1. Interpret $S_1$ geometrically

Given

$$S_1=\left\{z_1\in\mathbb C:\ |z_1-3|=\frac12\right\}.$$

This is the circle in the Argand plane with:

• center at $3$ i.e. $(3,0)$,
• radius $\frac12$.

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2. Simplify the condition for $S_2$

We are given

$$S_2=\left\{z_2\in\mathbb C:\ \left|z_2-|z_2+1|\right|=\left|z_2+|z_2-1|\right|\right\}.$$

Let

$$z_2=x+iy, \qquad r_1=|z_2+1|, \qquad r_2=|z_2-1|.$$

Then the condition becomes

$$|z_2-r_1|=|z_2+r_2|.$$

Now square both sides:

$$|z_2-r_1|^2=|z_2+r_2|^2.$$

Using $|w|^2=w\bar w$,

$$|z_2-r_1|^2=|z_2|^2-2r_1\operatorname{Re}(z_2)+r_1^2,$$ $$|z_2+r_2|^2=|z_2|^2+2r_2\operatorname{Re}(z_2)+r_2^2.$$

So

$$-2r_1x+r_1^2=2r_2x+r_2^2.$$

Hence

$$r_1^2-r_2^2=2x(r_1+r_2).$$

But

$$r_1^2=|z_2+1|^2=(x+1)^2+y^2,$$ $$r_2^2=|z_2-1|^2=(x-1)^2+y^2.$$

Therefore

$$r_1^2-r_2^2=(x+1)^2-(x-1)^2=4x.$$

So the equation becomes

$$4x=2x(r_1+r_2).$$

That is,

$$2x=x(r_1+r_2).$$

Thus

$$x\bigl((r_1+r_2)-2\bigr)=0.$$

So either

• $x=0$, or
• $r_1+r_2=2$.

Now $r_1+r_2=|z_2+1|+|z_2-1|$ is the sum of distances from $(x,y)$ to $(-1,0)$ and $(1,0)$. By triangle inequality,

$$|z_2+1|+|z_2-1|\ge 2,$$

and equality holds exactly for points on the line segment joining $-1$ and $1$, i.e.

$$y=0, \quad -1\le x\le 1.$$

But these points are already included in $x=0$ only when $x=0$; otherwise they also satisfy the second case.

Hence

$$S_2=\{(x,y):x=0\}\ \cup\ \{(x,0):-1\le x\le 1\}.$$

So geometrically, $S_2$ is:

• the imaginary axis, and
• the real segment from $-1$ to $1$.

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3. Find the least distance between $S_1$ and $S_2$

The circle $S_1$ has center $(3,0)$ and radius $\frac12$.

We first find the minimum distance from the center $(3,0)$ to $S_2$.

(i) Distance from $(3,0)$ to the imaginary axis $x=0$

This distance is

$$3.$$

(ii) Distance from $(3,0)$ to the segment $[-1,1]$ on real axis

The nearest point on this segment to $(3,0)$ is $(1,0)$, so distance is

$$3-1=2.$$

Thus the minimum distance from the center of the circle to $S_2$ is

$$2.$$

Since $S_1$ is a circle of radius $\frac12$, the least distance from a point of $S_2$ to the circle is

$$2-\frac12=\frac32.$$

This is attained along the real axis: take

$$z_2=1\in S_2,$$

and the nearest point on $S_1$ is

$$z_1=\frac52,$$

for which

$$|z_2-z_1|=\left|1-\frac52\right|=\frac32.$$

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4. Evaluate options

• A: $0$ — impossible, since $S_1$ lies near $x=3$ while $S_2$ is at $x=0$ or between $-1$ and $1$ on real axis.
• B: $\frac12$ — too small.
• C: $\frac32$ — correct.
• D: $\frac52$ — not minimum.

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5. Final answer

The least value of $|z_2-z_1|$ is

$$\boxed{\frac32}.$$