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Question: If \[{z_1}\] and \[{z_2}\] are lying on \[\left| {z - 3} \right| \leqslant 4\] and \[\left| {z - 1} ...

If z1{z_1} and z2{z_2} are lying on ∣z−3∣⩽4\left| {z - 3} \right| \leqslant 4 and ∣z−1∣+∣z+1∣=3\left| {z - 1} \right| + \left| {z + 1} \right| = 3 respectively then range of ∣z1−z2∣\left| {{z_1} - {z_2}} \right| is
(A) [0,∞]\left[ {0,\infty } \right]
(B) [0,1]\left[ {0,1} \right]
(C) [0,172]\left[ {0,\dfrac{{17}}{2}} \right]
(D) [0,32]\left[ {0,\dfrac{3}{2}} \right]

Explanation

Solution

The modulus of a quantity is opened up such that the quantity is greater than its negative value and smaller than its positive value
i.e.−z<∣z∣<z - z < \left| z \right| < z, z can take any value.
z1{z_1}is lying on ∣z−3∣⩽4\left| {z - 3} \right| \leqslant 4and z2{z_2}is lying on ∣z−1∣+∣z+1∣=3\left| {z - 1} \right| + \left| {z + 1} \right| = 3 so, we will first find the range of z1{z_1}and z2{z_2}from given equation then take∣z1−z2∣\left| {{z_1} - {z_2}} \right|and find the range of ∣z1−z2∣\left| {{z_1} - {z_2}} \right|.

  • Range of the solution is the set of values having a maximum value and a minimum value in between which all the values satisfy our equation.

Complete step by step answer:
z1{z_1} is lying on ∣z−3∣⩽4\left| {z - 3} \right| \leqslant 4
Now we know that,the modulus of a quantity is opened up such that the quantity is greater than its negative value and smaller than its positive value.
i.e.−z<∣z∣<z - z < \left| z \right| < z, z can take any value.
Therefore using the above formula we open the mod terms into inequality form as,
−4⩽z1−3⩽4- 4 \leqslant {z_1} - 3 \leqslant 4
Adding 3 on each side we get:-
⇒−4+3⩽z1−3+3⩽4+3\Rightarrow - 4 + 3 \leqslant {z_1} - 3 + 3 \leqslant 4 + 3
⇒−1⩽z1⩽7\Rightarrow - 1 \leqslant {z_1} \leqslant 7
Now,z2{z_2}is lying on ∣z−1∣+∣z+1∣=3\left| {z - 1} \right| + \left| {z + 1} \right| = 3
So, we get four possibilities to open up the modulus .
Case 1:- If both z2−1{z_2} - 1 and z2+1{z_2} + 1are positive then
(z2−1)+(z2+1)=3\left( {{z_2} - 1} \right) + \left( {{z_2} + 1} \right) = 3
Solving it further we get:-
⇒2z2=3\Rightarrow 2{z_2} = 3
⇒z2=32\Rightarrow {z_2} = \dfrac{3}{2}
From this possibility we get,
⇒z2=32\Rightarrow {z_2} = \dfrac{3}{2}

Case 2:- If both z2−1{z_2} - 1 and z2+1{z_2} + 1are negative then
⇒−(z2−1)−(z2+1)=3\Rightarrow - \left( {{z_2} - 1} \right) - \left( {{z_2} + 1} \right) = 3
⇒−2z2=3\Rightarrow - 2{z_2} = 3
z2=−32{z_2} = \dfrac{{ - 3}}{2}

Case 3:- If z2−1{z_2} - 1 is positive and z2+1{z_2} + 1 is negative then
(z2−1)−(z2+1)=3\left( {{z_2} - 1} \right) - \left( {{z_2} + 1} \right) = 3
⇒−1−1=3\Rightarrow - 1 - 1 = 3
⇒−2=3\Rightarrow - 2 = 3
Which is not possible So this possibility doesn’t exist

Case 4: - If both z2−1{z_2} - 1 is negative and z2+1{z_2} + 1is positive then
⇒−(z2−1)+(z2+1)=3\Rightarrow - \left( {{z_2} - 1} \right) + \left( {{z_2} + 1} \right) = 3
⇒1+1=3\Rightarrow 1 + 1 = 3
⇒2=3\Rightarrow 2 = 3
Which is not possible so this possibility doesn’t exist
The minimum value for ∣z1−z2∣\left| {{z_1} - {z_2}} \right| is 0.
And for maximum value take the maximum value of z1{z_1} and minimum value z2{z_2}of and subtract both of them
z1−z2=7−(−32){z_1} - {z_2} = 7 - \left( {\dfrac{{ - 3}}{2}} \right)
⇒z1−z2=7+32\Rightarrow {z_1} - {z_2} = 7 + \dfrac{3}{2}
Taking the LCM of right hand side of the equation we get
⇒z1−z2=14+32\Rightarrow {z_1} - {z_2} = \dfrac{{14 + 3}}{2}
⇒z1−z2=172\Rightarrow {z_1} - {z_2} = \dfrac{{17}}{2}

The range for ∣z1−z2∣\left| {{z_1} - {z_2}} \right| is [0,172][0,\dfrac{{17}}{2}]. Therefore, option C is correct.

Note:
When you take any value’s mod then it gives positive value. In this question mod is given so whenever we remove the mod we take both positive and negative values, some students may forget to take a negative value and get an error in answer.