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√-4-3i

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Let, (a + ib)2 = - 4 - 3i

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

⇒ a2 + (bi)2 + 2abi = -4 -3i

Since i2 = -1

⇒ a2 - b2 + 2abi = - 4 - 3i

Now, separating real and complex parts, we get

⇒ a2 - b2 = - 4…………..eq.1

⇒ 2ab = - 3…….. eq.2

⇒ a = - 3/2b

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

⇒ \((-\frac{3}{2b})^2\) – b2 = - 4

⇒ 9 – 4b4 = -16b2

⇒ 4b4 - 16b2 - 9 = 0

Simplify and get the value of b2, we get,

⇒ b2 = 9/2 or b2 = - 2

As b is real no. so, b2 = 9/2

b = \(\frac{3}{\sqrt2}\) or b = -\(\frac{3}{\sqrt2}\)

Therefore, a = -\(\frac{1}{\sqrt2}\) or a = \(\frac{1}{\sqrt2}\)

Hence the square root of the complex no. is -\(\frac{1}{\sqrt2}\) + \(\frac{3}{\sqrt2}\)i and \(\frac{1}{\sqrt2}\) - \(\frac{3}{\sqrt2}\)i.

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