Let C be the set of all complex numbers. Let
$S_1 = \{z \in C : |z-2| \le 1\}$ and
$S_2 = \{z \in C : z(1+i) + \bar{z}(1-i) \ge 4\}$.
Then, the maximum value of $|z-\frac{5}{2}|^2$ for $z \in S_1 \cap S_2$ is equal to :
Show Hint
When maximizing or minimizing a distance (or its square) from a fixed point to a region, the extremum value will always occur at a point on the boundary of the region. Check all parts of the boundary (lines, arcs, corners).
Let $z = x + iy$.
Set $S_1$:
\[
|z-2| \le 1 \;\Rightarrow\; (x-2)^2 + y^2 \le 1
\]
This represents a closed disc with centre $(2,0)$ and radius $1$.
Set $S_2$:
\[
z(1+i) + \bar z(1-i) \ge 4
\]
Substituting $z=x+iy$ and $\bar z=x-iy$,
\[
(x+iy)(1+i)+(x-iy)(1-i) \ge 4
\]
\[
2x - 2y \ge 4 \;\Rightarrow\; x-y \ge 2
\]
This represents the half-plane below the line $y=x-2$.
Hence, $S_1 \cap S_2$ is the part of the disc lying below the line $y=x-2$.
We need to maximise:
\[
|z-\tfrac52|^2 = (x-\tfrac52)^2 + y^2
\]
The maximum distance from a fixed point to a closed region occurs on the boundary.
The boundary here consists of the circular arc and the line segment.
Substituting $y^2 = 1-(x-2)^2$ from the circle equation,
\[
|z-\tfrac52|^2 = (x-\tfrac52)^2 + 1 - (x-2)^2
= -x + \tfrac{13}{4}
\]
This is maximised when $x$ is minimum on the boundary.
Intersection of the circle and the line:
\[
(x-2)^2 + (x-2)^2 = 1
\Rightarrow (x-2)^2 = \tfrac12
\Rightarrow x = 2 - \tfrac{1}{\sqrt2}
\]
Thus,
\[
\max |z-\tfrac52|^2
= -\left(2-\tfrac{1}{\sqrt2}\right) + \tfrac{13}{4}
= \frac{5+2\sqrt2}{4}
\]
\[
\boxed{\text{Maximum value } = \frac{5+2\sqrt2}{4}}
\]
Was this answer helpful?
0
0
Top Questions on Complex Numbers and Quadratic Equations
