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Step-by-Step Solution
Step 1: Interpret the geometric constraints
We are given a set
S = \{\,z = x + i\,y \in \mathbb{C} : |z - 2| \le 1,\; z(1+i) + \overline{z}\,(1-i) \le 2\}.
Writing z = x + i\,y :
The condition |z - 2| \le 1 represents all points (x,y) in the closed circle of radius 1 centered at (2,0) in the complex plane, i.e.
(x - 2)^2 + y^2 \le 1.
The expression z(1 + i) + \overline{z}\,(1 - i) simplifies to 2(x - y) . Indeed:
\[
z(1 + i) = (x + i\,y)\,(1 + i) = (x-y) + i(x+y),
\]
\[
\overline{z}\,(1 - i) = (x - i\,y)\,(1 - i) = (x-y) + i(-\,x - y),
\]
so their sum is
\[
(x - y + x - y) + i\bigl[(x + y) + (-\,x - y)\bigr] = 2(x - y).
\]
Hence the second condition z(1 + i) + \overline{z}\,(1 - i) \le 2 is
2(x-y) \le 2 \quad\Longrightarrow\quad x - y \le 1,
or equivalently,
y \;\ge\; x - 1.
Therefore, the set S is the set of points inside (or on) the circle (x-2)^2 + y^2 \le 1 that lie in the region y \ge x - 1.
Step 2: Express the distance we need to minimize or maximize
We want the points in S where |z - 4i| is minimized and maximized. In Cartesian form,
\[
z - 4i = (x + i\,y) - 4i = x + i\,(y - 4),
\]
so
\[
|z - 4i|^2 = x^2 + (y - 4)^2.
\]
We will find the points z_1, z_2 \in S that minimize and maximize |z - 4i| , respectively.
Step 3: Parametrize the circle and locate the feasible arc
The circle (x-2)^2 + y^2 = 1 can be parametrized as
\[
x = 2 + \cos t,\quad y = \sin t,\quad t \in [0, 2\pi].
\]
The line y = x - 1 becomes \sin t = (2 + \cos t) - 1 , or \sin t = 1 + \cos t , which rearranges to
\[
\sin t - \cos t = 1 \quad\Longrightarrow\quad \sqrt{2}\,\sin\Bigl(t - \frac{\pi}{4}\Bigr) = 1.
\]
Solving this within the circle’s intersection shows that t ranges between \frac{\pi}{2} and \pi on that arc. Thus, the feasible portion (where y \ge x - 1 ) is the arc of the circle for t \in \left[\frac{\pi}{2}, \pi\right] .
Step 4: Set up the function to find minima and maxima
Define
\[
f(t) = |z - 4i|^2 = x^2 + (y - 4)^2 = (2 + \cos t)^2 + (\sin t - 4)^2,
\]
for t \in \Bigl[\frac{\pi}{2}, \pi\Bigr] . Let us expand:
\[
f(t) = (2 + \cos t)^2 + (\sin t - 4)^2
= (4 + 4\cos t + \cos^2 t)\;+\;( \sin^2 t - 8\sin t + 16).
\]
Since \cos^2 t + \sin^2 t = 1, we get
\[
f(t) = 4 + 4\cos t + \cos^2 t + \sin^2 t - 8\sin t + 16
= 20 + 4\cos t - 8\sin t + (\cos^2 t + \sin^2 t)
= 21 + 4\cos t - 8\sin t.
\]
Step 5: Find critical points and evaluate at endpoints
Differentiate:
\[
f'(t) = \frac{d}{dt}\bigl(21 + 4\cos t - 8\sin t\bigr)
= -4\sin t - 8\cos t.
\]
Setting f'(t)=0 gives
\[
-4\sin t - 8\cos t = 0
\quad\Longrightarrow\quad \sin t + 2\cos t = 0
\quad\Longrightarrow\quad \sin t = -2\cos t.
\]
In the interval t \in [\tfrac{\pi}{2}, \pi], the solution is
\[
t_c = \pi - \alpha,
\]
where \alpha = \arctan(2). This lies in [\tfrac{\pi}{2}, \pi].
Evaluate f(t) at the boundaries:
\[
t = \frac{\pi}{2}\; \Longrightarrow\; \cos \frac{\pi}{2} = 0,\; \sin \frac{\pi}{2} = 1,
\]
\[
f\Bigl(\frac{\pi}{2}\Bigr) = 21 + 4(0) - 8(1) = 21 - 8 = 13.
\]
\[
t = \pi\; \Longrightarrow\; \cos \pi = -1,\; \sin \pi = 0,
\]
\[
f(\pi) = 21 + 4(-1) - 8(0) = 21 - 4 = 17.
\]
Evaluate at the critical point t_c = \pi - \alpha with \tan \alpha = 2 :
\[
\sin(\pi - \alpha) = \sin \alpha,\quad \cos(\pi - \alpha) = -\cos \alpha.
\]
Since \tan \alpha = 2 , we have
\[
\sin \alpha = \frac{2}{\sqrt{1 + 2^2}} = \frac{2}{\sqrt{5}},
\quad
\cos \alpha = \frac{1}{\sqrt{1 + 2^2}} = \frac{1}{\sqrt{5}}.
\]
Thus
\[
\sin\bigl(\pi - \alpha\bigr) = \frac{2}{\sqrt{5}},
\quad
\cos\bigl(\pi - \alpha\bigr) = -\,\frac{1}{\sqrt{5}}.
\]
Hence
\[
f(t_c) = 21 + 4\!\Bigl(-\frac{1}{\sqrt{5}}\Bigr) \;-\; 8\!\Bigl(\frac{2}{\sqrt{5}}\Bigr)
= 21 - \frac{4}{\sqrt{5}} - \frac{16}{\sqrt{5}}
= 21 - \frac{20}{\sqrt{5}}.
\]
Numerically, 20/\sqrt{5} is about 20 / 2.236 \approx 8.94 , so f(t_c) \approx 21 - 8.94 = 12.06, which is less than 13. Thus the minimum value of |z - 4i|^2 occurs at the critical point t_c, and the maximum in the allowed arc is at t = \pi.
Step 6: Identify z_1 and z_2
\displaystyle z_1 = the point giving the minimum distance to 4i.
At t = t_c = \pi - \alpha,
\[
x_1 = 2 + \cos t_c = 2 - \frac{1}{\sqrt{5}},
\quad
y_1 = \sin t_c = \frac{2}{\sqrt{5}}.
\]
Thus
\[
z_1 = \Bigl(2 - \frac{1}{\sqrt{5}}\Bigr) + i\,\Bigl(\frac{2}{\sqrt{5}}\Bigr).
\]
\displaystyle z_2 = the point giving the maximum distance on the feasible arc to 4i.
At t = \pi,
\[
x_2 = 2 + \cos \pi = 2 - 1 = 1,
\quad
y_2 = \sin \pi = 0.
\]
Hence
\[
z_2 = 1 + 0\,i = 1.
\]
Step 7: Compute |z_1|^2 and |z_2|^2
Recall |z_1|^2 = (x_1)^2 + (y_1)^2 and |z_2|^2 = (1)^2 + (0)^2 = 1.
For z_1 :
\[
x_1 = 2 - \frac{1}{\sqrt{5}},
\quad
y_1 = \frac{2}{\sqrt{5}}.
\]
Thus
\[
|z_1|^2 = \Bigl(2 - \frac{1}{\sqrt{5}}\Bigr)^2 + \Bigl(\frac{2}{\sqrt{5}}\Bigr)^2.
\]
Compute each term:
\[
\Bigl(2 - \frac{1}{\sqrt{5}}\Bigr)^2
= 4 \;-\; \frac{4}{\sqrt{5}} \;+\; \frac{1}{5},
\quad
\Bigl(\frac{2}{\sqrt{5}}\Bigr)^2
= \frac{4}{5}.
\]
Summing gives
\[
|z_1|^2 = 4 - \frac{4}{\sqrt{5}} + \frac{1}{5} + \frac{4}{5}
= 4 - \frac{4}{\sqrt{5}} + \frac{5}{5}
= 5 - \frac{4}{\sqrt{5}}.
\]
Therefore,
\[
|z_2|^2 = 1,
\]
and so
\[
|z_1|^2 + |z_2|^2 = \Bigl(5 - \frac{4}{\sqrt{5}}\Bigr) + 1
= 6 - \frac{4}{\sqrt{5}}.
\]
Step 8: Evaluate 5\bigl(|z_1|^2 + |z_2|^2\bigr) and identify \alpha, \beta
\[
5 \Bigl(|z_1|^2 + |z_2|^2\Bigr)
= 5 \Bigl(6 - \frac{4}{\sqrt{5}}\Bigr)
= 30 - \frac{20}{\sqrt{5}}.
\]
Rationalize:
\[
\frac{20}{\sqrt{5}}
= \frac{20}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}
= \frac{20\sqrt{5}}{5}
= 4 \sqrt{5}.
\]
Thus
\[
30 - \frac{20}{\sqrt{5}} = 30 - 4\sqrt{5}.
\]
Hence in the form \alpha + \beta \sqrt{5} , we have
\[
\alpha = 30,\quad \beta = -\,4.
\]
Therefore
\[
\alpha + \beta = 30 + (-4) = 26.
\]
Final Answer
The value of \alpha + \beta is \boxed{26}.
