Complex Numbers 2 Question 22

23.

Let s,t,r be non-zero complex numbers and L be the set of solutions z=x+iy(x,yR,i=1) of the equation sz+tz¯+r=0, where z¯=xiy. Then, which of the following statement(s) is (are) TRUE?

(2018 Adv.)

(a) If L has exactly one element, then |s||t|

(b) If |s|=|t|, then L has infinitely many elements

(c) The number of elements in Lz:|z1+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:

  1. We have,

sz+tz¯+r=0

On taking conjugate

s¯z¯+t¯z+r¯=0

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

z=r¯trs¯|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¯trs¯ may or may not be zero.

So, z may have no solutions.

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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