Question

If $x=9^{1/3}{9}^{1/9}{9}^{1/27}...\infty ,y=4^{1/3}{4}^{-1/9}{4}^{1/27}{4}^{-1/81}...\infty ,$ 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. $-{\tan }^{-1}\left(\frac{2}{\sqrt{3}}\right)$

Answer 1

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

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

$y=4^{\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}...\infty }$
$\Rightarrow y=4^{\frac{\frac{1}{3}}{1+\frac{1}{3}}}$
$\Rightarrow y=4^{\frac{1}{4}}$
$\Rightarrow y=2^{\frac{1}{2}}$
$\Rightarrow y=\sqrt{2}$

$z=\frac{1}{(1+i{)}^1}+\frac{1}{(1+i{)}^2}+\frac{1}{(1+i{)}^3}...\infty$
Since it is a G.P with a common ratio of $\frac{1}{(1+i)}$ we get the sum as
$=\frac{\frac{1}{1+i}}{1-\frac{1}{1+i}}$
$=\frac{1}{i}$
$=-i$.
Hence $x+yz=3-\sqrt{2}i$
$\tan \theta =\frac{-\sqrt{2}}{3}$
$\theta ={\tan }^{-1}\left(\frac{-\sqrt{2}}{3}\right)$
$=-{\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)$
Hence, option 'C' is correct.