Question

Let $s,t,r$ be non-zero complex numbers and $L$ be the set of solutions $z=x+iy(x,y\in \mathbb{R},i=\sqrt{-1})$ of the equation $sz+t\bar{z}+r=0$, where $\bar{z}=x-iy$. 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 at most 2
• D. If $L$ has more than one element, then $L$ has infinitely many elements

Answer 1

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

Given
$sz+t\bar{z}+r=0$ .....(1)
on taking conjugate $\bar{s}\bar{z}+\bar{t}z+\bar{z}+\bar{r}=0$ ........(2)
from (1) and (2) elliminating $\bar{z}$
$z(|s{|}^2-|t{|}^2)=\bar{r}t-r\bar{s}$
(A) If $|s|\ne |t|$ then z has unique value
(B) If $|s|=|t|$ then $\bar{r}t-r\bar{s}$ may or may not be zero so L may be empty set
(C) locus of z is noll set or singleton set or a line in all cases it will intersect given circle at most two points.
(D) In this case locus of z is a line so L has infinite elements