Concluding the solutions of two complex integrals from one. I am trying to solve the following question:
Compute the following integrals:
$$ 
\int_{|z| = 1} \frac{|dz|}{|z - \frac{i}{2}|^2}\;,\quad
 \int_{|z| = 1} \frac{dz}{(z - \frac{i}{2})^2}\;,\quad
 \int_{|z| = 1} \frac{dz}{(z - \frac{i}{2})}
$$
I am wondering, how can finding the solution of one of them leads to finding the solutions of the others? Or each integration should be solved separately? But I think there are some relations among these three integrals.
What is the importance of the absolute value in $|dz|$ in the first integral?
Any details about how to tackle this problem will be greatly appreciated!
 A: *

*(a) Applying substitution
$$\;z=e^{it},\quad t\in(0,2\pi),\quad|\text dz|=|ie^{it}\,\text dt|=2\,\text dt,\tag1$$
(pointed in the OP comments), one can get
$$I_a=\int\limits_0^{2\pi}\dfrac{\text dt}{\left(\frac12-\sin2t \right)^2+\cos^22t}
=4\int\limits_0^{2\pi}\dfrac{\text dt}{5-4\sin t}
=4\int\limits_0^\pi\dfrac{10\,\text dt}{25-16\sin^2 t}$$
$$=\dfrac{80}9\int\limits_0^{\large\frac\pi2}\dfrac{\text dt}{\left(\tan^2t+\dfrac{25}9\right)\cos^2t}\,
=\dfrac{16}{3}\arctan\left(\dfrac{3\tan t}5\right)\bigg|_0^{\large\frac\pi2},$$
$$I_a=\dfrac83\,\pi.\tag2$$


*(b) Applying substitution
$$z=\dfrac12w,\quad \text dz=\dfrac12\,\text dw.\tag3, $$
one can get
$$I_b(n)=\int\limits_{|z|=1} \dfrac{\text dz}{\left(z-\frac i2\right)^n}
=2^{n-1}\int\limits_{|w|=2} \dfrac{\text dw}{\left(w-i\right)^n},$$
wherein
$$f_n(w)=\dfrac1{\left(w-i\right)^{n}}=c_{n,0}+\dfrac{c_{n,-1}}{w-i}+\dfrac{c_{n,-2}}{(w-i)^2}+\dots.$$
Then
$$I_b(n)=2^{n-1}\cdot2\pi i c_{n,-1},$$
$$I_b(1)=2\pi i,\quad I_b(2)=0.\tag4$$
A: For any smooth curve $\gamma:[a,b]\to\mathbb{C}$ and a holomorphic function $f:U\to\mathbb{C}$ where $U\subset\mathbb{C}$ is an open connected set that contains the image of the curve, the integral $\int_\gamma f(z)|dz|$ is defined as:
$$
\int_\gamma f(z)|dz|:=\int_a^b f(\gamma(t))|\gamma'(t)|dt
$$
So your first integral is
$$
\begin{align}
&\int_{0}^{2\pi}\frac{1}{|e^{it}-\frac{i}{2}|^2}|ie^{it}|dt
=\int_0^{2\pi}\frac{1}{|e^{it}-\frac{i}{2}|^2}dt \\
&=\int_{0}^{2\pi}\frac{1}{\cos^2t+(\sin t+\frac12)^2}dt
=\int_{0}^{2\pi}\frac1{\frac54+\sin t}\;dt
=\frac{8\pi}{3}
\end{align}
$$
where you can calculate by the substitution $u=\tan(t/2)$.
The second and the third ones are trivial by the residue theorem because they are already in the form of the Laurent series (only one term!) and you can read the coefficient of $\dfrac{1}{z-i/2}$ in the expansion directly. So
$$
\int_{|z|=1}\frac{dz}{(z-i/2)^2}=2\pi i\cdot 0=0\quad
\int_{|z|=1}\frac{dz}{z-i/2}=2\pi i\cdot 1=2\pi i
$$
