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

$\begin{array}{l}\mathrm{x}=9^{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\dots }=9^{\frac{\frac{1}{3}}{1-\frac{1}{3}}}=9^{\frac{1}{2}}=3\\ \mathrm{y}=4^{\frac{1}{3}-\frac{1}{9}+\frac{1}{27}+\cdots }=4^{\frac{1}{3}}1+\frac{1}{3}=4^{\frac{1}{4}}=\sqrt{2}\end{array}$

$\begin{array}{c}\mathrm{z}=\sum _{\mathrm{r}=1}^{\mathrm{\infty }} (1+\mathrm{i}{)}^{-\mathrm{r}}=\frac{1}{1+\mathrm{i}}+\frac{1}{(1+\mathrm{i}{)}^2}+\frac{1}{(1+\mathrm{i}{)}^3}+\dots \\ =\frac{\frac{1}{1+\mathrm{i}}}{1-\frac{1}{1+\mathrm{i}}}=\frac{1}{\mathrm{i}}=-\mathrm{i}\end{array}$

Let $\mathrm{\alpha }=\mathrm{x}+\mathrm{yz}=3-\mathrm{i}\sqrt{2}$

$\therefore \arg \mathrm{\alpha }=-{\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)$