√-4-3i

1.	√-4-3i
Answer» 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.

Answer»

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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