If  $\mathrm{x}=9^{1/3}.9^{1/9}.9^{1/27}.....\mathrm{\infty },\mathrm{y}=4^{1/3}.4^{-1/9}.4^{1/27}....\mathrm{\infty }$ and $\mathrm{z}=\sum _{\mathrm{r}=1}^{\mathrm{\infty }}{(1+\mathrm{i})}^{-\mathrm{r}}$, where $\mathrm{i}=\sqrt{-1}\text{\hspace{0.17em}\hspace{0.17em}then\hspace{0.17em}\hspace{0.17em}arg\hspace{0.17em}}(\mathrm{x}+\mathrm{yz})$ is equal to

$\begin{array}{l}\because \mathrm{x}=9^{1/3}\cdot 9^{1/9}\cdot 9^{1/27}\cdots \dots \mathrm{\infty }\\ =9^{1/3+1/9+1/27+\dots \mathrm{\infty }}=9^{\frac{1/3}{1-1/3}}=9^{1/2}=3\\ =4^{\frac{1/3}{1+1/3}}=4^{1/4}=\sqrt{2}\end{array}$
and $\mathrm{z}=\sum _{\mathrm{\tau }=1}^{\mathrm{\infty }} (1+\mathrm{i}{)}^{-\mathrm{\tau }}=\frac{1}{(1+\mathrm{i})}+\frac{1}{(1+\mathrm{i}{)}^2}+\frac{1}{(1+\mathrm{i}{)}^3}+\dots \mathrm{\infty }=-\mathrm{i}$
Now, $\mathrm{x}+\mathrm{yz}=3-\mathrm{i}\sqrt{2}$
$\therefore \arg (\mathrm{x}+\mathrm{yz})=\arg (3-\mathrm{i}\sqrt{2})=-{\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)$