A trick to simplify $z\ln z+\overline{z}\ln\overline{z}$? 
Q: Given complex conjugates $z = a+bi$ and $\overline z = a-bi$, what would be substitution needed to find $R$,
  $$R= z\ln z+\overline{z}\ln\overline{z}\tag1 $$
  such that $R$ is an expression without imaginary numbers?

This question arose as a special case of this post. Given $x^3-x-1=0$, let $x_1\approx1.3247$ be its real root (the plastic constant) and $x_2,x_3$ its complex roots. Then
$$\begin{aligned}\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}
&=  x_1\ln(1+x_1)+x_2\ln(1+x_2)+x_3\ln(1+x_3)\\ 
&= \frac32\,x_1\ln(1+x_1)-\frac{1}{2}\sqrt{\frac{3-x_1}{x_1}}\arccos\Big(\frac{x_1-6}{6x_1+2}\Big)\\[2mm]
&= 0.5179778\dots\end{aligned}$$
The first is by yours truly, while the second is a simplified version of a result by Reshetnikov. How do we show the equality of the first two lines? In general, how do we get rid of the imaginary unit in $(1)$, and can we always express it as the sum of $\ln$ and $\arccos$ of real numbers?

With a notation that should be obvious,
$$R=(x+iy)(\log r+i\theta)+(x-iy)(\log r-i\theta)=2(x\log r-y\theta)=2\left(x\log\sqrt{x^2+y^2}-y\,\arccos\frac xr\right).$$
