A line integral equals zero implies a real integral also is zero

Question

I'm asked to check that the following line integral is zero:

$$\int_{C(0,r)} \frac {\log(1+z)}z dz=0$$

(where $C(0,r)$ is the circle of radius $r$ centered at $0$) and then to conclude that for every $r\in (0,1)$ the following integral is also $0$:

$$\int_0^{\pi}\log(1+r^2+2r\cos t)dt=0$$

I've managed to calculated the first integral expanding in Taylor series (we didn't do any residue yet in class) and I can imagine that it is connected in some way with the second, however I cannot find out exactly how.

Answer 1

Parametrizing the contour integral by setting $z = re^{it}$, $-\pi \le t \le \pi$, we write $$0 =\int_{C(0,r)} \frac{\log(1 + z)}{z}\, dz = \int_{-\pi}^\pi \frac{\log(1 + re^{it})}{re^{it}} ire^{it}\, dt = i \int_{-\pi}^\pi \log(1 + re^{it})\, dt$$ Taking imaginary parts, $0 = \int_{-\pi}^\pi \log|1 + re^{it}|\, dt$, or $$0 = \frac{1}{2}\int_{-\pi}^\pi \log|1 + re^{it}|^2\, dt$$ Expanding $|1 + re^{it}|^2$ and using symmetry, the result is obtained.

Answer 2

As long as $D(0,r)$ is a simply connected domain and the function $f(z)=1+z\neq 0, \forall z \in D(0,r)$ then a holomorphic branch of $\log f$ is defined well, and you proceed as the others said. We must be very careful with branches of logarithms and how we use them