√(-2+2√3i)

1.	√(-2+2√3i)
Answer» Let, (a + ib)² = - 2 + 2√3 i Now using, (a + b)² = a² + b² + 2ab ⇒ a² + (bi)²+ 2abi = - 2 + 2√3 i Since i² = -1 ⇒ a² - b² + 2abi = - 2 + 2√3 i Now, separating real and complex parts, we get ⇒ a² - b² = - 2…………..eq.1 ⇒ 2ab = 2√3 ……..eq.2 ⇒ a = √3/b Now, using the value of a in eq.1, we get ⇒ (√3/b)² – b² = -2 ⇒3 – b⁴ = -2b² ⇒ b⁴ - 2b² - 3= 0 Simplify and get the value of b², we get, ⇒ b² = -1 or b² = 3 As b is real no. so, b² = 3 b = √3 or b = -√3 Therefore, a = 1 or a = -1 Hence the square root of the complex no. is 1 + √3 i and -1 - √3 i.

Answer»

Let, (a + ib)² = - 2 + 2√3 i

Now using, (a + b)² = a² + b² + 2ab

⇒ a² + (bi)²+ 2abi = - 2 + 2√3 i

Since i² = -1

⇒ a² - b² + 2abi = - 2 + 2√3 i

Now, separating real and complex parts, we get

⇒ a² - b² = - 2…………..eq.1

⇒ 2ab = 2√3 ……..eq.2

⇒ a = √3/b

Now, using the value of a in eq.1, we get

⇒ (√3/b)² – b² = -2

⇒3 – b⁴ = -2b²

⇒ b⁴ - 2b² - 3= 0

Simplify and get the value of b², we get,

⇒ b² = -1 or b² = 3

As b is real no. so, b² = 3

b = √3 or b = -√3

Therefore, a = 1 or a = -1

Hence the square root of the complex no. is 1 + √3 i and -1 - √3 i.

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