Let x+iy=4ab−2a2−b2i ⇒x+iy2=4ab−2a2−b2i Comparing the real and imaginary terms , x2-y2=4ab ...(i) xy=a2-b2...(ii) Comparing the magnitude , x2+y2=4a4+4b4+8a2b2 ⇒x2+y2=2a2+2b2 ...(iii) Solving (i) and (iii) ⇒x2=a2+b2+2ab and y2=a2+b2-2ab Hence , x=a+b , y=a-b or x=-a+b , y=-a-b

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$\sqrt{4 ab - 2 (a^{2} - b^{2}) i} =$

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$\pm {(i + 5 b) + (a - 4 b)}$

$\pm {(a + b) - i (a - b)}$

$\pm \{\frac{(a - b) + i (a - b)}{2}\}$

$\pm {(i + b) - i (a - b)}$

answer is C.

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Detailed Solution

$L e t (x + i y) = \sqrt{4 ab - 2 (a^{2} - b^{2}) i} \Rightarrow {(x + i y)}^{2} = 4 ab - 2 (a^{2} - b^{2}) i Comparing the real and imaginary terms, x^{2} - y^{2} = 4 ab . . . (i) xy = a^{2} - b^{2} . . . (ii) Comparing the magnitude, x^{2} + y^{2} = \sqrt{4 a^{4} + 4 b^{4} + 8 a^{2} b^{2}} \Rightarrow x^{2} + y^{2} = 2 a^{2} + 2 b^{2} . . . (iii) Solving (i) and (iii) \Rightarrow x^{2} = a^{2} + b^{2} + 2 ab and y^{2} = a^{2} + b^{2} - 2 ab Hence, x = a + b, y = a - b or x = - (a + b), y = - (a - b)$