Question

If $x=9^{1/3}\cdot 9^{1/9}\cdot 9^{1/27}\cdots \infty ;y=4^{1/3}\cdot 4^{-1/9}\cdot 4^{1/27}\cdots \infty$ and $z=\sum _{r=1}^{\infty }(1+i{)}^{-r}$ and the principal argument of the complex number $p=x+yz$ is $-{\tan }^{-1}\frac{\sqrt{a}}{b}$, then the value of $a^2+b^2$ is ($a$ and $b$ are coprime natural numbers)

Answer 1

$x=9^{1/3}\cdot 9^{1/9}\cdot 9^{1/27}\cdots \infty$
$\Rightarrow x=9^{\frac{1/3}{1-1/3}}=9^{1/2}=3$

$y=4^{1/3}\cdot 4^{-1/9}\cdot 4^{1/27}\cdots \infty$
$\Rightarrow y=4^{\frac{1/3}{1+1/3}}=4^{1/4}=\sqrt{2}$
Also $z=\frac{1}{1+i}+\frac{1}{(1+i{)}^2}+\frac{1}{(1+i{)}^3}+\cdots$
$=\frac{1/(1+i)}{1-1/(1+i)}=\frac{1}{i}=-i$

$\therefore p=x+yz=3+\sqrt{2}(-i)$

Let $\text{Argument}=\theta$
Since $p$ is lying in $4^{th}$ Quadrant,
we can say $\tan \alpha =\frac{|\sqrt{2}|}{|3|}$
$\Rightarrow \alpha ={\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)$
$\Rightarrow \text{Argument}=-\alpha =-{\tan }^{-1}\frac{\sqrt{2}}{3}$
Comparing we have, $a=2;b=3$
$\Rightarrow a^2+b^2=13$