=\frac1{a^2-1}ℑ\int_{γ(0,1)}\frac{z^n}{z-a^{-1}}-\frac{z^n}{z-a}\mathrm{d}z\quad\text{Use Cauchy's Integral Formula }I(γ,a^{-1})=1,I(γ,a)=0,\\&=\frac{2πa^{-n}}{a^2-1}\end{align*}
-
Suppose that \(D\) is a domain bounded by a contour \(C\), which we assume can be parameterized by a function \(γ_1:[0,1]→ℂ\) (that is, \(C=γ_1^*\) ). Let \(z_0 \in D\) and let \(r>0\) be small enough so that \(\bar{B}\left(z_0, r\right) \subset D\). The region \(D ∖ \bar{B}\left(z_0, r\right)\) is thus bounded by \(C \cup \partial B\left(z_0, r\right)\). Use the result of question 3 to show that if \(f\) is holomorphic on \(D ∖\left\{z_0\right\}\) then
\[
\int_{γ_1} f(z)\mathrm{\ d}z=\int_{γ_2} f(z)\mathrm{\ d}z
\]
where \(γ_2(t)=z_0+r e^{i t},(0 \leq t \leq 2 π)\).
Use this and question 6 to calculate
\[
\int_0^{2π}\frac{\mathrm{d}t}{a^2\cos^2t+b^2\sin^2t} (a,b>0)
\]
Proof 1.
The region \(D∖\bar{B}\left(z_0, r\right)\) is bounded by $γ_1,γ_2^-$, by Green's theorem $\int_{γ_1} f(z)\mathrm{\ d}z+\int_{γ_2^-} f(z)\mathrm{\ d}z=0⇒\int_{γ_1} f(z)\mathrm{\ d}z=\int_{γ_2} f(z)\mathrm{\ d}z$.
Proof 2.
There exists a piecewise $C^1$ path $ℓ:[0,1]→D∖\bar{B}\left(z_0, r\right)$ such that $ℓ(0)=γ_1(0)=γ_1(1)$ and $ℓ(1)=γ_2(0)=γ_2(1)$, then $γ=ℓ^-*γ_2^-*ℓ*γ_1$ is a simple closed piecewise $C^1$ path, and $f$ is holomorphic on interior of γ, by Cauchy's theorem, $∫_γf(z)\mathrm{\ d}z=0$. Since the integral on ℓ and $ℓ^-$ cancel, we have $\int_{γ_1} f(z)\mathrm{\ d}z=\int_{γ_2} f(z)\mathrm{\ d}z$.
Solution 1. Consider the (positively-oriented ellipse) contour $γ(t)=a\cos t+ib\sin t,t∈[0,2π]$ \[\int_γ\frac{\mathrm{d}z}z=I(γ,0)=2πi\] On the other hand,\begin{align*}\int_\gamma\frac{\mathrm dz}z&=\int_0^{2\pi}\frac{\gamma'(t)}{\gamma(t)}\,\mathrm dt\\&=\int_0^{2\pi}\frac{-a\sin(t)+b\cos(t)i}{a\cos(t)+b\sin(t)i}\,\mathrm dt\\&=\int_0^{2\pi}\frac{(b^2-a^2)\cos(t)\sin(t)}{a^2\cos^2(t)+b^2\sin^2(t)}\,\mathrm dt+i\int_0^{2\pi}\frac{ab}{a^2\cos^2(t)+b^2\sin^2(t)}\,\mathrm dt\end{align*} Taking imaginary parts, \[\int_0^{2π}\frac{\mathrm{\ d}t}{a^2\cos^2t+b^2\sin^2t}=\frac{2π}{ab}\] Solution 2.
Since $\cos^2t=\cos^2(t+π),\sin^2t=\sin^2(t+π)$, it suffices to prove \[ \int_0^π\frac{\mathrm{d}t}{a^2\cos^2t+b^2\sin^2t}=\fracπ{ab} \] Note the trigonometric identity: \[\sum_{k=0}^{n-1}\frac1{a^2\cos^2\left(\frac kn⋅π\right)+b^2\sin^2\left(\frac kn⋅π\right)}=\frac{n}{ab}⋅\frac{(a+b)^n+(b-a)^n}{(a+b)^n-(b-a)^n}\] By definition of Riemann integral:\begin{align*} \int_0^π\frac{\mathrm{d}t}{a^2\cos^2t+b^2\sin^2t}&=\lim_{n→∞}\fracπn\sum_{k=0}^{n-1}\frac1{a^2\cos^2\left(\frac kn⋅π\right)+b^2\sin^2\left(\frac kn⋅π\right)}\\ &=\lim_{n→∞}\fracπ{ab}⋅\frac{(a+b)^n+(b-a)^n}{(a+b)^n-(b-a)^n}\\ &=\fracπ{a b} \end{align*} Solution 3.
Substitute $u=\frac ba\tan t,\,\frac{\mathrm du}{\mathrm dt}=\frac ba\sec^2t$ \begin{align*} \int_0^{2π}\frac{\mathrm{d}t}{a^2\cos^2t+b^2\sin^2t}&=4\int_0^{\pi/2}\frac{\mathrm{d}t}{a^2\cos^2t+b^2\sin^2t}\\ &=4\int_0^{\pi/2}\frac{\sec^2t}{a^2+b^2\tan^2t}\,\mathrm{d}t\\&=\frac4{ab}\int_0^{\pi/2}\frac{\frac ba\sec^2t}{1+\left(\frac ba\tan t\right)^2}\,\mathrm dt\\&=\frac4{ab}\int_0^\infty\frac1{1+u^2}\,\mathrm du\\&=\frac{2\pi}{ab}\end{align*}
In the following questions, for $a \in ℂ, r \in ℝ_{>0}$ we let $γ(a, r)$ denote the positively oriented circle centred at $a$ of radius $r>0$.