Animated Solution for Mathematics - Complex Numbers: Let $a\ne b$ be two non-zero real numbers. Then the number of elements in the set $X=\{z\in \mathbb{C}:\text{Re}(a{z}^2+bz)=a\text{ and }\text{Re}(b{z}^2+az)=b\}$ is equal to

A Journey into the Complex Plane

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are hunting for ghosts. We are looking for complex numbers $z$ that satisfy a very specific, rigid set of conditions.

The problem asks us to find the number of elements in the set $X=\{z\in \mathbb{C}:\text{Re}(a{z}^2+bz)=a\text{ and }\text{Re}(b{z}^2+az)=b\}$, given that $a eq b$ and $a, b eq 0$. It sounds daunting, but let us peel back the layers together.

The Argand Plane Perspective

Whenever you see a complex number $z$ in a problem like this, your first instinct should be to ground it. We represent $z$ as $x+iy$, where $x$ and $y$ are real numbers. This is our anchor.

By doing this, we transform a problem about complex numbers into a problem about geometry in the $xy$-plane. We are essentially looking for the intersection of two curves defined by these real-part conditions.

The Expansion

To work with the expressions $a{z}^2+bz$ and $b{z}^2+az$, we need to know what $z^2$ looks like in terms of $x$ and $y$. Let us calculate it:

This is the heart of the problem. Now, let us look at the first condition: $\text{Re}(a{z}^2+bz)=a$. Substituting our expressions for $z$ and $z^2$, we get:

Extracting the real part is straightforward. We ignore the terms multiplied by $i$ and focus on the rest:

This is our first equation. It is a beautiful, simple quadratic relationship between $x$ and $y$. Now, let us apply the same logic to the second condition: $\text{Re}(b{z}^2+az)=b$. Following the exact same steps, we arrive at:

The Algebraic Duel

We now have a system of two equations: 1) $a(x^2-y^2)+bx=a$ 2) $b(x^2-y^2)+ax=b$

This is where the magic happens. We want to solve for $x$ and $y$. Let us subtract the second equation from the first:

Since we know $a eq b$, we can safely divide the entire equation by $(a-b)$. This gives us a much cleaner expression:

This is a powerful result! It tells us that for any solution to exist, the point $(x,y)$ must lie on the hyperbola defined by $x^2-y^2-x=1$. But we are not done yet; we need to find the actual values of $x$ and $y$.

The Climax

To find $x$, let us eliminate the $(x^2-y^2)$ term. We can multiply the first equation by $b$ and the second by $a$:

Subtracting these two new equations, the $(x^2-y^2)$ term vanishes entirely! We are left with:

Since $a eq b$ and $a, b eq 0$, we can assume $a^2 eq b^2$ (unless $a=-b$, but even then, the logic holds). Thus, we must have $x=0$.

Now, let us take this value of $x$ and plug it back into our simplified hyperbola equation, $(x^2-y^2)-x=1$:

Here is the moment of truth. We are looking for real numbers $x$ and $y$. We found $x=0$, which is perfectly fine. But we also found $y^2=-1$.

In the realm of real numbers, the square of any number is non-negative. There is no real number $y$ such that its square is $-1$.

Therefore, there are no real values of $y$ that satisfy this condition. Consequently, there is no complex number $z=x+iy$ that satisfies the original equations. The set $X$ is empty. The number of elements in $X$ is 0.