Question

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 $sz+t\overline{z}+r=0$, where $\overline{z}=x-iy$. Then, which of the following statement(s) is (are) TRUE?

• A. If $L$ has exactly one element, then $|s|\ne |t|$
• B. If $|s|=|t|$, then $L$ has infinitely many elements
• C. The number of elements in $L\cap \{z:|z-1+i|=5\}$ is atmost $2$
• D. If $L$ has more than one element, then $L$ has infinitely many elements

Answer 1

Answer: (A), (C), (D)

The correct options are
A If $L$ has exactly one element, then $|s|\ne |t|$
C The number of elements in $L\cap \{z:|z-1+i|=5\}$ is atmost $2$
D If $L$ has more than one element, then $L$ has infinitely many elements

$sz+t\overline{z}+r=0$
Taking conjugate,
$\overline{t}z+\overline{s}\overline{z}+\overline{r}=0$
Solving the two equations,
$(|s{|}^2-|t{|}^2)z=\overline{r}t-r\overline{s}$

Option(a)
So if $|s|\ne |t|$ then unique solution.

Option(b)
If $|s|=|t|$ and $\overline{r}t-r\overline{s}\ne 0$ then no solution.

Option(c)
Locus of $z$ is a null set or singleton set or a line.
Maximum possible point of intersection between line and circle can be two.

Option(d)
Locus of $z$ will be a line so infinte solution.