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√-15-8i

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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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