Notes - Complex Analysis MT23, Cauchy's theorem
Flashcards
What three steps do we use to show Cauchy’s theorem (in the context of complex analysis)?
- Show Cauchy’s theorem for triangular domains
- Show Cauchy’s theorem for star-like domains
- Show Cauchy’s theorem for any domain
Can you state the full version of Cauchy’s theorem?
Suppose
- $f : U \to C$
- $U$ is a simply-connected domain
- $f$ holomorphic
- $\gamma$ closed path laying entirely in interior of $U$
Then
\[\int_\gamma f(z) \text dz = 0\]What does it mean for a set $X$ to be star-like with respect to some point $z _ 0 \in X$?
For every $w \in X$, the line segement $[z _ 0, w]$ joinging $z _ 0$ and $w$ lies in $X$.
How can you relate convex and star-like domains?
Convex domains are star-like with respect to every point.
Homotopy forms of Cauchy’s theorem
Can you state the homotopy form of Cauchy’s theorem for parths $\gamma$ and $\eta$, and why this is significantly more powerful than the versions for triangular and star-like domains?
Suppose
- $U$ is a domain in $\mathbb C$
- $f : U \to \mathbb C$ holomorphic
- $a, b \in U$
- $\gamma, \eta$ paths from $a$ to $b$
- $\gamma, \eta$ are homotopic
Then
\[\int_\gamma f(z) \text dz = \int_\eta f(z) \text{d}z\]This is better since $U$ doesn’t even have to be a domain where $f$ has a primitive (all previous proofs have involved finding a primitive for $f$ on the domain).
Can you state the simply-connected domain version of Cauchy’s theorem for paths $\gamma$ and $\eta$?
Suppose
- $U$ is a simply connected domain in $\mathbb C$
- $f : U \to \mathbb C$ is holomorphic
- $a, b$ in $U$
- $\gamma, \eta$ are paths from $a$ to $b$
Then
\[\int_\gamma f(z) \text dz = \int_\eta f(z) \text{d}z\]Can you state the simply-connected domain version of Cauchy’s theorem for a single closed path $\gamma$?
Suppose
- $U$ is a simply connected domain in $\mathbb C$
- $f : U \to \mathbb C$ is holomorphic
- $\gamma$ closed path
Then
\[\int_\gamma f(z) \text dz = 0\]Quickly prove, by appealing to the more general homotopy form of Cauchy’s theorem, the simply-connected domain version of Cauchy’s theorem for paths $\gamma$ and $\eta$, i.e. suppose
- $U$ is a simply connected domain in $\mathbb C$
- $f : U \to \mathbb C$ is holomorphic
- $a, b$ in $U$
- $\gamma, \eta$ are paths from $a$ to $b$
then
\[\int_\gamma f(z) \text dz = \int_\eta f(z) \text{d}z\]
$U$ is simply connected, hence any two paths $\gamma$ and $\eta$ are homotopic. Then by Cauchy’s theorem for homotopic paths, the integrals must be equal.
What correspondence is there between primitive domains (there exist antiderivatives for holomorpic functions) and simply-connected domains?
They are the same, i.e.
\[U \text{ primitive} \iff U \text{ simply-connected}\]Suppose
- $U \subseteq \mathbb C$ open
- $T$ triangle whose interior is contained in $U$
- $f : U \to \mathbb C$ holomorphic
Quickly prove Cauchy’s theorem for a triangular domain, i.e. that then:
\[\int_T f(z) \text dz = 0\]
Suppose for a contradiction that
\[I := \left| \int_T f(z) \text d z \right| > 0\]The idea is to build a sequence of triangles around which the integral is still large despite the triangles getting smaller and smaller.
Let $T^0 = T$, and suppose we have constructed $T^i$ for all $0 \le i < k$. Then define $T^k$ by taking $T^{k-1}$ and joining the midpoints to form 4 smaller triangles.
Then note that
\[\int_{T^{k-1}\\,} f(z) \text dz = \sum^4_{i = 1} \int_{S_i} f(z) \text dz\]since the paths about the middle triangle cancel out.
Then, defining
\[\begin{aligned} I_k &= \left| \int_{T_{k-1}\\,} f(z) \text dz\right| \\\\ &= \left| \sum^4_{i = 1} \int_{S_i} f(z) \text dz \right| \\\\ &\le \sum^4_{i = 1} \left|\int_{S_i} f(z) \text dz\right| \end{aligned}\]Then $\exists i$ such that
\[\left| \int_{S_i} f(z) \text dz \right| \ge \frac{I_{k-1}\\,}{4}\]Set $T _ k$ to be this triangle, i.e. $T _ k = S _ i$. Then by induction:
\[\begin{aligned} \ell(T^k) &= 2^{-k} \ell(T) \\\\ \text{diam}(\mathcal T^k) &= 2^{-k} \text{diam}(\mathcal T) \end{aligned}\]where $\mathcal T^i$ represents the solid triangle bounded by the region $T^i$.
Then note that $\text{diam}(\mathcal T^k) \to 0$, and since the sets are nested, Cantor’s intersection theorem implies there is a unique point in every $\mathcal T^i$, say $z _ 0$.
Since $f$ is holomorphic at $z _ 0$,
\[f(z) = f(z_0) + f'(z_0) (z - z_0) + \varepsilon(z)(z-z_0)\]where $\varepsilon(z) \to 0$ as $z \to z _ 0$. Since $\varepsilon$ is continuous (it is in fact holomorphic), it is integrable on $U$. Note then that
\[\begin{aligned} I_k &= \left|\int_{T_k} f(z) \text dz\right| \\\\ &= \left|\int_{T_k} f(z_0) + f'(z_0) (z - z_0) + \varepsilon(z)(z-z_0) \text dz \right| \\\\ &= \left|\int_{T_k} f(z_0) + f'(z_0) z - z_0f'(z_0) + \varepsilon(z) (z-z_0) \text dz \right|\\\\ &= \left|\int_{T_k} \varepsilon(z)(z-z_0) \text d z \right| \end{aligned}\](where we can remove the preceeding terms since they are just linear and so have a primitive). Then, by the estimation lemma,
\[\begin{aligned} I_k &= \left| \int_{T_k} (z-z_0) \varepsilon(z) \text dz \right| \\\\ &\le \ell(T^k) \cdot \sup_{z \in T_k} |(z-z_0)\varepsilon(z)| \\\\ &= \ell(T^k) \cdot \text{diam}(\mathcal T^k) \cdot \sup_{z \in T^k} |\varepsilon(z)| \\\\ &= (2^{-k} \ell(T)) \cdot (2^{-k} \text{diam}(\mathcal T)) \cdot \sup_{z \in T^k} |\varepsilon(z)| \\\\ &= 4^{-k} \ell(T) \text{diam}(\mathcal T) \cdot \sup_{z \in T^k} |\varepsilon(z)| \end{aligned}\]Since $\varepsilon(z) \to 0$ as $z \to z _ 0$, $\sup _ {z \in T^k} \vert \varepsilon(z) \vert \to 0$ as $k \to \infty$. Hence,
\[\begin{aligned} 4^kI^k &= \ell(T) \text{diam}(\mathcal T) \cdot \sup_{z \in T^k} |\varepsilon(z)| \\\\ &\to 0 \end{aligned}\]as $k \to \infty$. But then
\[4^k I_k \ge I > 0\]a contradiction.
Suppose:
- $f : U \to \mathbb C$ is holomorphic
- $U$ star-like domain
- $\gamma$ closed path in $U$
Quickly prove Cauchy’s theorem for star-like domains, i.e. that then:
\[\int_\gamma f(z) \text dz = 0\]
(you can assume Cauchy’s theorem for triangular domains).
The approach is to show that $f$ has a primitive on $U$ (which implies the integral is zero, by the Fundemantal Theorem of Calculus).
Let $a$ be a point for which $\forall z \in U$, $[z, a] \subseteq U$ (i.e. the line segement joining $z$ and $a$ doesn’t leave $U$). This point must exist, by the assumption that $U$ is star-like.
Define
\[\gamma_z(t) := a + t(z-a)\]and we aim to show that a primitive for $f$ is given by
\[F(z) = \int_{\gamma_z} f(w) \text dw\]We want to be able to evaluate the limit $\lim _ {z \to z _ 0} \frac{F(z _ 0) - F(z)}{z _ 0 - z}$, so we need some information about $f$ close to $z _ 0$. Take $\varepsilon > 0$ such that $B(z _ 0, \varepsilon) \subseteq U$. Then if $z \in B(z _ 0, \varepsilon)$, note that the triangle $T$ with vertices $a, z, z _ 0$ lies entirely in $U$, by assumption that $U$ is star-like with respect to $a$.
(this image was from before I relabelled to match the proof in
[[Notes - Complex Analysis MT23, Complex integration]]U, if bold variables are the ones in the diagram, then $\pmb{z _ 0}$ should be $a$, $\pmb z$ should be $z _ 0$ and $\pmb w$ should be $z$).
Define
\[\eta(t) = z + t(z_0 - z)\](i.e. the straight line path from $z$ to $z _ 0$), we have that
\[T = \gamma_{z_0}^- \star \gamma_z \star \eta\]is a triangle contained in $U$. Then
\[\int_T f(w) \text dw = 0\]so
\[\begin{aligned} &-\int_{\gamma_{z_0} } f(w) \text dw + \int_{\gamma_z} f(w) \text dw + \int_{\eta} f(w) \text dw = 0 \\\\ \implies&\int_{\gamma_z} f(w) \text dw -\int_{\gamma_{z_0} } f(w) \text dw = \int_{\eta} f(w) \text dw \\\\ \implies&F(z) - F(z_0) = \int_\eta f(w) \text dw\end{aligned}\](this looks like it is the wrong way around/there is a sign error, but this is actually right, since when we expand out the parameterisation, the derivative of $\eta$ is actually $z - z _ 0$, not $z - z _ 0$).
So
\[\begin{aligned} \left| \frac{F(z_0) - F(z)}{z_0 - z} - f(z_0) \right| &= \left| \frac{-1}{z_0-z} \int_\eta f(w) \text dw - f(z_0) \right| \\\\ &= \left| \frac{-(z-z_0)}{z_0 - z} \int^1_0 f(z_0 + t(z-z_0)) \text dt - f(z_0) \right| \\\\ &= \left| \int^1_0 f(z_0 + t(z - z_0)) - f(z_0) \text dz \right| \\\\ &\le \sup_{t \in [0, 1]} |f(z_0 + t(z - z_0)) - f(z_0) | \end{aligned}\]since $f$ is continuous at $z _ 0$, this tends to zero as $z \to z _ 0$, so
\[F'(z_0) = f(z_0)\]Hence $f$ is primitive on $U$, so any integral on a closed curve will be $0$.