Let A,B,C be three sets of complex numbers as defined below
A={z:Im(z)≥1}
B={z:|z-2-i|=3}
C={z:Re((1-i)z)=2}
Q. Let z be any point A∩B∩C and let w be any point satisfying |w-2-i|<3. Then, |z|-|w|+3 lies between [2008]
(4)
Given: A={z:Im(z)≥1}={(x,y):y≥1}
Clearly A is the set of all points lying on or above the line y=1 in Cartesian plane.
B={z:|z-2-i|=3}={(x,y):(x-2)2+(y-1)2=9}
⇒ B is the set of all points lying on the boundary of the circle with centre (2, 1) and radius 3.
C={z:Re[(1-i)z]=2}={(x,y):x+y=2}
⇒ C is the set of all points lying on the straight line represented by x+y=2.
Graphically, the three sets are represented as shown below:
Given |w-2-i|<3
⇒Distance between w and (2+i), i.e. S is smaller than 3.
⇒w is a point lying inside the circle with centre S and radius 3.
⇒Distance between z (i.e. the point P) and w should be smaller than 6 (the diameter of the circle)
i.e. |z-w|<6
But we know that ||z|-|w||≤|z-w|
⇒||z|-|w||<6⇒-6<|z|-|w|<6
⇒-3<|z|-|w|+3<9