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√4i

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Let, (a + ib)2 = 0 + 4i

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

⇒ a2 + (bi)2 + 2abi = 0 + 4i

Since i2 = -1

⇒ a2 - b2 + 2abi = 0 + 4i

Now, separating real and complex parts, we get

⇒ a2 - b2 = 0 …………..eq.1

⇒ 2ab = 4…….. eq.2

⇒ a = 2/b

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

⇒ \((\frac{2}{b})^2\) – b2 = 0

⇒ 4 – b4 = 0

⇒ b4 = 4

Simplify and get the value of b2, we get,

⇒ b2 = - 2 or b2 = 2

As b is real no. so, b2 = 2

b = √2 or b = -√2

Therefore , a = √2 or a = -√2

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

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