Animated Solution for Mathematics - Complex Numbers: Let $S=\{z\in \mathbb{C}:|z-1|=1\text{ and }(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2\sqrt{2}\}$. Let $z_1,z_2\in S$ be such that $|z_1|={\max }_{z\in S}|z|$ and $|z_2|={\min }_{z\in S}|z|$. Then $|\sqrt{2}{z}_1-z_2{|}^2$ equals :

Welcome, future engineer. Today, we are not just solving an equation; we are exploring the elegant dance between geometry and algebra. When you look at a problem involving complex numbers, I want you to stop seeing $z$ as just a variable. See it as a point, a traveler wandering through the Argand plane.

Our goal is to find where this traveler is allowed to exist. The problem presents us with a set $S$, defined by two conditions. The first is $|z-1|=1$.

If you have spent any time with complex numbers, your intuition should immediately scream 'Circle!' This is the locus of all points at a distance of $1$ from the point $(1,0)$.

What about the second condition? $(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2\sqrt{2}$. It looks intimidating, doesn't it? This is where we use our most reliable tool: the Cartesian substitution.

Let $z=x+iy$. We know that $z+\bar{z}=2x$ and $z-\bar{z}=2iy$. Substituting these into our equation, we get:

Since $i^2=-1$, the term $-i(2iy)$ becomes $+2y$. Dividing the entire equation by $2$, we arrive at the beautiful, clean equation of a straight line:

We have successfully translated the abstract complex constraints into the language of coordinate geometry: a circle and a line.

Now, we need to find where these two paths cross. We substitute $y=\sqrt{2}-(\sqrt{2}-1)x$ into the circle equation $(x-1{)}^2+y^2=1$. This leads us to a quadratic equation in $x$.

After careful expansion and grouping, we arrive at:

If you divide by $2$, you get $(2-\sqrt{2})x^2-(3-\sqrt{2})x+1=0$. Here is the secret: always check the sum of the coefficients.

Here, $(2-\sqrt{2})-(3-\sqrt{2})+1=0$. When the sum of coefficients is zero, $x=1$ is always a root! This is a classic shortcut that saves you from the quadratic formula.

The other root, by the product of roots, is $x=\frac{1}{2-\sqrt{2}}$, which simplifies to $1+\frac{1}{\sqrt{2}}$.

We have our $x$-coordinates: $x=1$ and $x=1+\frac{1}{\sqrt{2}}$. Plugging these back into our line equation gives us the corresponding $y$-coordinates.

For $x=1$, we get $z_2=1+i$. For $x=1+\frac{1}{\sqrt{2}}$, we get $z_1=(1+\frac{1}{\sqrt{2}})+\frac{i}{\sqrt{2}}$.

Now, we calculate the modulus to identify $z_1$ (maximum) and $z_2$ (minimum). A quick check shows that $|z_2|=\sqrt{2}$ and $|z_1|=\sqrt{2+\sqrt{2}}$.

Finally, we evaluate the expression $|\sqrt{2}{z}_1-z_2{|}^2$. Watch closely as we multiply $z_1$ by $\sqrt{2}$:

Now, subtract $z_2=1+i$:

The $1$ and $i$ terms vanish, leaving us with just $\sqrt{2}$. Squaring this gives us 2.

This is the essence of JEE Advanced mathematics. It is not about brute force; it is about finding the path of least resistance, trusting your algebraic manipulations, and watching as the complexity dissolves into a simple, elegant integer.