Complex integrals of $1/(z+a)$ over the unit circle

Question

In quite a few questions here, it has been settled that $$ \int_C \frac{1}{z} dz = 2\pi i $$ where $C$ is the unit circle with the origin at its center. I also understand how to think about this, using that the complex logarithm has a branch cut on the positive real axis, so $\log{1}$ takes different values in the two limits.

My problem arises with integrals like $$ \int_C \frac{1}{z+3} dz = 0 $$ From the Cauchy Residue theorem it clear that the integral is $0$, but I can't get that result when doing the integral explicitly:

$$ \int_C \frac{1}{z+3} dz = \left[ \log{(e^{i\phi} + 3)} \right]_0^{2\pi} $$

Using the argument from the first integral above, the answer should be $2\pi i$. Obviously that's not right, but why?

Answer 1

No. Using the residue theorem, the integral is also $0$, because that theorem says that the integral is the product of $3$ numbers:

• $2\pi i$
• the residue at $-3$ (which is $1$);
• the winding number of the loop around the origin $t\mapsto e^{it}$ ($t\in[0,2\pi]$), which is $0$.

Since the last number is $0$, the product is $0$.