Key Concepts and Formulas

• Complex Number Representation: A complex number $z$ can be represented as $z = x + iy$, where $x = \operatorname{Re}(z)$ and $y = \operatorname{Im}(z)$ are real numbers.
• Real Part of a Complex Expression: If $z = x + iy$, then $\operatorname{Re}(z) = x$.
• Squaring a Complex Number: $(x+iy)^2 = x^2 - y^2 + 2ixy$.

Step-by-Step Solution

Step 1: Express $z$ and $z^2$ in terms of real and imaginary parts.

We represent the complex number $z$ as $z = x + iy$, where $x$ and $y$ are real numbers. We need to express $z^2$ in terms of $x$ and $y$ to find the real parts of the given expressions. $z = x + iy$ $z^2 = (x + iy)^2 = x^2 + 2ixy - y^2 = (x^2 - y^2) + i(2xy)$

Step 2: Calculate $\operatorname{Re}(az^2 + bz)$ and set it equal to $a$.

We substitute $z = x + iy$ and $z^2 = (x^2 - y^2) + i(2xy)$ into the expression $az^2 + bz$ and find its real part. This real part must be equal to $a$ as given in the problem statement. $az^2 + bz = a[(x^2 - y^2) + i(2xy)] + b(x + iy) = a(x^2 - y^2) + 2axyi + bx + byi$ $az^2 + bz = [a(x^2 - y^2) + bx] + i[2axy + by]$ $\operatorname{Re}(az^2 + bz) = a(x^2 - y^2) + bx$ Since $\operatorname{Re}(az^2 + bz) = a$: $a(x^2 - y^2) + bx = a$ Since $a \neq 0$, we can divide by $a$: $x^2 - y^2 + \frac{b}{a}x = 1 \quad \text{(Equation 1)}$

Step 3: Calculate $\operatorname{Re}(bz^2 + az)$ and set it equal to $b$.

Similarly, we substitute $z = x + iy$ and $z^2 = (x^2 - y^2) + i(2xy)$ into the expression $bz^2 + az$ and find its real part. This real part must be equal to $b$ as given in the problem statement. $bz^2 + az = b[(x^2 - y^2) + i(2xy)] + a(x + iy) = b(x^2 - y^2) + 2bxyi + ax + ayi$ $bz^2 + az = [b(x^2 - y^2) + ax] + i[2bxy + ay]$ $\operatorname{Re}(bz^2 + az) = b(x^2 - y^2) + ax$ Since $\operatorname{Re}(bz^2 + az) = b$: $b(x^2 - y^2) + ax = b$ Since $b \neq 0$, we can divide by $b$: $x^2 - y^2 + \frac{a}{b}x = 1 \quad \text{(Equation 2)}$

Step 4: Solve the system of equations.

We have the following system of equations: $x^2 - y^2 + \frac{b}{a}x = 1 \quad \text{(Equation 1)}$ $x^2 - y^2 + \frac{a}{b}x = 1 \quad \text{(Equation 2)}$ Subtracting Equation 2 from Equation 1, we get: $\left(x^2 - y^2 + \frac{b}{a}x\right) - \left(x^2 - y^2 + \frac{a}{b}x\right) = 1 - 1$ $\frac{b}{a}x - \frac{a}{b}x = 0$ $x\left(\frac{b}{a} - \frac{a}{b}\right) = 0$ $x\left(\frac{b^2 - a^2}{ab}\right) = 0$ Since $a \neq 0$ and $b \neq 0$, $ab \neq 0$. Also, $a \neq b$. Thus, either $x = 0$ or $b^2 - a^2 = 0$. The latter implies $b^2 = a^2$, so $b = a$ or $b = -a$. Since $a \neq b$, we can have $x=0$ or $b=-a$.

Step 5: Analyze the case $x=0$.

If $x = 0$, substituting into Equation 1 gives: $0^2 - y^2 + \frac{b}{a}(0) = 1$ $-y^2 = 1$ $y^2 = -1$ Since $y$ must be a real number, there are no real solutions for $y$.

Step 6: Analyze the case $b=-a$.

If $b = -a$, substituting this into Equation 1 gives: $x^2 - y^2 + \frac{-a}{a}x = 1$ $x^2 - y^2 - x = 1$ Similarly, substituting $b=-a$ into Equation 2 also yields $x^2 - y^2 - x = 1$. In this case, we have infinitely many solutions since we have one equation with two unknowns.

Step 7: Conclusion.

The problem asks for a definite number of elements in the set $X$. While the case $b=-a$ yields infinite solutions, it is a specific condition not generally true for any non-zero $a \neq b$. In the general case, $x=0$, which leads to no real solution for $y$. Therefore, there are no solutions for $z$.

Common Mistakes & Tips

• Remember to consider both cases when solving for $x$, i.e., $x=0$ and $b=-a$.
• Be careful when dividing by variables; ensure they are not zero.
• The problem asks for the number of elements, so make sure to check if the solutions for $x$ and $y$ are real.

Summary

We represented the complex number $z$ as $x + iy$ and substituted it into the given equations. By equating the real parts and solving the resulting system of equations, we found that in the general case where $a \neq b$ and $a \neq -b$, there are no real solutions for $x$ and $y$. This implies that there are no complex numbers $z$ that satisfy the given conditions.

The final answer is $\boxed{0}$. This corresponds to option (A).