Question

Let $z$ be any point in $A\cap B\cap C$ and let $w$ be any point satisfying $|w-2-i|<3$. Then, $|z|-|w|+3$ lies between

• A. $-6$ and $3$
• B. $-3$ and $6$
• C. $-6$ and $6$
• D. $-3$ and $9$

Answer 1

Answer: (D)

$-3$ and $9$

Solution:

$A=$ Set of points on and above the line $y=1$ in the Argand plane.

$B=$ Set of points on the circle $(x-2{)}^2+(y-1{)}^2=3^2.$

$C=Re\left(\right(1-i\left)z\right)=Re(1-i)(x+iy)$

$⟹x+y=\sqrt{2}$

$(A\cap B\cap C)$ has only one point of intersection and that is $z.$

$|w-2-i|<3$ denotes set of points in the circle with centre at $(2,1)$ and radius $3$ units.

Now,

$||z|-|w||<|z-w|$

and $|z-w|=$ Distance between $z$ and $w$

Since $z$ is fixed. Hence distance between $w$ and $z$ would be maximum for diametrically opposite points.

or, $|z-w|<6$

or, $-6<|z|-|w|<6$

or, $-6+3<|z|-|w|+3<6+3$

or, $-3<|z|-|w|+3<9$

Therefore, $|z|-|w|+3$ lies between $-3$ and $9.$