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

- \(\alpha \in \C:\) \(\ds \int_C \left(\alpha \, f \right) \left(z\right) dz = \alpha \int_C f \left(z\right) dz \).
- \(\ds\int_C \left(\, f + g\right) \left(z\right) dz = \int_C f \left(z\right) dz + \int_C g \left(z\right) dz\).
- \(C_1+ C_2\) defines a contour when end point of \(C_1\) is equal to start point of \(C_2\).

**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

- \(f\) has an anti-derivative on \(D\);
- For any \(z_1,z_2 \in D\) and any contour \(C\) from \(z_1\) to \(z_2\) in \(D\), we have that \(\int_C f\left(z\right) dz\) is independent of \(C\);
- For any closed contour \(C\) in \(D\) there holds \[\int_C f\left(z\right)dz = 0.\]

**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 \[ \int_C z^n \,dz = 0, \quad \forall n\in \Z\setminus \{-1\}. \]