√(5+12i)

1.	√(5+12i)
Answer» Let, (a + ib)² = 5 + 12i Now using, (a + b)² = a² + b² + 2ab ⇒ a² + (bi)² + 2abi = 5 + 12i Since i²= -1 a² - b² + 2abi = 5 + 12i Now, separating real and complex parts, we get ⇒ a² - b² = 5…………..eq.1 ⇒ 2ab = 12……..eq.2 ⇒ a = 6/b Now, using the value of a in eq.1, we get ⇒ (6/b)² – b² = 5 ⇒ 36 – b⁴ = 5b² ⇒ b⁴ + 5b²- 36 = 0 Simplify and get the value of b², we get, b² = - 9 or b² = 4 As b is real no. so, b² = 4 b = 2 or b = - 2 Therefore, a = 3 or a = - 3 Hence the square root of the complex no. is 3 + 2i and - 3 -2i.

Answer»

Let, (a + ib)² = 5 + 12i

Now using, (a + b)² = a² + b² + 2ab

⇒ a² + (bi)² + 2abi = 5 + 12i

Since i²= -1

a² - b² + 2abi = 5 + 12i

Now, separating real and complex parts, we get

⇒ a² - b² = 5…………..eq.1

⇒ 2ab = 12……..eq.2

⇒ a = 6/b

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

⇒ (6/b)² – b² = 5

⇒ 36 – b⁴ = 5b²

⇒ b⁴ + 5b²- 36 = 0

Simplify and get the value of b², we get,

b² = - 9 or b² = 4

As b is real no. so, b² = 4

b = 2 or b = - 2

Therefore, a = 3 or a = - 3

Hence the square root of the complex no. is 3 + 2i and - 3 -2i.

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