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Ex 5.3, 1 - Chapter 5 Class 11 Complex Numbers (Term 1)

Last updated at Jan. 14, 2022 by

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Ex 5.3, 1 Solve the equation π‘₯2 + 3 = 0 Now, π‘₯2 + 3 = 0 π‘₯2 = βˆ’3 𝒙 = Β± √(βˆ’πŸ‘) π‘₯ = Β± √(βˆ’1 Γ— 3) 𝒙 = Β± √(βˆ’πŸ) Γ— βˆšπŸ‘ π‘₯ = Β± i Γ— √3 𝒙 = Β± βˆšπŸ‘ i

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