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B
π4
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C
π3
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D
π2
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Solution
The correct option is Aπ6 Let z=1(√3−i)25 then arg(z)=arg(1)−arg((√3−i)25)+2kπ,k∈Z ⇒arg(z)=0−25arg(√3−i)+2kπ,k∈Z⇒arg(z)=0−25(−π6)+2kπ,k∈Z =25×π6+2kπ,k∈Z If k=−2 arg(z)=π6∈(−π,π]