Question

Let S={z=x+iy:|z-1+i|≥|z|,|z|<2,|z+i|=|z-1|}.
Then the set of all values of x, for which w = 2x + iy ∈ S for some y ∈ R is

• A. (-$\sqrt2$,$\frac{1}{2\sqrt2}$)
• B. (-$\frac{1}{\sqrt2}$,$\frac{1}{4}$)
• C. (-$\sqrt2$,$\frac{1}{2}$)
• D. ($\frac{1}{\sqrt2}$,$\frac{1}{2\sqrt2}$)

Answer 1

Answer: (B)

(-$\frac{1}{\sqrt2}$,$\frac{1}{4}$)

Answer 2

S:{z=x+iy:|z–1+i|≥|z|,|z|<2,|z–i|=|z–1|}|z–1+i|≥|z|
|z| < 2
|z–i|=|z–1|
∵ w∈S and w=2x+iy
2x<$\frac{1}{2}$ ∴x<$\frac{1}{4}$
(2x)2+(−2x)2<4
4x2+4x2<4
x2<$\frac{1}{2}$
⇒x∈(−$\frac{1}{\sqrt2}$,$\frac{1}{\sqrt2}$)
∴x∈(−$\frac{1}{2}$,$\frac{1}{4}$)
So, the correct option is (B).