Integrate $\int_{C} \frac{1}{r-\bar{z}}dz$ - conflicting answers In an homework exercise, we're asked to integrate $\int_{C} \frac{1}{k-\bar{z}}dz$ where C is some circle that doesn't pass through $k$.
I tried solving this question through two different approaches, but have arrived at different answers.
Idea: use the fact that in C, $z\bar{z}=r^2$, where $r$ is the radius of C, to get 
$$\int_{C} \frac{1}{k-\frac{r^2}{z}}dz$$
We then apply Cauchy to get that the answer is $2πi r^2/k^2$ when C contains $r^2/k$, and 0 otherwise.
Another idea: Intuitively, since C is a circle we get ${\int_{C} \frac{1}{k-\bar{z}}dz} = \int_{C} \frac{1}{k-z}dz$ (since $\bar{z}$ belongs to C iff $z$ belongs to C) and we can then use Cauchy's theorem and Cauchy's formula (depending on what C contains) to arrive at a different answer.
Which answer (if any) is correct?

The original function is not holomorphic since $\frac{d}{d\overline{z}}\frac{1}{k-\overline{z}}=\frac{-1}{(k-\overline{z})^{2}}$. So you cannot apply Cauchy's integral formula. 
Let $C$ be centered at $c$ with radius $r$, then we have $z=re^{i\theta}+c=c+r\cos(\theta)+ri\sin(\theta)$, and its conjugate become $c+r\cos(\theta)-ri\sin(\theta)=c+re^{-i\theta}$. Therefore we have $k-\overline{z}=k-c-re^{-i\theta}$. To noramlize it we mutiply $(k-c-r\cos(\theta))-ri\sin(\theta)=k-c-re^{i\theta}$. The result is $(k-c-r\cos(\theta))^{2}+r^{2}\sin^{2}\theta$. Rearranging gives $$k^{2}+c^{2}+r^{2}-2kc-2(k-c)r\cos(\theta)=A-2Br\cos(\theta),A=k^{2}+c^{2}+r^{2}-2kc,B=k-c$$
So we have $$\int_{C}\frac{1}{k-\overline{z}}=\int^{2\pi}_{0}\frac{k-c-re^{i\theta}}{(k-c-re^{-i\theta})(k-c-re^{-i\theta})}dre^{i\theta}=ri\int^{2\pi}_{0}\frac{Be^{i\theta}-re^{2i\theta}}{A-2Br\cos(\theta)}d\theta$$
So it suffice to integrate $$\frac{e^{i\theta}}{1-C\cos(\theta)}d\theta,C=constant$$ since the above integral is a linear combination of integrals of this type. But this should be possible in general as we can make trigonometric substitutions. 
