Let S_{1}={z_{1} in mathbb{C}:|z_{1}-3|=frac{ | vedclass

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(C) Given $|z_{2}-|z_{2}+1||=|z_{2}+|z_{2}-1||$. Squaring both sides,we get:$|z_{2}-|z_{2}+1||^2 = |z_{2}+|z_{2}-1||^2$$(z_{2}-|z_{2}+1|)(\bar{z}_{2}-

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$\frac{3}{2}$

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(C) Given $|z_{2}-|z_{2}+1||=|z_{2}+|z_{2}-1||$. Squaring both sides,we get:$|z_{2}-|z_{2}+1||^2 = |z_{2}+|z_{2}-1||^2$$(z_{2}-|z_{2}+1|)(\bar{z}_{2}-|z_{2}+1|) = (z_{2}+|z_{2}-1|)(\bar{z}_{2}+|z_{2}-1|)$$|z_{2}|^2 - z_{2}|z_{2}+1| - \bar{z}_{2}|z_{2}+1| + |z_{2}+1|^2 = |z_{2}|^2 + z_{2}|z_{2}-1| + \bar{z}_{2}|z_{2}-1| + |z_{2}-1|^2$$-(z_{2}+\bar{z}_{2})(|z_{2}+1| + |z_{2}-1|) = |z_{2}-1|^2 - |z_{2}+1|^2$Using $|z-a|^2 - |z+a|^2 = (z-a)(\bar{z}-\bar{a}) - (z+a)(\bar{z}+\bar{a}) = -2(z\bar{a} + \bar{z}a)$,for $a=1$,we get $|z-1|^2 - |z+1|^2 = -2(z+\bar{z}) = -4	ext{Re}(z_{2})$.So,$-(z_{2}+\bar{z}_{2})(|z_{2}+1| + |z_{2}-1|) = -2(z_{2}+\bar{z}_{2})$.$(z_{2}+\bar{z}_{2})(|z_{2}+1| + |z_{2}-1| - 2) = 0$.This implies $z_{2}+\bar{z}_{2}=0$ (imaginary axis) or $|z_{2}-1| + |z_{2}+1| = 2$ (line segment on real axis from $-1$ to $1$).$S_{1}$ is a circle with center $3$ and radius $\frac{1}{2}$.The distance from the set $S_{2}$ to the center of the circle is the minimum distance minus the radius.The closest point in $S_{2}$ to the circle is $z_{2}=1$. The distance is $|3-1| - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}$.

Let $S_{1}=\{z_{1} \in \mathbb{C}:|z_{1}-3|=\frac{1}{2}\}$ and $S_{2}=\{z_{2} \in \mathbb{C}:|z_{2}-|z_{2}+1||=|z_{2}+|z_{2}-1||\}$. Then,for $z_{1} \in S_{1}$ and $z_{2} \in S_{2}$,the least value of $|z_{2}-z_{1}|$ is:

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