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(B) Given $z_0 = \frac{1}{2} + \frac{3}{2}i$ and $z_1 = 1 + i$.Calculate $|z_1 - z_0| = |(1 - \frac{1}{2}) + (1 - \frac{3}{2})i| = |\frac{1}{2} - \fra

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$\frac{5}{2}$

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(B) Given $z_0 = \frac{1}{2} + \frac{3}{2}i$ and $z_1 = 1 + i$.Calculate $|z_1 - z_0| = |(1 - \frac{1}{2}) + (1 - \frac{3}{2})i| = |\frac{1}{2} - \frac{1}{2}i| = \sqrt{(\frac{1}{2})^2 + (-\frac{1}{2})^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}}$.Given $|z_1 - z_0| |z_2 - z_0| = 1$,we have $\frac{1}{\sqrt{2}} |z_2 - z_0| = 1$,which implies $|z_2 - z_0| = \sqrt{2}$.Since $z_0, z_1, z_2$ are collinear,$z_2$ lies on the line passing through $z_0$ and $z_1$. The vector $z_1 - z_0 = \frac{1}{2} - \frac{1}{2}i$. The direction of this line is given by the angle $	heta$ where $	an 	heta = \frac{-1/2}{1/2} = -1$,so $	heta = 135^{\circ}$ or $315^{\circ}$.Thus,$z_2 = z_0 + \sqrt{2} e^{i 	heta} = (\frac{1}{2} + \frac{3}{2}i) + \sqrt{2} (\cos 	heta + i \sin 	heta)$.For $	heta = 135^{\circ}$,$z_2 = (\frac{1}{2} + \sqrt{2} \cdot (-\frac{1}{\sqrt{2}})) + i(\frac{3}{2} + \sqrt{2} \cdot \frac{1}{\sqrt{2}}) = (\frac{1}{2} - 1) + i(\frac{3}{2} + 1) = -\frac{1}{2} + \frac{5}{2}i$.Then $|z_2|^2 = (-\frac{1}{2})^2 + (\frac{5}{2})^2 = \frac{1}{4} + \frac{25}{4} = \frac{26}{4} = \frac{13}{2}$.For $	heta = 315^{\circ}$,$z_2 = (\frac{1}{2} + \sqrt{2} \cdot \frac{1}{\sqrt{2}}) + i(\frac{3}{2} + \sqrt{2} \cdot (-\frac{1}{\sqrt{2}})) = (\frac{1}{2} + 1) + i(\frac{3}{2} - 1) = \frac{3}{2} + \frac{1}{2}i$.Then $|z_2|^2 = (\frac{3}{2})^2 + (\frac{1}{2})^2 = \frac{9}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2}$.The smaller value of $|z_2|^2$ is $\frac{5}{2}$.

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