We know by Cauchy's integral formula that $$\int\limits_{|z|=1} \frac{1}{z-a} \ dz = 2 \pi i$$ for $|a| < 1$.
We can try to calculate this directly. If $a = 0$ then simply substituting in $z = e^{i\theta}$ works. However if $a \neq 0$ say $a = \frac{1}{2}$ then upon substituting we get $$\int\limits_{|z|=1} \frac{1}{z-\frac{1}{2}} \ dz = i\int\limits_{0}^{2\pi} \frac{e^{i\theta}}{e^{i\theta}-\frac{1}{2}} \ d\theta = i\int\limits_{0}^{2\pi} 1+\frac{1}{2e^{i\theta}-1} \ d\theta = 2 \pi i + i\int\limits_{0}^{2\pi} \frac{1}{2e^{i\theta}-1} \ d\theta$$
From the integral formula we know that the last integral must be zero. Is there an elementary way of seeing this?
$$\frac1{2e^{it}-1}=\frac1{2e^{it}}\frac1{1-e^{-it}/2} =\sum_{n=1}^\infty\frac{e^{-int}}{2^n}$$ Integrate termwise, using $$\int_0^{2\pi} e^{-int}\,dt=0.$$