Why is $\int_{|z|=1} \dfrac{\log(1-az)}{z}dz = 0$ for $|a|<1$?

Question

As the title suggests, I am a bit unsure why that is so? Log is not holomorphic so Cauchy seems a bit off.

If anyone could provide a hint that would be very appreciated.

Edit: I have tagged the question with "cauchy-integral formula" as I suspect it has something to do with it. If it is clearly wrong however, I will remove it if so.

Answer 1

The principal branch of logarithm, which I denote by $\log z$, is analytic in the disk $D(1,1)$. So $\log (1-az)$ is analytic in $D(0,\frac 1 {|a|})$ which includes the closed unit disk. Hence, Cauchy's integral formula is applicable and the value of the integral is $2\pi i$ times the value of $\log (1-az)$ at $0$ which is $0$.

Answer 2

If $z=x+iy$, then \begin{align} log(1-az)&=log(1-a(x+iy))\\ &=log(1-ax-i(ay))\\ &\Rightarrow \mathfrak{R}(f(z))=1-ax\geq 0 \Rightarrow x\leq\frac{1}{a} \\ &\Rightarrow \mathfrak{I}m(f(z))=0\Rightarrow y=0\\ \end{align}

then with conditions it'll verify that the domain analyticity of $f(z)$ like this in $z=0$ is inside the curve $\gamma:=|z|=1$, so yo can apply the Cauchy integral formule.