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MathematicsComplex NumberGeneral(Modulus,Argument,Conjugate)Hard2 minPYQ_2023
MathematicsHardstatement

Fora, letA={z:Re(a+z¯)>Im(a¯+z)}andB={z:Re(a+z¯)<Im(a¯+z)}. Then among the two statements:
(S1): IfRe(a),Im(a)>0, then the setAcontains all the real numbers
(S2): IfRe(a),Im(a)<0, then the setBcontains all the real numbers,

Options:

Answer:
D
Solution:

Let a=x1+iy1, a¯=x1-iy1

And z=x2+iy2, z¯=x2-iy2      

Now,

a+z¯=x1+x2+iy1-y2

and a¯+z=x1+x2+iy2-y1

Now according to the question,

Re(a+z¯)=x1+x2 and

Im(a¯+z)=y2-y1

A=z:x1+x2>y2-y1=z:x1+y1+x2>y2
  B=z:x1+x2<y2-y1=z:x1+y1+x2<y2

If  y2=0 and x1,y1>0 then

A={z:x2>-x1+y1}

A covers a part of negative real axis and therefore, does not contain whole real axis

Similarly if y2=0, and x1,y1<0, then 
B={z:x2<-x1+y1}
 B covers part of positive real axis and therefore does not cover whole real axis.

Hence, both are false.

Stream:JEESubject:MathematicsTopic:Complex NumberSubtopic:General(Modulus,Argument,Conjugate)
2mℹ️ Source: PYQ_2023

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