MTH2003 Complex Analysis Pre-Lecture 6

• Lecture: MTH2003 Complex Analysis Lecture 6 (Provided Notes).
• Practical: MTH2003 Complex Analysis Practical 3 (Practical Sheet).

Finding Complex Integrals

Let $f$ is a continuous complex function, $L$ the length of the contour $C$ and $M$ the upper bound for $|f|$ on $C$, then…

$$\left|{\int }_{c}f(z)dz\right|\le ML$$

This is called the Estimation Lemma and can be proven directly using properties of integrals.

Further than just estimation, we can evaluate contour integrals using paths…

$${\int }_a^{b}f(\gamma (t))\cdot {\gamma }^{\prime }(t)dt$$

or, by converting $z\to u+iv$…

$$\begin{array}{rl}{\int }_{\gamma }f(z)dz& ={\int }_{\gamma }[u+iv](dx+idy)\\ & ={\int }_{\gamma }[(udx-vdy)+i(vdx+udy)]\\ & ={\int }_{\gamma }(udx-vdy)+i{\int }_{\gamma }(vdx+udy)\end{array}$$

For example, to find the integral of $f(z)=z^2$ around the contour $c=c_1+c_2+c_3$, where $c_1:0\to 1$, $c_2:1\to 1+i$, and $c_3:1+i\to 0$…

insert the two aforementioned methods.

Similarly, we can integrate $\bar{z}$ around a circular contour of radius $R$ centred on the origin, $|z|=R$…

insert polar method, stating dependence on radius.

And finally, we could integrate $z^{-1}$ around a circular contour of radius $R$ centred on the origin, $|z|=R$…

insert method, stating lack of dependence on radius.

These properties insert explanation of why they’re interesting with relation to Cauchy’s Theorem.

This Theorem however relies on another from real analysis…

Green’s Theorem

Let the real functions $P(x,y)$ and $Q(x,y)$, along with all their partial derivatives, be continuous in the closed set $R$ consisting of a simple closed contour $\gamma$ plus its interior. Let $\gamma$ be described in the positive (anti-clockwise) direction. Then,

$${\oint }_{\gamma }(Pdx+Qdx)={\iint }_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy$$

which is a relationship between the line integral (${\oint }_{\gamma }$) and the are integral ${\iint }_R$. Simple closed contour simply means that the curve intersect itself only at its endpoints, i.e. not a figure-eight.

This theorem basically means that the integral along the perimeter is equal to the integral of the area. With non-simple contours, we can split them into multiple separate simple contours, as any interior lines would be traversed in both directions, hence cancelling.

insert proof of Green’s Theorem, starting by focusing on the 2nd RHS term.