If `x=9^(1//3)xx9^(1//9)xx 9^(1//27)xx….,y=4^(1//3)xx-4^(1//9)xx 4^(1//27)x….,` and `z=Sigma_(r=1)^(oo) (1+i)^(r)` then arg (x+yz) is equal to

Options

A. 0
B. `pi-tan^(-1)(sqrt(2)/(3))`
C. `-tan^(-1)(sqrt(2)/(3))`
D. `-tan^(-1)((2)/(sqrt(3)))`

Answer

C

Explanation

`x=9^(1/3+1/9+1/27+..)`=`9^((1/3)/(1-1/3))=9^(1/2)=3`
`y=4^(1/3-1/9+1/27+..)=4^((1/3)/(1+1/3))=4^(1/4)=sqrt2`
`z=sum_(r=1)^(oo)(1+i)^(-r)=1/(1+i)+1/((1+i)^(2))+1/((1+i)^(3))+..`
`=(1/(1+i))/(1-1/(1+i))=1/i=-i`
Let `alpha=x+yz=3-isqrt2`
`thereforeargalpha=-tan^(-1)(sqrt2/3)`