Lecture 19
Consider \(C\) a contour \(z(t)\), \(\,t\in [a,b]\), \(\,z_1=z(a)\), \(\,z_2=z(b)\);
and \(f\) is pwc on \(C\).
Properties:
\((*)\) Linearity
Properties (cont):
\((*)\) Given a contour \(C\), we define \(-C\) as follows: \[ w(t) = z(-t), \quad -b\leq t \leq -a. \]
Then (check with change of parameter formula 📝): \[ \int_{-C} f\left(z\right) dz = - \int_{C} f\left(z\right) dz. \]
Hence \(C_1 - C_2 = C_1 + \left(-C_2\right)\) is defined when end point of \(C_1\) is equal to start point of \(-C_2\), which is equal to end point of \(C_2\).
Example 1: Evaluate \(I=\ds \int_C \conj{z}\, dz\) where \[C: z=2e^{i\theta}, -\pi/2\leq \theta \leq \pi /2\]
\(C\) is pwc with respect to this parametrisation (indeed differentiable) and \(f\) is continuous on \(C\).
Example (cont)
Note that \(z'(\theta) = 2 i e^{i\theta}\). Then
\(I = \ds \int_{-\pi/2}^{\pi/2} f\left( z\left(\theta \right)\right) z'\left(\theta \right) d\theta\)
\(\;\;= \ds \int_{-\pi/2}^{\pi/2} \conj{\left( 2 e^{i\theta}\right)} \cdot 2i e^{i\theta} d\theta\)
\(\;\;= 4i \ds \int_{-\pi/2}^{\pi/2} e^{-i\theta} e^{i\theta} d\theta\) \(=4\pi i\).
On \(C\), \(z\conj{z} = 4\). That is \(\conj{z} = \dfrac{4}{z}\) Then \[ \int_C \frac{dz}{z}= \pi i. \]
Let \(D\) be a domain in \(\C\) (i.e. an open, connected subset of \(\C\)).
An anti-derivative of \(f\) on \(D\) is \(F\) such that \[F'(z) = f(z)\text{ on } D.\]
Theorem: The following are equivalent
Proof: (i) \(\Rightarrow\) (ii) follows from the fundamental theorem.
(ii) \(\Rightarrow\) (iii): Let \(C\) be a closed contour in \(D\) with \(z(a) = z(b) = z_1\). Fix \(\gamma \in (a,b)\) such that \(z(\gamma)\neq z_1\) and define contours in \(D\) \[ C_1: \;\; w(t) = z(t), \quad a\leq t \leq \gamma\\ C_2: \;\; w(t) = z(t), \quad \gamma\leq t \leq b \]
(ii) \(\Rightarrow\) (iii) (cont): Then \(C_1+C_2 = C\) and
\(\ds \int_C f\) \(=\ds \int_{C_1+C_2} f\) \(=\ds \int_{C_1} f + \int_{C_2} f\)
\(\qquad=\ds \int_{C_1} f - \int_{-C_2} f\) \(=0 \), by (ii). \(\square\)
(iii) \(\Rightarrow\) (ii) \(\Rightarrow\) (i) see B/C.
Note in particular for \(C: z_1\ra z_2\) in \(D\) under (i) \(\Rightarrow\) (iii),
there folds:
\[
\int_Cf\left(z\right) dz = F(b)-F(a),
\]
where \(F\) is ANY anti-derivative of \(f\).
Further examples
Example 2: Evaluate \(I=\ds \int_{0}^{1+i} z^2\, dz\).
\(f(z)= z^2\) is continuous on \(\C\) and \(F(z) = \dfrac{z^3}{3}\) is an anti-derivative on \(\C\).
Theorem \(\Rightarrow I = F(1+i) -F(0)\) \(=\frac{2}{3} (-1+i)\).
Further examples
Example 3: Evaluate \(I=\ds \int_C \frac{dz}{z^2}\), where \(C= 2e^{i\theta}\), \(0\leq \theta \leq 2 \pi\).
\(f(z)= \dfrac{1}{z^2}\) has an anti-derivative on \(\C_*\), namely \(-\dfrac{1}{z}\), and \(C\) is a closed contour lying completely in \(\C_*\),
Theorem \(\Rightarrow I = 0\).
Indeed, same argument works to show \[ \int_C z^n \,dz = 0, \quad \forall n\in \Z\setminus \{-1\}. \]