JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 5)
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?
If L has exactly one element, then |s|$$ \ne $$|t|
If |s| = |t|, then L has infinitely many elements
The number of elements in $$L \cap \{ z:|z - 1 + i| = 5\} $$ is at most 2
If L has more than one element, then L has infinitely many elements
Explanation
We have,
$$sz + t\overline z + r = 0$$ ...(i)
On taking conjugate,
$$\overline {sz} + \overline t z + \overline r = 0$$ ... (ii)
On solving Eqs. (i) and (ii), we get
$$z = {{\overline r t - r\overline s } \over {|s{|^2} - |t{|^2}}}$$
(a) For unique solutions of z
$$|s{|^2} - |t{|^2} \ne 0 \Rightarrow |s|\, \ne \,|t|$$
It is true
(b) If |s| = |t|, then $${\overline r t - r\overline s }$$ 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.
$$sz + t\overline z + r = 0$$ ...(i)
On taking conjugate,
$$\overline {sz} + \overline t z + \overline r = 0$$ ... (ii)
On solving Eqs. (i) and (ii), we get
$$z = {{\overline r t - r\overline s } \over {|s{|^2} - |t{|^2}}}$$
(a) For unique solutions of z
$$|s{|^2} - |t{|^2} \ne 0 \Rightarrow |s|\, \ne \,|t|$$
It is true
(b) If |s| = |t|, then $${\overline r t - r\overline s }$$ 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.
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