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

$x=9^{1/3}\cdot 9^{1/9}\cdot 9^{1/27}\cdots \infty \Rightarrow x=9^{\frac{1/3}{1-1/3}}=9^{1/2}=3y=4^{1/3}\cdot 4^{-1/9}\cdot 4^{1/27}\cdots \infty \Rightarrow y=4^{\frac{1/3}{1+1/3}}=4^{1/4}=\sqrt{2}$
Also $z=\frac{1}{1+i}+\frac{1}{(1+i{)}^2}+\frac{1}{(1+i{)}^3}+\cdots$
As $\frac{1}{|1+i|}=\frac{1}{\sqrt{2}}$, so
$z=\frac{1/(1+i)}{1-1/(1+i)}=\frac{1}{i}=-i\therefore p=x+yz=3-i\sqrt{2}$

Let $\text{Arg}\left(p\right)=\theta$
Since $p$ is lying in $4^{th}$ Quadrant, we know
$\theta =-\alpha$
Where
$\tan \alpha =\frac{|\sqrt{2}|}{|3|}\Rightarrow \alpha ={\tan }^{-1}\left(\frac{\sqrt{2}}{3}\right)\Rightarrow \text{Arg}\left(p\right)=-\alpha =-{\tan }^{-1}\frac{\sqrt{2}}{3}$
On comparing, we have
$a=2;b=3\therefore a^2+b^2=13$