I'm trying to determine if $\oint_C \frac{dz}{z-3}=0$ where $|z-2|=5$ and if it contradicts Cauchy's theorem. I assume it won't be $0$, and then it won't contradict Cauchy's theorem because $C$ isn't centered at the origin, it's centered at $(2,0)$. I'm stuck on evaluating the integral though. Here is what I have:
$$z=2+5e^{i\theta}, f(z)=\frac{1}{2+5e^{i\theta}-3}, dz=5ie^{i\theta}d\theta$$
$$\int_0^{2\pi} \frac{5ie^{i\theta}}{5e^{i\theta}-1}d\theta$$ I don't know what to do with this.
Convert to real integrals.
\begin{align*}\DeclareMathOperator{\i}{i\!} \DeclareMathOperator{\rme}{e} \DeclareMathOperator{\diff}{d} I &= \int_0^{2\pi} \frac {\i 5 \rme^{\i t}} {5\rme^{\i t} - 1} \diff t \\ &= \i \int_0^{2\pi} \frac {5 \rme^{\i t} (5\rme ^{-\i t} - 1)} {(5\rme^{\i t} -1)(5\rme^{-\i t} - 1)} \diff t\\ &= \i \int_0^{2\pi} \frac {25 - 5 \rme^{-\i t}}{26-10\cos(t)}\diff t \\ & = \i \int_0^{2\pi} \frac {25-5\cos(t) + 5\i \sin(t)}{26-10\cos(t)}\diff t \\ &= \i \int_0^{2\pi} \frac {25 - 5\cos(t)}{26-10\cos(t)}\diff t - \int_0^{2\pi} \frac {5\sin(t)} {26-10\cos(t)} \diff t \\ &=: \i I_1 + I_2. \end{align*} For $I_2$, note that $\diff(-\cos(t)) = \sin(t)$, thus $$ I_2 = \frac 12 \int_0^{2\pi} \frac {\diff(13-5\cos(t))}{13-5\cos(t)} =\frac 12 \log(13 -5 \cos(t))|_0^{2\pi} = 0. $$ For $I_1$: $$ I_1 = \pi + \frac 12\left(\int_0^{\pi} + \int_\pi^{2\pi} \right)\frac {12 \diff t}{13-5\cos(t)}, $$ while $$ \int_\pi^{2\pi} \frac {12\diff t}{13-5\cos(t)} = \int_\pi^0 \frac {12}{13 -5 \cos(2\pi -u)} \diff (2\pi - u) = \int_0^\pi \frac {12\diff u}{13- 5\cos(u)}, $$ thus \begin{align*} I_1 &= \pi + 12\int_0^\pi \frac {\diff t}{13-5\cos(t)} \\ &= \pi +24 \int_0^\pi \frac {\sin(t/2)^2 + \cos(t/2)^2} {8\cos(t/2)^2 + 18\sin(t/2)^2} \diff (t/2)\\ &= \pi + 24 \int_0^{\pi/2} \frac {\diff (\tan(u))}{8 + 18 \tan(u)^2} \\ &= \pi + 3 \cdot \frac 23\int_0^{\pi/2} \frac {\diff (3\tan(u)/2)}{1 + (3\tan(u)/2)^2}\\ &= \pi + 2 \cdot \frac \pi 2 = 2\pi. \end{align*}
Therefore $I= \i I_1 + I_2= \i 2\pi$. This does not contradict Cauchy theorem, because if we let $\varGamma$ be $\{z \colon |z - 3| = 1\}$ with the clockwise orientation then $\varGamma \cup C$ encloses an region on which $1/(z-3)$ is analytic, so the integral is actually $I= \int_{\varGamma^-} \diff z/(z -3)$ by Cauchy theorem, which is $$ I = \int_0^{2\pi} \frac {\i \rme^{\i t}}{\rme^{\i t}} \diff t = \i 2\pi $$ that agree with the result from direct computation.