Question

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

• A. $0$
• B. $\pi -{\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)$
• C. $-{\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)$
• D. $\pi -{\tan }^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Answer 1

The correct option is C $-{\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)$

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

sum of $3^{-1}+3^{-2}+3^{-3}.....\infty$

$⟹a=3^{-1},r=3^{-1}$

sum$=\frac{3^{-1}}{1-3^{-1}}=\frac{1}{2}$

$x=9^{1/2}=3$

$y=4^{1/3},4^{-1/9},4^{1/27}.........\infty$

sum of $3^{-1}-3^{-2}+3^{-3}.....\infty$

$⟹a=3^{-1},r=-3^{-1}$

sum$=\frac{3^{-1}}{1+3^{-1}}=\frac{1}{4}$

$y=4^{1/4}=\sqrt{2}$

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

$z=(\sqrt{2}{e}^{(-\pi /2)}{)}^{-1}$$+(\sqrt{2}{e}^{(-\pi /2)}{)}^{-2}+.....\infty$

$a=(\sqrt{2}{e}^{(-\pi /2)}{)}^{-1}$$r=(\sqrt{2}{e}^{(-\pi /2)}{)}^{-1}$

sum $=\frac{\frac{1}{1+i}}{1-\frac{1}{1+i}}=1/i$

$⟹\frac{i}{i^2}=-i$

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

arg$=-{\tan }^{-1}\left(\sqrt{2}/3\right)$