Test Index

Complex Numbers Part 2

Section: Mathematics

© examsnet.com

Question : 90 of 106

Marks: +1, -0

Let

S_1=\{z_1 \in \mathbb{C} : |z_1-3|=\frac{1}{2}\}

S_2=\{z_2 \in \mathbb{C} : |z_2 - |z_2+1|| = |z_2 + |z_2-1||\}

z_1 \in S_1

z_2 \in S_2

, the least value of

|z_2 - z_1|

[28-Jul-2022-Shift-1]

0
$\;\frac{1}{2}$
$\;\frac{3}{2}$
$\;\frac{5}{2}$

Solution:

\because |Z_2 + |Z_2-1||^2 = |Z_2 - |Z_2+1||^2

\Rightarrow (Z_2 + |Z_2-1|)(\overline{Z}_2 + |Z_2-1|) = (Z_2 - |Z_2+1|)(\overline{Z}_2 - |Z_2+1|)

\Rightarrow Z_2(|Z_2-1| + |Z_2+1|) + \overline{Z}_2(|Z_2-1| + |Z_2+1|) = |Z_2+1|^2 - |Z_2-1|^2

\Rightarrow (Z_2 + \overline{Z}_2)(|Z_2+1| + |Z_2-1|) = 2(Z_2 + \overline{Z}_2)

\Rightarrow \;\text{Either}\; Z_2 + \overline{Z}_2 = 0 \;\text{or}\; |Z_2+1| + |Z_2-1| = 2

So,

Z_2

lies on imaginary axis or on real axis within

[-1,1]

Also

|Z_1-3| = \frac{1}{2} \Rightarrow Z_1

lies on the circle having center 3 and radius

\;\frac{1}{2}

.

Clearly

|Z_1-Z_2|_{\min} = \frac{3}{2}

© examsnet.com

Go to Question:

Prev Question

Next Question