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Questions  

 Least positive argument of the 4th  root of the complex number 2i12 is: 

a
π6
b
5π12
c
7π12
d
11π12

detailed solution

Correct option is B

z4=2(1−3i)=412−32i=4cos⁡−π3+isin⁡−π3z=2cos⁡2mπ−π34+isin⁡2mπ−π34For m=1, z=2cos⁡5π12+isin⁡5π12∴The least positive argument of z is 5π12

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