Question

Let s, t, r be the non-zero complex numbers and L be the set of solution $z = x + iy$ $\left( {x,y \in R,i = \sqrt { - 1} } \right)$ of the equation $sz + t\overline z + r = 0$, where $\overline z = x - iy$. Then which of the following statement(s) is (are) true?
This question has multiple correct options.
A. If L has exactly one element, then $\left| s \right| \ne \left| t \right|$.
B. $\left| s \right| = \left| t \right|$, then L has infinitely many elements.
C. The number of elements in L \cap \left\\{ {z:\left| {z - 1 + i} \right| = 5} \right\\} is at most 2.
D. If L has more than one element, then L has infinitely many elements.

Answer 1

To solve this question, we have to remember some basic concepts of complex numbers. We will use the concept that for a complex number, $z = a + ib$ is a and b both are equals to 0, then $2\operatorname{Re} \left( z \right) = z + \overline z$ and $2\operatorname{Im} \left( z \right) = z - \overline z$, where Re(z) is real part and Im(z) is imaginary part of the complex number z.

Complete step by step answer:
A number in the form of $a + ib$, where a and b are real numbers, is called a complex number.
For the complex number $z = a + ib$, a is the real part, denoted by Re z and b is the imaginary part, denoted by Im z of the complex number z.
According to the question,
s, t, r be the non-zero complex numbers and their equation is given as:
$sz + t\overline z + r = 0$.
So , we can say that $sz + t\overline z + r = 0$ is also in the form of a complex number and a complex number is of the form $a + ib$. Therefore,
$\Rightarrow sz + t\overline z + r = a + ib$.
We are given that this value is equal to 0.
$\Rightarrow sz + t\overline z + r = a + ib = 0$.
We know that, if a complex number is equals to 0, then its real part and imaginary part both will be equals to 0, i.e.
$\Rightarrow a = 0,b = 0$.
We also know that, if a complex number is equals to 0, then
$2\operatorname{Re} \left( z \right) = z + \overline z$ and $2\operatorname{Im} \left( z \right) = z - \overline z$.
Let us say, $Z' = sz + t\overline z + r$.
Now we have to find the conjugate of $Z'$.
$\Rightarrow \overline {Z'} = \overline s \overline z + \overline t z + \overline r$.
Now,
$2\operatorname{Re} \left( {Z'} \right) = Z' + \overline {Z'}$.
Putting the values, we will get

$$\Rightarrow 2\operatorname{Re} \left( {Z'} \right) = sz + t\overline z + r + \overline s \overline z + \overline t z + \overline r \\\ \Rightarrow 2\operatorname{Re} \left( {Z'} \right) = z\left( {s + \overline t } \right) + \overline z \left( {\overline s + t} \right) + r + \overline r \\\ \Rightarrow 2\operatorname{Im} \left( {Z'} \right) = sz + t\overline z + r - \overline s \overline z - \overline t z - \overline r \\\ \Rightarrow 2\operatorname{Im} \left( {Z'} \right) = z\left( {s - \overline t } \right) + \overline z \left( {t - \overline s } \right) + r - \overline r \\\$$

Now, we know that these are two equations of line and they will be parallel and coincident, if the ratio of coefficients of $Z'$ and $\overline {Z'}$ are equal. i.e.
$\Rightarrow \dfrac{{s + \overline t }}{{s - \overline t }} = \dfrac{{t + \overline s }}{{t - \overline s }}$.
Cross-multiplication on both sides,
$\Rightarrow \left( {s + \overline t } \right)\left( {t - \overline s } \right) = \left( {t + \overline s } \right)\left( {s - \overline t } \right) \\\ \Rightarrow st - {\left| s \right|^2} + {\left| t \right|^2} - \overline s \overline t = st + {\left| s \right|^2} - {\left| t \right|^2} - \overline s \overline t \\\ \Rightarrow 2{\left| s \right|^2} = 2{\left| t \right|^2} \\\ \Rightarrow \left| s \right| = \left| t \right| \\\$
Here, we clearly see that, if $\left| s \right| \ne \left| t \right|$, then the lines will intersect and exactly one solution will form.
Therefore, option (A) is the correct answer.
We know that if L has more than one element, the L will have infinitely many elements.
Therefore, the option (D) is also correct.
Here, we have L.
L can be either a null set or a point or it can be a line.
If L is a null set, then it will have 0 elements.
If L is a point, then it will have only one element.
And if L is a line, then it will have two elements.
So, we can say that L will have either zero, one or two elements.
That is, L has at most two elements.
Therefore, option (C) is also correct.
Now, we know that if a line has infinitely many solutions, then the ratio of all of its coefficients must be equal.
Here, if $\left| s \right| = \left| t \right|$, then it will have no solutions.
So, the option (B) is incorrect.

Hence, option (A), (C) and (D) are the correct answers.

Note: Whenever we ask such types of questions, first we have to remember the concept of solutions of a linear equation. We have to form the equations with the help of the given complex number using some basic properties. Then we will find out the solutions of those equations by comparing their coefficients. After that we will check for the options one by one and through this, we will get the required answer.