1. Interpret the set (S)

Let $w=2x+iy.$ We are given that (w\in S), where $S=\{z=X+iY: |z-1+i|\ge |z|,\ |z|<2,\ |z+i|=|z-1|\}.$

So for (w=2x+iy), we must satisfy all three conditions:

• (|w-(1-i)|\ge |w|), i.e. (|w-1+i|\ge |w|)
• (|w|<2)
• (|w+i|=|w-1|)

We need the set of all real (x) such that there exists some real (y) for which these hold.

2. Use the condition (|w+i|=|w-1|)

Put $w=2x+iy.$ Then $w+i=2x+i(y+1),$ $w-1=(2x-1)+iy.$

So $|w+i|=|w-1|$ becomes $\sqrt{(2x)^2+(y+1)^2}=\sqrt{(2x-1)^2+y^2}.$

Squaring, $4x^2+(y+1)^2=(2x-1)^2+y^2.$

Expand: $4x^2+y^2+2y+1=4x^2-4x+1+y^2.$

Thus $2y=-4x \implies y=-2x.$

Hence every admissible point must lie on the line $y=-2x.$

So $w=2x+i(-2x)=2x(1-i).$

3. Apply the condition (|w|<2)

Now $|w|=|2x(1-i)|=2|x|\cdot |1-i|=2|x|\sqrt2.$

Thus $2\sqrt2|x|<2$ which gives $|x|<\frac{1}{\sqrt2}.$

So far, $-\frac{1}{\sqrt2}<x<\frac{1}{\sqrt2}.$

4. Apply the inequality (|w-1+i|\ge |w|)

Since (y=-2x), $w=2x-2ix.$ Then $w-1+i=(2x-1)+i(1-2x).$

Hence $|w-1+i|^2=(2x-1)^2+(1-2x)^2=2(2x-1)^2.$

Also, $|w|^2=(2x)^2+(-2x)^2=8x^2.$

Condition (|w-1+i|\ge |w|) is equivalent to $|w-1+i|^2\ge |w|^2,$ so $2(2x-1)^2\ge 8x^2.$

Divide by 2: $(2x-1)^2\ge 4x^2.$

Expand: $4x^2-4x+1\ge 4x^2,$ so $-4x+1\ge 0,$ which gives $x\le \frac14.$

5. Combine all conditions

From step 3: $-\frac{1}{\sqrt2}<x<\frac{1}{\sqrt2}.$

From step 4: $x\le \frac14.$

Therefore, $x\in \left(-\frac{1}{\sqrt2},\ \frac14\right].$

Note that (x=\frac14) is allowed because the inequality is non-strict, and (|w|<2) is still satisfied there.

6. Match with options

This corresponds to $\boxed{\left(-\frac{1}{\sqrt2},\ \frac14\right]}$ which is Option B.

7. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.