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(B) For statement $I$: $\sqrt{-a} 	imes \sqrt{-b} = (i\sqrt{a}) 	imes (i\sqrt{b}) = i^2 \sqrt{ab} = -\sqrt{ab}$. Thus,statement $I$ is false.For sta

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Only $II$ is true

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(B) For statement $I$: $\sqrt{-a} 	imes \sqrt{-b} = (i\sqrt{a}) 	imes (i\sqrt{b}) = i^2 \sqrt{ab} = -\sqrt{ab}$. Thus,statement $I$ is false.For statement $II$: Let $z = \frac{1+i\sqrt{3}}{1-i\sqrt{3}}$. Multiplying numerator and denominator by the conjugate of the denominator $(1+i\sqrt{3})$:$z = \frac{(1+i\sqrt{3})^2}{1^2 + (\sqrt{3})^2} = \frac{1 - 3 + 2i\sqrt{3}}{4} = \frac{-2 + 2i\sqrt{3}}{4} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$.This complex number lies in the second quadrant. The argument is $\pi - 	an^{-1}(\frac{\sqrt{3}/2}{1/2}) = 180^{\circ} - 60^{\circ} = 120^{\circ}$. Thus,statement $II$ is true.

Consider the following statements:
$I$: If $a$ and $b$ are positive real numbers,then $\sqrt{-a} \times \sqrt{-b} = \sqrt{ab}$
$II$: The argument of $\frac{1+i\sqrt{3}}{1-i\sqrt{3}}$ is $120^{\circ}$
Then:

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Consider the following statements:$I$: If $a$ and $b$ are positive real numbers,then $\sqrt{-a} \times \sqrt{-b} = \sqrt{ab}$$II$: The argument of $\frac{1+i\sqrt{3}}{1-i\sqrt{3}}$ is $120^{\circ}$Then: