What I am given:
$log(z) = ln(r) + i\theta$ , where $(r>0, \alpha<\theta<\alpha+2\pi)$
$log(z)=\frac{1}{2}ln(x^2+y^2)+itan^{-1}(\frac{y}{x})$
What I am told: Use the Cauchy-Riemann equations to show that the given branch is analytic in its domain.
What I tried:
I let $u=\frac{1}{2}ln(x^2+y^2)$ and $v=tan^{-1}(\frac{y}{x})$
I found the partials: $u_x=v_y= \frac{x}{x^2+y^2}$ and $u_y=-v_x=\frac{y}{x^2+y^2}$
My guess is domain $D=x^2+y^2>0$ and $y\neq tan(\alpha)x$. I can see that the partials are defined for the first condition. I do not know how to check the second condition.
The domain is the complex plane with the branch cut removed. So
$$D=\Bbb C\setminus \{r(\cos\alpha+i\sin\alpha):r\ge 0\}.$$