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(2-i√(3) )/(2+i√(3) ) Solve in standard complex form, i = imaginary number
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(2-i√(3) )/(2+i√(3) )

Solve in standard complex form, i = imaginary number

User Kaustubh Bhagwat
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Answer:

Explanation:


(2-\iota√(3) )/(2+\iota√(3) ) =(2-\iota√(3) )/(2+\iota√(3) )* (2-\iota √(3) )/(2-\iota √(3) ) \\=((2-\iota√(3) )^2)/(2^2-(\iota√(3) )^2) \\=(4+3 (\iota)^2-4\iota √(3))/(4-3\iota^2) \\=(4-3-4\iota√(3))/(4+3) \\=(1)/(7) -(4 √(3))/(7) \iota

User Jim Nilsson
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