Question

Let \(s, t, r\) be non-zero complex numbers and \(L\) be the set of solutions \(z=x+i y\) \((x, y \in \mathbb{R}, i=\sqrt{-1})\) of the equation \(s z+t \bar{z}+r=0\), where \(\bar{z}=x-i y\). Then, which of the following statement \((\mathrm{s})\) is (are) TRUE? (A) If \(L\) has exactly one element, then \(|s| \neq|t|\) (B) If \(|s|=|t|\), then \(L\) has infinitely many elements (C) The number of elements in \(L \cap\\{z:|z-1+i|=5\\}\) is at most 2 (D) If \(L\) has more than one element, then \(L\) has infinitely many elements

Short answer

(A) True, (B) True, (C) True, (D) True. When L has exactly one element, then |s| eq |t|. If |s| = |t|, there are infinitely many solutions in L. The number of elements in \(L \cap\{z : |z - 1 + i| = 5\}\) is at most 2. If L has more than one element, then it must have infinitely many elements.

Step-by-step solution

Express the given equation in terms of x and y

Rewrite the equation using the relationship between z and its conjugate: substitute into the equation \(s z+t \bar{z}+r=0\) with \(z = x + iy\) and \(\bar{z} = x - iy\) to get \(s(x + iy) + t(x - iy) + r = 0\). Separate the real and imaginary parts to use in further steps.

Separate the real and imaginary parts of the equation

Separate the real and imaginary parts of the equation from the previous step to form two real equations: one equating all the real parts to zero and another equating all the imaginary parts to zero.

Analyze the condition for exactly one element in L

Determine the condition for the set L to have exactly one solution. If the coefficients of x and y in real and imaginary parts are not proportional, there will be a unique solution for x and y, so in this case, \(|s| eq |t|\). This addresses option (A).

Discuss the implications of |s| = |t|

If \(|s| = |t|\), the coefficient of x in the real part will have the same magnitude as the coefficient of x in the imaginary part, and similarly for y, making the equations for x and y proportional. Thus, there are infinitely many solutions for x and y, and hence for z. This confirms option (B).

Determine the number of intersection points with the circle

For a fixed circle \(\{z : |z - 1 + i| = 5\}\), we will analyze the intersection points with L by substituting \(z\) into the circle's equation and solving for x and y. Depending on the given conditions, the maximum number of intersections can be determined as at most 2 which corresponds to option (C).

Assess the number of elements in L when there's more than one

If L has more than one element, the system of equations derived for x and y must be dependent, meaning that there is not a unique solution but rather an infinite number of solutions, satisfying option (D).

Key concepts

Complex Numbers in Equations

Understanding how complex numbers can be integrated into equations is crucial for solving many mathematical problems involving complex variables. Here, a given equation involves a complex number z and its conjugate \bar{z}. The concept of complex conjugate is pivotal, where for any complex number z = x + iy, its conjugate is \bar{z} = x - iy.

In the given exercise, separating the real and imaginary parts transforms the equation into a system of linear equations. If these equations are independent, then a unique solution exists for the real x and imaginary y components of z. However, this is often not straightforward, as we must consider the magnitude of the complex coefficients s and t. When |s| \neq |t|, we can infer that the system of equations will yield a unique solution, implying a single solution for z in set L. Conversely, if |s| = |t|, then the system of equations has infinite solutions, leading to an infinite set of solutions for z.

Set of Solutions for Complex Equations

When it comes to analyzing the set of solutions, L, for complex equations, we are essentially looking for values of z that satisfy the given conditions. It's helpful to think of this in geometric terms: each solution represents a point in the complex plane, and the set L forms a collection of these points.

For instance, if an equation has exactly one solution, the set L corresponds to a single point on the complex plane. When |s| = |t|, the equation forms a line—either straight or circular—populated with an infinite number of points, representing infinite solutions for z. Additionally, when combining sets, such as L and a circle in the complex plane, the intersection represents common solutions, leading to specific scenarios like a maximum of two intersection points. Such scenarios underline the rich and varied nature of solution sets in complex-number equations.

Intersection of Complex Solution Sets

The intersection of complex solution sets involves finding common solutions between two or more sets of complex numbers. This often requires combining geometric and algebraic methods.

In our exercise, we consider the intersection of set L with a circle defined by \{z : |z - 1 + i| = 5\}. Here, \(|z - 1 + i| = 5\) represents all points that are exactly 5 units from the complex number 1 - i on the complex plane. To find the intersection, we compare the conditions that define L and the circular set, which might involve substituting values of z from L into the circle's equation and solving the resulting system. Depending on how these sets are positioned relative to each other, there can be zero, one, or two points of intersection, which is tied to the geometric shapes these sets represent.