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√-8i

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Let, (a + ib)2 = 0 - 8i

Now using, (a + b)2 = a2 + b2 + 2ab

⇒ a2 + (bi)2 + 2abi = 0 - 8i

Since i2 = -1

⇒ a2 - b2 + 2abi = 0 - 8i

Now, separating real and complex parts, we get

⇒ a2 - b2 = 0 …………..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 = 0

⇒ 16 – b4 = 0

⇒ b4 = 16

Simplify and get the value of b2, we get,

⇒ b2 = - 4 or b2 = 4

As b is real no. so, b2 = 4

b = 2 or b = -2

Therefore , a = -2 or a = 2

Hence the square root of the complex no. is -2 + 2i and 2 - 2i.

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