The problem asks you for what values of $a,b,c\in\mathbb R$ you can find a domain in $\mathbb C$ where $$f(z)=\log(x^2+y^2)-a(\arg z)^2+ib\arg\bar z+\dfrac{c}{\bar z}$$ is analytic.
Due to the nature of the function, it is most preferable to work in polar coordinates: $$f(z)=2\log r-a\theta^2-ib\theta+\dfrac{c}{r}e^{i\theta}.$$
We now impose that it satisfies Cauchy-Riemann in polars for the function to be analytic; namely, $$\dfrac{\partial f}{\partial r}=\dfrac{1}{ir}\dfrac{\partial f}{\partial\theta}.$$
In brief, doing the operations I get $$\dfrac{2}{r}-\dfrac{c}{r^2}e^{i\theta}=i\dfrac{2a\theta}{r}-\dfrac{b}{r}+\dfrac{c}{r^2}e^{i\theta}.$$
Finally, comparing both real and imaginary parts, we get $$ \left\{ \begin{aligned} &\mathfrak{Re}: \dfrac{2+b}{r}=\dfrac{2c}{r^2}\cos\theta\\ &\mathfrak{Im}: \dfrac{2a\theta}{r}=-\dfrac{2c}{r^2}\sin\theta . \end{aligned} \right. $$
Solutions says $a,c=0$ and $b=-2$. How can I deduce this from the previous system of equations if it'd have $1$ degree of freedom?
$$\log z^2 = \log |z|^2 + i \arg z^2 = \log (x^2+y^2)+ 2i \arg z. $$
This, the polar form, and the compact form of the Cauchy Riemann equations: $$\frac{df}{d\overline{z}}=0$$ (https://mathworld.wolfram.com/Cauchy-RiemannEquations.html)
yields: $$b=-2, a=c=0.$$
The domain is $$\mathbb{C} \backslash \{0\}.$$