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

The Correct Option is B

To solve this problem, we need to determine the set of all values of \( x \) for which \( w = 2x + iy \in S \) for some \( y \in \mathbb{R} \), where \( S = \{ z = x + iy : |z - 1 + i| \geq |z|, |z| < 2, |z + i| = |z - 1| \} \). 

1. Let's analyze the given conditions for \( z = x + iy \).
2. The first condition, \( |z - 1 + i| \geq |z| \), represents the region outside or on the perpendicular bisector of the line segment joining the point \( (1, -1) \) and the origin. Simplifying this, it gives us:
   • \( |(x - 1) + (y + 1)i| \geq |x + yi| \) simplifies to \( (x - 1)^2 + (y + 1)^2 \geq x^2 + y^2 \).
   • This expands to \( x^2 - 2x + 1 + y^2 + 2y + 1 \geq x^2 + y^2 \).
   • After canceling terms, we get \( -2x + 2y + 2 \geq 0 \) or \( x \leq y + 1 \).
3. The second condition, \( |z| < 2 \), represents the region inside the circle centered at the origin with radius 2.
4. The third condition, \( |z + i| = |z - 1| \), is the perpendicular bisector of the line segment joining the points \(-i\) and \(1\). Simplifying this:
   • \( |x + (y + 1)i| = |x - 1 + yi| \).
   • Squaring both sides gives \( x^2 + (y+1)^2 = (x-1)^2 + y^2 \).
   • This simplifies to \( x^2 + y^2 + 2y + 1 = x^2 - 2x + 1 + y^2 \), leading to \( 2y = -2x \), or \( y = -x \).
5. Combine these results:
   • The condition \( y = -x \) and \( x \leq y + 1 \) means \( x \leq -x + 1 \), or \( 2x \leq 1 \). Thus, \( x \leq \frac{1}{2} \).
   • Additionally, since \( |z| < 2 \), \( \sqrt{x^2 + y^2} = \sqrt{x^2 + (-x)^2} = \sqrt{2x^2} < 2 \), or \( x^2 < 2 \), giving \( |x| < \sqrt{2} \).
6. Next, find the range of \( w = 2x + iy \). Substituting \( y = -x \) into \( w \), we have:
   • \( w = 2x + i(-x) = 2x - ix \).
   • This implies \( Re(w) = 2x \).
   • Using \( x \leq \frac{1}{2} \) and \( |x| < \sqrt{2} \), we find the possible range for \( 2x: -2\sqrt{2} < 2x < 1 \).
   • Thus, \( -\sqrt{2} < x < \frac{1}{2} \).
7. Finally, comparing these intervals with the given options, we determine the correct range of \( x \).

The correct answer is: \((- \frac{1}{\sqrt{2}}, \frac{1}{4})\).

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).