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√-11-60i

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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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