This integral is strange

Question

$$ \int_{C_1}\frac{dz}{z}=\int_0^{2\pi}\frac{-R\sin{t}+iR\cos{t}}{R\cos{t}+iR\sin{t}}dt=\int_0^{2\pi}i\text{ }dt=2\pi i\tag{24.36} $$ Shouldn't it simply be $$\left[\ln(R \cos t + iR \sin t)\right]_0^{2\pi} ~?$$

That gives an answer of 0.

Answer 1

You've forgotten about the branch cut. the value of $\ln z$ jumps by $2 \pi \mathbf{i}$ when you cross typical choices for the branch cut, which should give confidence that if you correctly compute the antiderivative with your method with enough care to get the behavior regarding the branch cut correct, you'll get the correct answer.

But to understand the listed simplification, what do you get when you multiply the denominator by $\mathbf{i}$?