In the lecture we have learned that we can define $f = \sqrt{z}$ analytic as $f = \exp(\frac{1}{2} \log z)$ in any domain where $\log z$ is analytic, for example by considering $\mathbb{C}\setminus (-\infty,0]$. However, how can I define and make the function $g = \sqrt{6z^2-5z+1}$ analytic in terms of $\log z$?
Our lecturer said it is defined and analytic on $\mathbb{C}\setminus (-\infty,\frac{1}{2}] $ and $(-\infty,\frac{1}{3}) $. I understand that I should consider branch points $\frac{1}{2}$ and $\frac{1}{3}$ but I could not show it. Any help would be appreciated.
The set $\Bbb C\setminus\left(\left(-\infty,\frac12\right]\cup\left[\frac13,\infty\right)\right)$ is simply connected and the function $\varphi(z)=6z^2-5z+1$ has no zero there. Therefore, there is some analytic function $g\colon\Bbb C\setminus\left(\left(-\infty,\frac12\right]\cup\left[\frac13,\infty\right)\right)\longrightarrow\Bbb C$ which is a logarithm of $\varphi$, that is, such that$$\left(\forall z\in\Bbb C\setminus\left(\left(-\infty,\frac12\right]\cup\left[\frac13,\infty\right)\right)\right):e^{g(z)}=\varphi(z)=6z^2-5z+1.$$So, $e^{g(z)/2}$ is an analytic square root of $6z^2-5z+1$.