Q.

Let s, t, r be non-zero complex numbers and L be the set of solutions z=x+iy (x,y,i=-1) of the equation

 sz+tz¯+r=0,  where z¯=x-iy. Then, which of the following statement(s) is (are) TRUE               [2018]

1 If L has exactly one element, then |s||t|  
2 If |s|=|t|, then L has infinitely many elements  
3 The number of elements in L{z:|z-1+i|=5} is at most 2  
4 If L has more than one element, then L has infinitely many elements  

Ans.

(1, 3, 4)

We have, 

       sz+tz¯+r=0                 (i)

On taking conjugate   

       s¯z¯+tz¯+r¯=0                   (ii)

On solving Eqs. (i) and (ii), we get

z=r¯t-rs¯|s|2-|t|2

(a) For unique solutions of z

       |s|2-|t|20|s||t|

       It is true.

(b) If |s|=|t|, then r¯t-rs¯ may or may not be zero.

       So, z may have no solution.

         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 points.

        Hence, it is true.

(d) In the case locus of z is a line, so L has infinite elements.

        Hence, it is true.