Complex integration using the FTC? How would you integrate to show $$\int_{\partial B(1,2)} \frac{1}{(z-2)^3}dz = 0$$ where $B(1,2)$ is a ball centre 1, radius 2 in the complex plane. Apparently you're meant to use the FTC but I can't see how to use the FTC in the complex plane? 
Thanks!
 A: First way: Using Cauchy's differentiation formula $$f^{(n)}(a)=\frac{n!}{2\pi i}\oint_\gamma\frac{f(z)}{(z-a)^{n+1}}\,dz$$ we get that$$\oint_{|z-1|=2}\frac{1}{(z-2)^3}\,dz=\left.\frac{2\pi i}{2!}\,\frac{\mathcal d^2}{\mathcal d z^2}(1)\,\,\right|_{z=2}=0$$
Second way: Making first the translation $\,z\to z+1\,$ , we take the circle of radius $\,2\,$ centered at the origin and $\,(z-1)^3\,$ in the denominator, getting $$I:=\oint_{|z|=2}\frac{1}{(z-1)^3}\,dz$$
Now, we put $\,z=2e^{i\theta}\,\,,\,0\leq\theta\leq 2\pi\Longrightarrow dz=2ie^{i\theta}\,$ , so we get$$I=\int_0^{2\pi}\frac{2ie^{i\theta}d\theta}{\left(2e^{i\theta}-1\right)^3}=\int_0^{2\pi}\frac{(2e^{i\theta})}{\left(2e^{i\theta}-1\right)^3}=\left.-\frac{1}{2}\frac{1}{(2e^{i\theta}-1)^2}\right|_0^{2\pi}=$$$$=-\frac{1}{2}\left(\frac{1}{(2-1)^2}-\frac{1}{(2-1)^2}\right)=0$$
A: We are told to compute the given integral $Q$ without the benefits of the Cauchy theorem. The circle in question can be parametrized as
$$\gamma:\quad \phi\mapsto 1+2e^{i\phi}\qquad(0\leq\phi\leq2\pi)\ ,$$
which gives $dz=2i\,e^{i\phi}\,d\phi$. Therefore we have
$$Q=\int_0^{2\pi}{2i e^{i\phi}\over \bigl(2e^{i\phi}-1\bigr)^3}\ d\phi\ .\qquad(*)$$
Note that the denominator in the last integral is $\ne0$ throughout.
The function ${1\over(z-2)^3}$ appearing in the definition of $Q$ has a primitive, namely $-{1\over 2(z-2)^2}$. As a consequence the complex function of the real variable $\phi$ appearing in $(*)$ also has a simple primitive. In fact it is easily checked that
$$Q=-{1\over2\bigl(2e^{i\phi}-1\bigr)^2}\Biggr|_0^{2\pi}\ ,$$
and since $e^{i\phi}$ assumes the same value at $0$ and at $2\pi$ the  difference on the right hand side is $0$.
A: FTC way:
$$
\frac{1}{(z-2)^3} = \frac{d}{dz} \frac{-1}{2(z-2)^2}
$$
everywhere on your curve, so to get the value of the integral along the curve, evaluate $-1/(2(z-2)^2))$ at the two endpoints and subtract.  Of course if it is a closed curve, that difference is zero.
A: No one has mentioned residues, so I will throw that in, too.
Let $w=z-2$, and the integral becomes
$$
\oint_{\partial B(-1,2)}\frac{1}{w^3}\,\mathrm{d}w
$$
The coefficient of the $\dfrac1w$ term in the Laurent series is $0$, so the residue at the singularity at $w=0$ is $0$ even though the path of integration, $|w+1|=2$, circles it. Thus,
$$
\oint_{\partial B(1,2)}\frac{1}{(z-2)^3}\,\mathrm{d}z=0
$$
