If $\mathrm{x}=9^{1/3}{9}^{1/9}{9}^{1/27}\cdots \mathrm{\infty },\mathrm{y}=4^{1/3}{4}^{-1/9}{4}^{1/27}\cdots \mathrm{\infty },$and  $\mathrm{z}=\sum _{\mathrm{r}=1}^{\mathrm{\infty }} (1+\mathrm{i}{)}^{-\mathrm{r}}$ then arg(x + yz) is equal to

$\mathrm{x}=9^{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots }=9^{\frac{\frac{1}{3}}{1-\frac{1}{3}}}=9^{\frac{1}{2}}=3$
$\begin{array}{l}\mathrm{y}=4^{\frac{1}{3}-\frac{1}{9}+\frac{1}{27}+\cdots }=4^{\frac{\frac{1}{3}}{1+\frac{1}{3}}}=4^{\frac{1}{4}}=\sqrt{2}\\ \mathrm{z}=\sum _{\mathrm{r}=1}^{\mathrm{\infty }} (1+\mathrm{i}{)}^{-\mathrm{r}}=\frac{1}{1+\mathrm{i}}+\frac{1}{(1+\mathrm{i}{)}^2}+\frac{1}{(1+\mathrm{i}{)}^3}+\cdots \\ =\frac{\frac{1}{1+\mathrm{i}}}{1-\frac{1}{1+\mathrm{i}}}=\frac{1}{\mathrm{i}}=-\mathrm{i}\end{array}$

Let $a=\mathrm{x}+\mathrm{yz}=3-\sqrt{2}\mathrm{i}$ (fourth quadrant). Then
$\arg \mathrm{\alpha }=-{\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)$