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Question

If z=(1+i)(1+2i)(1+3i)(1+ni)(1i)(2i)(3i)(ni), where i=1, nN, then principal argument of z can be -

A
0
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B
π2
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C
π2
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D
π
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Solution

The correct option is D π
z=(ii2)(2ii2)(nii2)(1i)(2i)(ni)=in

z can be 1,1, i,i

So, argz can be 0, π, π2,π2

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