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If x=9^(1/3) 9^(1/9) 9^(1/27) ...->∞ , ...

If `x=9^(1/3) 9^(1/9) 9^(1/27) ...->`∞ , `y= 4^(1/3) 4^(-1/9) 4^(1/27) ....->`∞ and `z= sum_(r=1)^oo (1+i)^-r` then , the argument of the complex number `w = x+yz` is

A

0

B

`pi-tan^(-1)(sqrt(2)/(3))`

C

`-tan^(-1)(sqrt(2)/(3))`

D

`-tan^(-1)((2)/(sqrt(3)))`

Text Solution

Verified by Experts

The correct Answer is:
C

`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)`
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