√-15-8i

1.	√-15-8i
Answer» Let, (a + ib)² = -15 - 8i Now using, (a + b)² = a² + b² + 2ab ⇒ a² + (bi)²+ 2abi = -15 - 8i Since i² = -1 ⇒ a² - b² + 2abi = -15 - 8i Now, separating real and complex parts, we get ⇒ a² - b² = -15…………..eq.1 ⇒ 2ab = -8…….. eq.2 ⇒ a = - 4/b Now, using the value of a in eq.1, we get ⇒ \((-\frac{4}{b})^2\) – b² = -15 ⇒ 16 – b⁴ = -15b² ⇒ b⁴ - 15b² - 16 = 0 Simplify and get the value of b², we get, ⇒ b² = 16 or b² = -1 As b is real no. so, b² = 16 b = 4 or b = - 4 Therefore, a = -1 or a = 1 Hence the square root of the complex no. is -1 + 4i and 1 - 4i.

Answer»

Let, (a + ib)² = -15 - 8i

Now using, (a + b)² = a² + b² + 2ab

⇒ a² + (bi)²+ 2abi = -15 - 8i

Since i² = -1

⇒ a² - b² + 2abi = -15 - 8i

Now, separating real and complex parts, we get

⇒ a² - b² = -15…………..eq.1

⇒ 2ab = -8…….. eq.2

⇒ a = - 4/b

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

⇒ \((-\frac{4}{b})^2\) – b² = -15

⇒ 16 – b⁴ = -15b²

⇒ b⁴ - 15b² - 16 = 0

Simplify and get the value of b², we get,

⇒ b² = 16 or b² = -1

As b is real no. so, b² = 16

b = 4 or b = - 4

Therefore, a = -1 or a = 1

Hence the square root of the complex no. is -1 + 4i and 1 - 4i.

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