Q.

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 ABC and let w be any point satisfying |w-2-i|<3. Then, |z|-|w|+3 lies between         [2008]

1 - 6 and 3    
2 - 3 and 6    
3 - 6 and 6    
4 - 3 and 9  

Ans.

(4)

Given:  A={z:Im(z)1}={(x,y):y1}

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