√-11-60i

1.	√-11-60i
Answer» Let, (a + ib)²= -11 - 60i Now using, (a + b)² = a²+ b² + 2ab ⇒ a² + (bi)² + 2abi = -11 - 60i Since i² = -1 ⇒ a² - b² + 2abi = -11 - 60i Now, separating real and complex parts, we get ⇒ a² - b² = -11…………..eq.1 ⇒ 2ab = - 60…….. eq.2 ⇒ a = -30/b Now, using the value of a in eq.1, we get ⇒ \((-\frac{30}{b})^2\) – b² = -11 ⇒ 900 – b⁴ = -11b² ⇒ b⁴ - 11b² - 900 = 0 Simplify and get the value of b² , we get, ⇒ b² = 36 or b² = -25 as b is real no. so, b² = 36 b = 6 or b = -6 Therefore , a = - 5 or a = 5 Hence the square root of the complex no. is - 5 + 6i and 5 – 6i.

Answer»

Let, (a + ib)²= -11 - 60i

Now using, (a + b)² = a²+ b² + 2ab

⇒ a² + (bi)² + 2abi = -11 - 60i

Since i² = -1

⇒ a² - b² + 2abi = -11 - 60i

Now, separating real and complex parts, we get

⇒ a² - b² = -11…………..eq.1

⇒ 2ab = - 60…….. eq.2

⇒ a = -30/b

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

⇒ \((-\frac{30}{b})^2\) – b² = -11

⇒ 900 – b⁴ = -11b²

⇒ b⁴ - 11b² - 900 = 0

Simplify and get the value of b² , we get,

⇒ b² = 36 or b² = -25

as b is real no. so, b² = 36

b = 6 or b = -6

Therefore , a = - 5 or a = 5

Hence the square root of the complex no. is - 5 + 6i and 5 – 6i.

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