The imaginary part of (3+2 sqrt{-54})^{1/2} - | vedclass

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(A) Let $z = (3+2 \sqrt{-54})^{1/2} - (3-2 \sqrt{-54})^{1/2}$.First,simplify the expression inside the square root: $\sqrt{-54} = \sqrt{54}i = 3\sqrt{

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$-2 \sqrt{6}$

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(A) Let $z = (3+2 \sqrt{-54})^{1/2} - (3-2 \sqrt{-54})^{1/2}$.First,simplify the expression inside the square root: $\sqrt{-54} = \sqrt{54}i = 3\sqrt{6}i$.So,$3+2\sqrt{-54} = 3 + 2(3\sqrt{6}i) = 3 + 6\sqrt{6}i$. This does not form a perfect square of the form $(a+bi)^2$ easily.Let $(3+2\sqrt{-54})^{1/2} = x+iy$. Then $x^2-y^2 = 3$ and $2xy = 2\sqrt{54} = 6\sqrt{6}$.$x^2+y^2 = \sqrt{3^2 + (6\sqrt{6})^2} = \sqrt{9 + 216} = \sqrt{225} = 15$.Adding the equations: $2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3$.Subtracting: $2y^2 = 12 \implies y^2 = 6 \implies y = \pm \sqrt{6}$.Thus,$(3+2\sqrt{-54})^{1/2} = \pm(3+\sqrt{6}i)$ and $(3-2\sqrt{-54})^{1/2} = \pm(3-\sqrt{6}i)$.Possible values for the expression are:$1) (3+\sqrt{6}i) - (3-\sqrt{6}i) = 2\sqrt{6}i$$2) (3+\sqrt{6}i) - (-(3-\sqrt{6}i)) = 6$$3) -(3+\sqrt{6}i) - (3-\sqrt{6}i) = -6$$4) -(3+\sqrt{6}i) - (-(3-\sqrt{6}i)) = -2\sqrt{6}i$.The imaginary parts are $2\sqrt{6}$ and $-2\sqrt{6}$.Comparing with options,the correct choice is $-2\sqrt{6}$.

The imaginary part of $(3+2 \sqrt{-54})^{1/2} - (3-2 \sqrt{-54})^{1/2}$ can be

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