Line integral with Mobius transformation 
Corollary: If $f:G\to \mathbb{C}$ is analytic and $\overline{B}(a;r)\subset G$
  then $$f^{(n)}(a)=\dfrac{n!}{2\pi i} \int
 \limits_{\gamma}\dfrac{f(w)}{(w-a)^{n+1}}dw,$$ where
  $\gamma(t)=a+re^{it}$, $0\leq t\leq 2\pi.$

Using the above corollary find the following integral: $$\int \limits_{\gamma}\dfrac{dz}{(z-\frac{1}{2})^n},$$ where $n\geq 1, \gamma(t)=e^{it}, t\in [0,2\pi]$.
Remark: I know that this problem can be solved easily using residues but the purpose of this problem is use the above corollary.
My approach: Define the mapping $w=T(z)$ with $T(z)=\dfrac{2z-1}{z-2}$ then we know that this maps unit disk onto unit disk with $T(1/2)=0$. Also we can show that  $z=T^{-1}(w)=\dfrac{2w-1}{w-2}$.
Let's make substitution in our integral and we get the following: $$\int \limits_{\gamma}\dfrac{d(\frac{2w-1}{w-2})}{\left(\dfrac{2w-1}{w-2}-\frac{1}{2}\right)^n}=\int \limits_{\gamma}\dfrac{2(w-2)-(2w-1)}{(w-2)^2} : \left( \dfrac{4w-2-w+2}{2(w-2)}\right)^ndw=$$$$=-2^n\int \limits_{\gamma} \dfrac{(w-2)^{n-2}}{w}dw =$$
If $f(w)=(w-2)^{n-2}$ then it clearly analytic on unit disk  and we can apply our corollary: $$=-2^n\times 2\pi i \times f(0)=(-1)^{n-1}2^{n+1}\pi i.$$
Can anyone please check it and say is my solution correct? Especially the answer.
Would be very grateful for help!
 A: This is really about complex integrals of homotopic curves. The following
is a proof based on a simple version of Cauchy's integral theorem 
applied to this specific case.
Excuse the rude picture. It is easier to describe the curves below using
a diagram than with symbols.
$\lambda$ is a circle of radius ${1}$ centered on ${0}$.
$\gamma$ is a circle of radius ${1\over 4}$ centered on ${1 \over 2}$.
Note: After drawing the pictures, I realised that I should have swapped
$\gamma,\lambda$.
Let $\epsilon$ be a small real number. Let $l_\epsilon$ be the line (curve) joining the points where the line $\operatorname{im} z = \epsilon$
intersects the circles $\lambda, \gamma$. Let $\gamma_\epsilon, \lambda_\epsilon$ be the portion of the circles $\lambda, \gamma$ that start and end where the lines $l_\epsilon, l_{-\epsilon}$ intersect
the circles $\lambda, \gamma$.

Suppose $h$ is a continuous function defined on some open $U$,
suppose $h$ has an antiderivative $H$, that is, $H' = h$, and
suppose $c$ is a closed, piecewise differentiable curve in $Y$,
then the fundamental theorem of calculus shows that $\int_c h dz  = 0$.
Let $g_n(z) = {1 \over (z-{1\over 2})^n}$, for $n =1,2...$. Note that
the $g_n$ are smooth on $\{ {1 \over 2} \}^c$.
Also note that
$z \mapsto \log (z-{1 \over 2})$ is an antiderivative of $g_1$ and it
is straightforward to compute antiderivatives of $g_2,g_3$, etc. Note that
these are all smooth on $U=\mathbb{C} \setminus (-\infty,{1 \over 2}]$.`
In particular, $\int_{\gamma_\epsilon+(-l_\epsilon)+(-\gamma_\epsilon)+l_{- \epsilon}} g_n dz = 0$.
A small amount of work shows that $\lim_{\epsilon \to 0} \int_{\gamma_\epsilon} g_n dz = \int_{\gamma} g_n dz$, similarly for $\lambda, \lambda_\epsilon$ and also
$\lim_{\epsilon \to 0} (\int_{-l_\epsilon} g_ndz + \int_{l_{-\epsilon}} g_ndz) = 0$.
Combining these shows that $\int_\gamma g_n dz = \int_\lambda g_n dz$.
Hence $\int_\lambda g_n dz = \int_\gamma {1 \over (z - {1\over 2})^n} dz = {2 \pi i \over (n-1)!} 1^{(n-1)}({1 \over 2})$, where I am abusing notation by using $1$ to represent the function $z \mapsto 1$.
In particular, the value is $2 \pi i$ for $n=1$ and zero otherwise.
