I've been considering:
$$ \int\limits_{|z|=1} \frac{dz}{(z-a)(z-b)} $$
when $|a|, |b| < 1$. You can work this out easily with partial fractions or by splitting up the areas so that one of the two terms in the denominator is holomorphic over that area (see here) - but these methods break down for $a=b$.
That's also easily resolvable (with the derivative form of Cauchy's Integral Formula) - but my question is if there's a way to unify the understanding of both of these, as they are fundamentally similar.
One idea I've had is consider it as a function of $b$ and then as it is zero everywhere except $a$, it must be equal to 0 to as $b \to a$? But how do we know this function has to be continuous?
Applying Cauchy's Integral Theorem, we assert that for $|a|<1$ and $|b|<1$ and any number $ R>1$
$$\oint_{|z|=1}\frac{1}{(z-a)(z-b)}\,dz=\oint_{|z|=R}\frac{1}{(z-a)(z-b)}\,dz$$
Then, we have the estimate
$$\begin{align} \left|\oint_{|z|=R}\frac{1}{(z-a)(z-b)}\,dz\right|\le \frac{2\pi R}{(R-|a|)(R-|b|)} \end{align}$$
Since $R>1$ is arbitrary, we let $R\to \infty$ and conclude that for all $|a|<1$ and $|b|<1$
$$\oint_{|z|=1}\frac{1}{(z-a)(z-b)}\,dz=0$$