Question

If z1 and z2 are lying on |z-3| < 4 and |z-1| + |z+1| = 3 respectively then range of |z1 - z2| is

Answer 1

Answer:

[0, 17/2)

Step-by-step explanation:

Hi,

The solution for this is more of graphical in nature.

Given z₂ lies on |z-1| + |z+1| = 3, hence the locus of z₂ is an ellipse whose foci

are (-1) and (1) and with major axis as 3.

Midpoint of foci is the center , hence (0) is the center of the ellipse.

Vertices are placed at distance of half the major axis on either sides of the

center, hence at distance of 3/2 from center (0).

Given z₁ lies on |z-3|<4, hence the locus of z₁ is the entire circular region

interior to  the circle with center (3) and radius a 4.

Hence locus of z₁ is ellipse

locus of z₂ is circular region.

Minimum of |z₁ - z₂|

From the diagram it is very clear that whenever z₁ lies on elliptical arc

BDE and z₂ also lies at the same position which is inside the circular

region, |z₁ - z₂| will be 0 , since both are at the same position.

Hence minimum value is 0

Maximum value of |z₁ - z₂|

From the diagram , when z₂ is near to point C(-7) and z₁ is at the opposite

vertex of the ellipse i.e., A(-3/2), we can see that z₁ and z₂ are separated by

the maximum distance which is 7 + 3/2 = 17/2

Hence , maximum value will be nearest to 17/2 since  z₂ can be only near to

C,it couldn't be on C exactly.

Hence the range of |z₁ - z₂| is [0, 17/2).

Hope, it helped !