Complex Numbers 2 Question 22
23.
Let $s, t, r$ be non-zero complex numbers and $L$ be the set of solutions $z=x+i y \quad(x, y \in R, i=\sqrt{-1})$ of the equation $s z+t \bar{z}+r=0$, where $\bar{z}=x-i y$. Then, which of the following statement(s) is (are) TRUE?
(2018 Adv.)
(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
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Answer:
Correct Answer: 23. (a,c,d)
Solution:
- We have,
$ s z+t \bar{z}+r=0 $
On taking conjugate
$ \bar{s} \bar{z}+\bar{t} z+\bar{r}=0 $
On solving Eqs. (i) and (ii), we get
$ z=\frac{\bar{r} t-r \bar{s}}{|s|^{2}-|t|^{2}} $
(a) For unique solutions of $z$
$|s|^{2}-|t|^{2} \neq 0 \quad \Rightarrow \quad|s| \neq|t|$
It is true
(b) If $|s|=|t|$, then $\bar{r} t-r \bar{s}$ may or may not be zero.
So, $z$ may have no solutions.
$\therefore L$ may be an empty set. It is false.
(c) If elements of set $L$ represents line, then this line and given circle intersect at maximum two point.
Hence, it is true.
(d) In this case locus of $z$ is a line, so $L$ has infinite elements. Hence, it is true.
