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Value of √i please explain step by step plzzzzzzzzzzzzzzzzzzzzzzzzz
i = Cosπ/2 +i Sinπ/2 -i = Cos (-π/2) + i Sin(-π/2) -i = e^ -iπ/2 sqrt(-i) = + or - (Cos(-π/4) + i Sin(-π/4)= (1-i)/sqrt(2)
note that i= e^i(pi/2)now root i= i^(1/2)= [e^i(pi/2)]^(1/2)use de moivres theorem= e^i(pi/4) or e^i(pi/4 + pi)= (1+i)/root2 or – (1+i)/root2 kindly approve :=)
√ ( √ -1 ) = √ ( √ -1 ) = ⁴√ -1 = (⁴√ -1 ) hope it helpsIn case of any difficulty, please feel free to qsk again RegardsArun
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