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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 |
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Answer» Correct Answer - 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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