Question

The square root of 4ab-2i(a^2-b^2)?

Answer 1

Given: The term 4ab - 2i(a^2 - b^2)

To find: The square root of the given term.

Solution:

• Now we have given the term: 4ab - 2i(a^2 - b^2)

• Let it be (x + iy)², then:

                 (x + iy)² = 4ab - 2i(a² - b²)

• Now comparing it, we get:

                 x² - y² = 4ab ..................(i)

                 xy = a² - b² .......................(ii)

• Now again, comparing the magnitude, we get:

                 x² + y² = √4a^4 + 4b^4 + 8a²b²

                 x² + y² = 2a² + 2b² .................(iii)

• Solving (i) and (ii), we get:

                 x² = a² + b² + 2ab = (a + b)²

                 x = (a + b) or -(a + b)

                 y² = a² + b² - 2ab = (a - b)²

                 y = (a - b) or -(a - b)

• So :

                 x + iy = ±{ (a + b) + i(a - b) }

Answer:

                 So the root is ±{ (a + b) + i(a - b) }

Answer 2

Step-by-step explanation:

$Let (x+iy) </p><p>2</p><p> =4ab−2i(a </p><p>2</p><p> −b </p><p>2</p><p> )$

$Comparing \: the \: real \: and \: \\ imaginary \: terms,$

$</p><p>x </p><p>2</p><p> −y </p><p>2</p><p> =4ab--------(i) \\ </p><p>xy=a </p><p>2</p><p> −b </p><p>2</p><p> ---------(ii)$

$Comparing \: the \: magnitude,$

$x {}^{2} + y {}^{2} = \sqrt[]{4a {}^{4} + 4b {}^{4} + 8a {}^{2 } \: b {}^{2} }$

$⇒ \times {}^{2} + y {}^{2} = 2a {}^{2} + 2b{}^{2} - (iii)$

$Solving \: (i) \: and \: (iii) \\ </p><p>x </p><p>2</p><p> =a </p><p>2</p><p> +b </p><p>2</p><p> +2ab \: and \: y \\ </p><p>2</p><p> =a </p><p>2</p><p> +b </p><p>2</p><p> −2ab$

$Hence, \\ </p><p>x=a+b , y=a−b \\ </p><p>or \: x=−(a+b) , y=b−a$