Evaluate the following complex integral $\frac{1}{2\pi i}\int_{|z|=1}\frac{\overline{f(z)}}{z-a}\,dz$ 
Let , $f$ be analytic for $|z|<2$. Show that $$\frac{1}{2\pi i}\int_{|z|=1}\frac{\overline{f(z)}}{z-a}\,dz=\begin{cases}\overline{f(0)} &\text{ if } |a|<1\\\overline{f(0)}-\overline{f(1/a)}&\text{ if }|a|>1\end{cases}$$

By putting $z=e^{i\theta}$ , $\displaystyle\int_{|z|=1}\frac{\overline{f(z)}}{z-a}\,dz=\int_0^{2\pi}\frac{\overline{f(e^{i\theta})}}{e^{i\theta}-a}ie^{i\theta}\,d\theta=i\int_0^{2\pi}\frac{\overline{f\left(\overline{e^{-i\theta}}\right)}}{1-ae^{-i\theta}}\,d\theta$.
Now putting , $e^{-i\theta}=t$ , it becomes $\displaystyle =-\int_{|t|=1}\frac{\overline{f(\bar t)}}{1-at}\,dt$. As $f(t)$ is analytic so $\overline{f(\bar t)}$ is also analytic. Now we can apply the residue theorem and finally I got the desire integral $=-\overline{f(0)}$ when $|a|<1$.

My problem is about the sign. I got an extra negative sign. Please verify my proof & detect where I made mistake ...

 A: $$\begin{align}
\oint_{|z|=1}\frac{\overline{ f(z)}}{z-a}\,dz&=\int_0^{2\pi}\frac{\overline{  f(e^{i\theta})}}{e^{i\theta}-a}ie^{i\theta}\,d\theta \tag 1\\\\
&=\int_{0}^{-2\pi}\frac{\overline{  f(e^{-i\theta})}}{e^{-i\theta}-a}ie^{-i\theta}\,(-1)\,d\theta \tag 2\\\\
&=\int_{-2\pi}^{0}\frac{\overline{  f(\overline{e^{i\theta}})}}{\left(e^{-i\theta}-a\right)e^{i2\theta}}ie^{i\theta}\,d\theta \tag 3\\\\
&=-\int_0^{2\pi}\frac{\overline{  f(\overline{e^{i\theta}})}}{\left(ae^{i\theta}-1\right)e^{i\theta}}ie^{i\theta}\,d\theta \tag 4\\\\ 
&=-\oint_{|z|=1}\frac{\overline{f(\overline{z})}}{z(az-1)}\,dz \tag 5\\\\
&=\oint_{|z|=1}\overline{f(\overline{z})}\left(\frac{1}{z}-\frac{a}{az-1}\right)\,dz \tag 6\\\\
&=2\pi i
\begin{cases}
\overline{f(0)}&,|a|<1\\\\
\overline{f(0)}-\overline{f(\overline{1/a})}&,|a|>1 \tag 7
\end{cases}
\end{align}$$
In arriving at $(1)$, we parametrized the unit circle letting $z=e^{i\theta}$.
In going from $(1)$ to $(2)$, we enforced the substitution $\theta \to -\theta$.
In going from $(2)$ to $(3)$, we absorbed the factor of $-1$ by transposing the integration limits.  We wrote $e^{-i\theta}=\overline{e^{i\theta}}$ in the argument of $f$. And we multiplied the numerator and denominator by $e^{i2\theta}$.
In going from $(3)$ to $(4)$, we multiplied $\left(e^{-i\theta}-a\right)e^{i\theta}=-(ae^{i\theta}-1)$ and then exploited the $2\pi$-periodicity of the integrand to transform the limits from $-2\pi$ to $0$ to $0$ to $2\pi$.
In going from $(4)$ to $(5)$, we moved from a parametric description back to the contour integral representation.
In going from $(5)$ to $(6)$, we used partial fraction expansion.
And in going from $(6)$ to $(7)$, we used the residue theorem.
