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(D) We know that $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + 2|z_1||z_2| \cos(	heta)$,where $	heta = |Arg(z_1) - Arg(z_2)|$.Given $|z_1| = \sqrt{2}$,$|z_2|

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$\frac{3\pi}{4}$

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(D) We know that $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + 2|z_1||z_2| \cos(	heta)$,where $	heta = |Arg(z_1) - Arg(z_2)|$.Given $|z_1| = \sqrt{2}$,$|z_2| = \sqrt{3}$,and $|z_1 + z_2| = \sqrt{5 - 2\sqrt{3}}$.Substituting these values into the formula:$5 - 2\sqrt{3} = (\sqrt{2})^2 + (\sqrt{3})^2 + 2(\sqrt{2})(\sqrt{3}) \cos(	heta)$$5 - 2\sqrt{3} = 2 + 3 + 2\sqrt{6} \cos(	heta)$$5 - 2\sqrt{3} = 5 + 2\sqrt{6} \cos(	heta)$$-2\sqrt{3} = 2\sqrt{6} \cos(	heta)$$\cos(	heta) = \frac{-2\sqrt{3}}{2\sqrt{6}} = -\frac{\sqrt{3}}{\sqrt{6}} = -\frac{1}{\sqrt{2}}$Since $\cos(	heta) = -\frac{1}{\sqrt{2}}$,we have $	heta = \frac{3\pi}{4}$.

List-$I$ (Complex number)	List-$II$ (Polar form)
$(i) \sqrt{3}-i$	$(a) 2 \operatorname{cis} \frac{\pi}{6}$
$(ii) \sqrt{3}+i$	$(b) 2 \operatorname{cis} \frac{5 \pi}{6}$
$(iii) -\sqrt{3}+i$	$(c) 2 \operatorname{cis}\left(-\frac{5 \pi}{6}\right)$
$(iv) -\sqrt{3}-i$	$(d) 2 \operatorname{cis}\left(-\frac{\pi}{6}\right)$

If complex numbers $z_1$ and $z_2$ are such that $|z_1| = \sqrt{2}$,$|z_2| = \sqrt{3}$ and $|z_1 + z_2| = \sqrt{5 - 2\sqrt{3}}$,then the value of $|Arg(z_1) - Arg(z_2)|$ is

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