Animated Solution for Mathematics - Complex Numbers: Let $S=\{z=x+iy:|z-1+i|\ge |z|,|z|<2,|z+i|=|z-1|\}$. Then the set of all values of $x$, for which $w=2x+iy\in S$ for some $y\in \mathbb{R}$, is

Welcome, future engineers! Today, we are going to peel back the layers of a beautiful complex numbers problem. Often, students see complex numbers and immediately think of algebraic manipulation.

But the secret to mastering JEE Advanced problems is to see the geometry hidden within the algebra. Let us embark on this journey together.

Decoding the Set S

We are given a set $S$ defined by three conditions on a complex number $z=X+iY$. Let us visualize these one by one.

The first condition is $|z+i|=|z-1|$. This is the locus of a point equidistant from two fixed points: $-i$ (or $(0,-1)$) and $1$ (or $(1,0)$). This is the perpendicular bisector of the segment joining these two points.

Algebraically, we square both sides:

Expanding this, we get $X^2+Y^2+2Y+1=X^2-2X+1+Y^2$. With a satisfying cancellation of $X^2$ and $Y^2$, we are left with $2Y=-2X$, or simply $Y=-X$. A beautiful, clean line passing through the origin!

Next, we have $|z-1+i|\ge |z|$. This represents the region on or above the line $Y-X+1\ge 0$.

Finally, $|z|<2$ is the interior of a circle centered at the origin with radius $2$. The set $S$ is the intersection of these three: the line segment of $Y=-X$ that lies inside the circle and on or above the line $Y=X-1$.

The Bridge to w

Now, we are given $w=2x+iy\in S$. This is where the magic happens. We are asked for the range of $x$.

Since $w$ is in $S$, it must satisfy the conditions we just derived. We map $w=X+iY$ to our coordinates: $X=2x$ and $Y=y$.

This is our bridge. We substitute these into our conditions. The first condition $Y=-X$ becomes $y=-2x$. This is the constraint that binds our imaginary part $y$ to our real part $x$.

The Intersection

Now, we substitute $X=2x$ and $Y=-2x$ into the remaining two conditions.

For the second condition, $Y-X+1\ge 0$, we get:

This simplifies to $-4x+1\ge 0$, or $x\le \frac{1}{4}$.

For the third condition, $X^2+Y^2<4$, we get:

This results in $4x^2+4x^2<4$, or $8x^2<4$. This simplifies to $x^2<\frac{1}{2}$, meaning:

We have two inequalities for $x$: $x\le \frac{1}{4}$ and $-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$.

Taking the intersection, we find the range of $x$ to be:

This is the set of all values of $x$ for which $w$ exists in $S$. You see? By breaking the problem into geometric pieces and then building a bridge with substitution, the solution reveals itself. Keep this mindset, and no problem will ever be too complex for you!