Let $f$ be holomorphic on $\mathbb{D}=\{z\in\mathbb{C} | |z|\leq 1\}$. Prove the following. I'm doing this exercise:

Let $f$ be holomorphic on $\mathbb{D}=\{z\in\mathbb{C} | |z|\leq 1\}$. Prove that:
a) $\frac{1}{2\pi i}\cdot \displaystyle\int_{|z|=1}{\frac{\bar{f}(z)}{z-a}dz}=\bar{f}(0)\ $ if $|a|<1$
b)  $\frac{1}{2\pi i}\cdot \displaystyle\int_{|z|=1}{\frac{\bar{f}(z)}{z-a}dz}=\bar{f}(0)-\bar{f}(1/\bar{a})\ $ if $|a|>1$

I know that if we call $\gamma=\partial D(0,1)$ the unit disk boundary, we have:
$$ \frac{1}{2\pi i}\cdot \int_{\gamma}{\frac{\bar{f}(z)}{z-a}dz}=\bar{f}(a)Ind(\gamma,a)$$
and we know that $Ind(\gamma,a)=1$, so
$$ \frac{1}{2\pi i}\cdot \int_{\gamma}{\frac{\bar{f}(z)}{z-a}dz}=\bar{f}(a)$$
But how can I prove that $\bar{f}(a)=\bar{f}(0)$?
Thanks for your time.
 A: Because $|a|<1$ then $\left|\frac{a}{z}\right|<1$ for $|z|=1$ and we can write
$$\frac{1}{z-a}=\frac{1}{z}\frac{1}{1-\frac{a}{z}}=
\frac{1}{z}\left(\sum\limits_{k=0}\left(\frac{a}{z}\right)^k\right)=
\sum\limits_{k=0}\frac{a^k}{z^{k+1}}$$
Now $\forall n>0$
$$\int\limits_{|z|=1}\frac{1}{z^{n}(z-a)}dz=
\sum\limits_{k=0}\left(a^k\int\limits_{|z|=1}\frac{1}{z^{n+k+1}}dz\right)=0 \tag{1}$$
which is easy to see from Cauchy's integral formula for $g^{(k)}(0)=\frac{k!}{2\pi i}\int\limits_{|z|=1}\frac{g(z)}{z^{k+1}}dz$ for $g(z)=1$ constant. As a result (from @glofatics comments)
$$\int\limits_{|z|=1}\frac{\overline{f(z)}}{z-a}dz=
\int\limits_{|z|=1}\left(\sum\limits_{n=0}\frac{\overline{a_n}}{z^n}\right)\left(\sum\limits_{k=0}\frac{a^k}{z^{k+1}}\right)dz \overset{(1)}{=}
\int\limits_{|z|=1}\overline{a_0}dz=
2\pi \overline{f(0)}$$

Because $|a|>1$ then $\left|\frac{z}{a}\right|<1$ for $|z|=1$ and we can write
$$\frac{1}{z-a}=-\frac{1}{a}\frac{1}{1-\frac{z}{a}}=
-\frac{1}{a}\left(\sum\limits_{k=0}\left(\frac{z}{a}\right)^k\right)=
-\sum\limits_{k=0}\frac{z^k}{a^{k+1}}$$
Now $\forall n>0$
$$\int\limits_{|z|=1}\frac{1}{z^{n}(z-a)}dz=
-\sum\limits_{k=0}\left(\frac{1}{a^{k+1}}\int\limits_{|z|=1}z^{k-n}dz\right)=
-2\pi \frac{1}{a^n}\tag{2}$$
(more details here) and it's $0$ for $n=0$ since $1=|z|<|a|$. As a result
$$\int\limits_{|z|=1}\frac{\overline{f(z)}}{z-a}dz=
-\int\limits_{|z|=1}\left(\sum\limits_{n=0}\frac{\overline{a_n}}{z^n}\right)\left(\sum\limits_{k=0}\frac{z^k}{a^{k+1}}\right)dz \overset{(2)}{=}
-2\pi \sum\limits_{n=1}\frac{\overline{a_n}}{a^n}=\\
2\pi \overline{f(0)}-2\pi \sum\limits_{n=0}\frac{\overline{a_n}}{a^n}=
2\pi \overline{f(0)}-2\pi \overline{\sum\limits_{n=0}\frac{a_n}{\overline{a}^n}}=
2\pi \overline{f(0)}-2\pi \overline{f\left(\frac{1}{\overline{a}}\right)}$$
A: No series developments are needed. Let
$$Q:={1\over 2\pi i}\int_\gamma{\overline{f(z)}\over z-a}\>dz\ .$$
Then
$$\bar Q={-1\over 2\pi i}\int_\gamma{f(z)\over\bar z-\bar a}\>d\bar z\ .$$
Now along $\gamma$ we have $\bar z={1\over z}$ and therefore $d\bar z=-{1\over z^2}\>dz\ .$ It follows that
$$\bar Q={1\over2\pi i}\int_\gamma{f(z)\over z(1-\bar a z)}\>dz={1\over2\pi i}\int_\gamma f(z)\left({1\over z}-{1\over z-1/\bar a}\right)\>dz\ .$$ From this we conclude that  $$|a|<1\ \Rightarrow \ \bar Q =f(0),\qquad |a|>1\ \Rightarrow \ \bar Q =f(0)-f(1/\bar a)\ .$$
A: Following up on a comment by @glofatics.
$f$ is holomorphic, so it is analytic and you can write $f(z)=\sum_n a_n z^n$ and so $\overline{f}(z)=\sum_n \overline{a_n} \cdot \overline{z}^n$.
By uniform convergence, $$\int_{|z|=1} \frac{\overline{f}}{z-a}=\sum_n \overline{a_n}\int_{|z|=1}\frac{\overline{z}^n}{z-a}dz$$
Take a look at $g_n(z)=\frac{\overline{z}^n}{z-a}$.
For $|z|=1$, $\overline{z}=\frac{1}{z}$, and so $\int_{|z|=1}g_n dz=\int_{|z|=1}h_n(z)dz$ for $h_n(z)=\frac{1}{z^n(z-a)}$.
$h_n$ is holomorphic, up to some poles, and so we can apply the residue theorem.
If $|a|<1$ then we have two poles in $B(0,1)$ which are $0,a$ of order $n,1$ respectively. 
Let us calculate residues.


*

*First, $Res_{z=a} h_n =a^{-n}$, for all $n\geq0$.

*It is not too difficult to see that for $n\geq1$ we have $$Res_{z=0} h_n =(-1)^{n-1}\cdot\frac{1}{(-a)^n}=-a^{-n}$$
This is because $$Res_{z=0} h_n=\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}(z^n h_n(z))$$ for pole of order $n$.
For $n=0$, it is clear that $\int_{|z|=1|}h_n(z)dz=2\pi i$ by Cauchy's formula.
So, to sum up, $\int_{|z|=1}h_n(z)=2\pi i (Res_{z=0}+Res_{z=a})$ which is $0$ if $n\geq1$ and $2\pi i$ if $n=0$.
Therefore, we obtained that the integral from the beginning is 
$$\sum_n \overline{a_n}\int_{|z|=1}h_n dz=\overline{a_0} \cdot 2\pi i=\overline{f(0)} \cdot 2\pi i$$
which is what you wanted in $1$. Similar solution will probably give $2$.
