How to apply the Cauchy-Goursat theorem for contour integration?

**Simply Connected:** A domain where any closed curve can be continuously shrunk to a point without leaving the domain. A disk is simply connected; an annulus (ring) is not — a curve around the hole cannot be shrunk to a point. **What happens with singularities inside $C$:** The Cauchy-Goursat theorem fails if singularities are inside the contour. Instead, we use: 1. **Deformation of contours:** If $C_1$ and $C_2$ are two closed contours in a domain where $f$ is analytic, and $C_1$ can be continuously deformed into $C_2$, then $\oint_{C_1} f = \oint_{C_2} f$. 2. **Cauchy's Integral Formula:** If $f$ is analytic inside and on $C$, then: $$f(a) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - a} \, dz$$ **Example: $\oint_C \frac{1}{z} \, dz$, $C$ is the unit circle.** Parametrise $z = e^{i\theta}$, $dz = ie^{i\theta} d\theta$: $$\oint_C \frac{1}{z} \, dz = \int_0^{2\pi} \frac{ie^{i\theta}}{e^{i\theta}} \, d\theta = \int_0^{2\pi} i \, d\theta = 2\pi i$$ This non-zero result tells us $1/z$ has a singularity (a simple pole) at $z = 0$ inside the contour, so Cauchy-Goursat doesn't apply. **Principle of Deformation of Contours:** If $f$ is analytic in a region between two closed curves $C_1$ and $C_2$ (where $C_2$ is inside $C_1$), then: $$\oint_{C_1} f(z) \, dz = \oint_{C_2} f(z) \, dz$$ This is extremely useful: we can deform any contour enclosing singularities into small circles around each singularity and sum the contributions.