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Question

If x=91/391/991/27......,y=41/3,41/9,41/27........, and z=r=1(1+i)r, then arg(x+yz) is equal to

A
0
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B
πtan1(23)
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C
tan1(23)
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D
πtan1(23)
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Solution

The correct option is C tan1(23)
x=931932933......
sum of 31+32+33.....
a=31,r=31
sum=31131=12
x=91/2=3
y=41/3,41/9,41/27.........
sum of 3132+33.....
a=31,r=31
sum=311+31=14
y=41/4=2
z=(1+i)1+(1+i)2.......
z=(2e(π/2))1+(2e(π/2))2+.....
a=(2e(π/2))1r=(2e(π/2))1
sum =11+i111+i=1/i
ii2=i
x+yz=32i
arg=tan1(2/3)

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