If $x=9^{\frac {1}{3}} 9^{\frac {1}{9}} 9^{\frac {1}{27}} .....\infty, y=4^{\frac {1}{3}} 4^{\frac {-1}{9}} 4^{\frac {1}{27}}....\infty,$ and $z=\sum_{r=1}^{\infty} (1+i)^{-r}$, then $arg (x+yz)$ is equal to

$x=9^{\dfrac{1}{3}+\dfrac{1}{3^{2}}+\dfrac{1}{3^{3}}...\infty}$

$x=9^{\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}}$

$x=9^{\dfrac{1}{2}}$

$x=3$

$y=4^{\dfrac{1}{3}-\dfrac{1}{3^{2}}+\dfrac{1}{3^{3}}...\infty}$

$y=4^{\dfrac{\dfrac{1}{3}}{1+\dfrac{1}{3}}}$

$y=4^{\dfrac{1}{4}}$

$y=2^{\dfrac{1}{2}}$

$y=\sqrt{2}$

$z=\dfrac{1}{(1+i)^{1}}+\dfrac{1}{(1+i)^{2}}+\dfrac{1}{(1+i)^{3}}...\infty$

Since it is a G.P with a common ratio of $\dfrac{1}{(1+i)}$ we get the sum as

$=\dfrac{\dfrac{1}{1+i}}{1-\dfrac{1}{1+i}}$

$=\dfrac{1}{i}$

$=-i$.

Hence $x+yz =3-\sqrt{2}i$

$\tan\theta=\dfrac{-\sqrt{2}}{3}$

$\theta=\tan^{-1}(\dfrac{-\sqrt{2}}{3})$

$=-\tan^{-1}(\dfrac{\sqrt{2}}{3})$