Animated Solution for Mathematics - Complex Numbers: Let $s,t,r$ be non-zero complex numbers and $L$ be the set of solutions $z=x+iy$ $(x,y\in R,i=\sqrt{-1})$ of the equation $sz+t\bar{z}+r=0$, where $\bar{z}=x-iy$. Then, which of the following statement(s) is (are) TRUE ?

Imagine standing on the complex plane, a vast, two-dimensional canvas where every point is a number. We are tackling the linear constraint $sz+t\bar{z}+r=0$.

This equation is a classic JEE Advanced trap because it hides its true geometric nature behind a veil of complex algebra. Let us pull back that veil together.

At first glance, this equation looks like a simple linear equation. But there is a catch: the presence of $\bar{z}$.

In the world of complex numbers, $z$ and $\bar{z}$ are not independent variables; they are reflections of each other across the real axis. If you try to solve for $z$ by moving terms around, you will find yourself stuck in a loop.

The secret is to create a system. We take the original equation, $sz+t\bar{z}+r=0$, and we take its conjugate:

Using the properties of conjugates—where the conjugate of a sum is the sum of conjugates and the conjugate of a product is the product of conjugates—we arrive at a second, perfectly symmetric equation:

Now, we have a system of two linear equations. We have successfully "untangled" the variables.

Now that we have our system, our goal is to isolate $z$ by eliminating $\bar{z}$. To do this, we multiply the first equation by $\bar{s}$ and the second by $t$.

This makes the coefficients of $\bar{z}$ identical, specifically $t\bar{s}\bar{z}$. When we subtract these two equations, the $\bar{z}$ terms vanish into thin air, leaving us with:

This is the moment of truth. This equation tells us everything we need to know about the set $L$.

We have arrived at the core of the problem. The behavior of our set $L$ depends entirely on the coefficient $(|s{|}^2-|t{|}^2)$.

1. The Unique Point: If $|s|^2 - |t|^2 eq 0$, then $|s| eq |t|$. We can divide by this non-zero coefficient to find a unique value for $z$. Geometrically, this means the set $L$ is just a single point. This confirms that Option A is true.

2. The Line or the Void: If $|s|=|t|$, the coefficient becomes zero. Our equation simplifies to:

If the right-hand side is zero, we get $0=0$, which is true for all $z$ on a line (infinitely many solutions). If the right-hand side is non-zero, we get a contradiction, meaning the set $L$ is empty. This is why Option B is false—it assumes there are always infinitely many solutions, ignoring the possibility of an empty set.

3. The Intersection: Option C asks about the intersection of $L$ with a circle $|z-1+i|=5$. Since $L$ can only be a point, a line, or empty, the intersection with a circle can have at most two points. Thus, Option C is true.

Finally, Option D is a logical consequence: if $L$ has more than one element, it cannot be a point or empty, so it must be a line, which contains infinitely many points. This is the elegance of complex geometry—what starts as a messy algebraic expression reveals itself to be a simple, beautiful geometric truth.