Let S1= z1 ∈ C:|z1-3|=(1/2) and S 2= z 2 ∈ C :| z 2-| z 2+1||=| z 2+| z 2-1|| . Then, for z1 ∈ S1 and z2 ∈ S2, the least value of |z2-z1| is :

Question

Let S1= z1 ∈ C:|z1-3|=(1/2) and S 2= z 2 ∈ C :| z 2-| z 2+1||=| z 2+| z 2-1|| . Then, for z1 ∈ S1 and z2 ∈ S2, the least value of |z2-z1| is : (A) 0 (B) (1/2) (C) (3/2) (D) (5/2)

Tardigrade Admin · Accepted Answer

| z 2+| z 2-.1|2=| z 2-| z 2+.1|2 ⇒ | z 2+| z 2-1||( overline z 2+| z 2-1|)=( z 2-| z 2+1|)( overline z 2-( z 2+1)) ⇒ z 2| overline z 2+122-1|-( overline z 2-| z 2+1|)+ overline z 2(| z 2-1|+| z 2+1|) =| z 2+1|2=| z 2-1|2 ⇒[ z 2+ overline z 2)(| z 2-1|)+( z 2+1 mid)=2( z 2+ overline z 2) ⇒( z 2+ overline z 2)(| z 2-1|+| z 2+1|-2)=0 ∴ z 2+ overline z 2=0 text or | z 2-1|+| z 2+1|-2=0 ∴ z 2 lie on imaginary axis. Or on real axis with in [-1,1] Also |z1-3|=(1/2) lie on circle having centre 3 and radius (1/2). <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/73aa926a16655f43ceff3d8fc7ab8c0b-.png" /> Clearly |z1-z2| min =(5/2)-1=(3/2)

Q. Let $S_{1} = {z_{1} \in C : ∣ z_{1} - 3 ∣ = \frac{1}{2}}$ and $S_{2} = {z_{2} \in C : ∣ z_{2} - ∣ z_{2} + 1∣∣ = ∣ z_{2} + ∣ z_{2} - 1∣∣}$ . Then, for $z_{1} \in S_{1}$ and $z_{2} \in S_{2}$ , the least value of $∣ z_{2} - z_{1} ∣$ is :

0

$\frac{1}{2}$

$\frac{3}{2}$

$\frac{5}{2}$

Q. Let S1​={z1​∈C:∣z1​−3∣=21​} and S2​={z2​∈C:∣z2​−∣z2​+1∣∣=∣z2​+∣z2​−1∣∣}. Then, for z1​∈S1​ and z2​∈S2​, the least value of ∣z2​−z1​∣ is :

0

21​

23​

25​

Q. Let $S_{1} = {z_{1} \in C : ∣ z_{1} - 3 ∣ = \frac{1}{2}}$ and $S_{2} = {z_{2} \in C : ∣ z_{2} - ∣ z_{2} + 1∣∣ = ∣ z_{2} + ∣ z_{2} - 1∣∣}$ . Then, for $z_{1} \in S_{1}$ and $z_{2} \in S_{2}$ , the least value of $∣ z_{2} - z_{1} ∣$ is :

$\frac{1}{2}$

$\frac{3}{2}$

$\frac{5}{2}$