Animated Solution for Mathematics - Complex Numbers: Comprehension Passage

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey through the complex plane. Many students fear complex numbers, treating them as abstract algebraic entities.

But I want you to see them for what they truly are: the language of geometry. When you look at the set $A=\{z:Im(z)\ge 1\}$, do not just see an inequality. See a region.

See a half-plane. Imagine standing on the horizontal line $y=1$ and looking up. Everything above that line, including the line itself, is your playground. That is Set $A$.

Now, let us look at Set $B=\{z:|z-(2+i)|=3\}$. This is the heartbeat of the problem. In the complex plane, the equation $|z-z_0|=r$ is the definition of a circle with center $z_0$ and radius $r$.

Here, our center is $z_0=2+i$, which corresponds to the point $(2,1)$ in Cartesian coordinates. The radius is $3$. If you draw this, you will notice something fascinating: the center $(2,1)$ lies exactly on the boundary of our Set $A$. This is not a coincidence; it is a hint from the examiner.

Finally, we have Set $C=\{z:Re((1-i)z)=\sqrt{2}\}$. This looks intimidating, doesn't it? Let us peel back the layers. Let $z=x+iy$.

Then $(1-i)(x+iy)=x+iy-ix-i^{2}y=(x+y)+i(y-x)$. The real part is simply $x+y$. So, the condition is $x+y=\sqrt{2}$. This is a straight line with intercepts at $(\sqrt{2},0)$ and $(0,\sqrt{2})$.

We are looking for the intersection $A\cap B\cap C$. We need a point that satisfies all three conditions. We have a line $x+y=\sqrt{2}$ and a circle $(x-2{)}^2+(y-1{)}^2=9$.

If we substitute $y=\sqrt{2}-x$ into the circle equation, we get a quadratic in $x$. Solving this will give us two intersection points. But remember, we have the constraint from Set $A$: $y\ge 1$.

When you solve for $y$, you will find that only one of these points satisfies the condition $y\ge 1$. This is our unique point $P$. We have successfully navigated the first hurdle!

Now, the second part of the problem asks us to evaluate $|z-(-1+i){|}^2+|z-(5+i){|}^2$. Let $Q=-1+i$ and $R=5+i$. The expression is $P{Q}^2+P{R}^2$.

Look at the coordinates: $Q$ is $(-1,1)$ and $R$ is $(5,1)$. The midpoint is $(\frac{-1+5}{2},1)=(2,1)$. This is the center of our circle! The distance $QR$ is $6$, which is exactly twice the radius. This means $QR$ is a diameter.

Here is where the magic happens. By the property of circles, any point $P$ on the circumference subtends a $90^{\circ }$ angle with the diameter. Therefore, $△PQR$ is a right-angled triangle.

By the Pythagoras Theorem, $P{Q}^2+P{R}^2=Q{R}^2$. Since $QR=6$, the sum is $36$. And $36$ lies perfectly between $35$ and $39$. Do you see the elegance? We didn't need to find the exact coordinates of $P$ to solve this. We only needed the geometry.

Finally, we consider $|z|-|w|+3$, where $w$ is any point satisfying $|w-2i|<3$. This describes the interior of a circle centered at $(0,2)$ with radius $3$.

By the Triangle Inequality, the absolute difference $||z|-|w||$ is bounded by the distance between the points. Through geometric analysis of the two regions, we find that the expression $|z|-|w|$ is bounded between $-6$ and $6$.

Adding $3$ to this range gives us the interval $(-3,9)$. My dear student, this problem was never about brute-force calculation. It was about seeing the geometry hidden within the algebra.

When you approach JEE Advanced problems, always ask yourself: "What is the shape? What is the constraint?" Once you see the picture, the math becomes a mere formality. Keep practicing, keep visualizing, and keep falling in love with the physics and math behind the problem.