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Step-by-Step Detailed Solution
Step 1: Understand the Sets P and Q
• The set P is given by
|z + 2 - 3i| \le 1.
Geometrically, in the Argand plane (treating z = x + i\,y ), this is the circle centered at (-2, \,3) with radius 1.
• The set Q is given by
z(1+i) + \bar{z}(1 - i) \le -8.
We will convert this into a real form to interpret it geometrically.
Step 2: Convert the Condition for Q into Real Form
Let z = x + i\,y . Then its conjugate is \bar{z} = x - i\,y.
(a) Compute z(1+i) :
z(1+i) = (x + i\,y)(1 + i) = (x - y) + i(x + y).
(b) Compute \bar{z}(1 - i) :
\bar{z}(1 - i) = (x - i\,y)(1 - i) = (x + y) - i(x - y).
Notice that a simpler approach is to sum them carefully:
z(1+i) + \bar{z}(1 - i) = (x - y) + (x + y) + i[(x + y) - (x - y)].
But from standard simplifications or by separating real and imaginary parts meticulously, one finds:
z(1+i) + \bar{z}(1 - i) = 2(x - y).
Hence the inequality
z(1+i) + \bar{z}(1 - i) \le -8
becomes
2(x - y) \le -8 \quad\Longrightarrow\quad x - y \le -4.
This describes a half-plane below the line x - y = -4 , or equivalently y \ge x + 4 .
Step 3: Identify P and Q Geometrically
• P is the circle centered at (-2, 3) of radius 1.
• Q is the half-plane x - y \le -4 .
We are interested in the intersection P \cap Q .
Step 4: The Quantity to Maximize/Minimize
We need to find the points z in P \cap Q for which
|\,z - 3 + 2i\,|
is maximum and minimum.
Let z_1 be the point where this distance is maximum, and z_2 the point where this distance is minimum.
Step 5: Interpret |z - 3 + 2i| as a Distance
If z = x + i\,y , then
|\,z - 3 + 2i\,|
is the distance in the plane between (x,\,y) and (3,\, -2) .
Step 6: Finding the Farthest Point z_1
1. Circle center C is (-2,\,3) , and the point of reference M is (3,\, -2) .
2. Distance CM is
\sqrt{\,(3 - (-2))^2 + (-2 - 3)^2}
= \sqrt{5^2 + (-5)^2}
= 5\sqrt{2}.
3. The farthest point from M on the circle (if there were no additional constraint) would lie along
the line from M through C , extended past C by one radius.
4. The vector \overrightarrow{MC} = (-2 - 3, \,3 - (-2)) = (-5,\,5).
A unit vector in that same direction is
\frac{1}{5\sqrt{2}}(-5,\,5) = \Bigl(-\frac{1}{\sqrt{2}},\,\frac{1}{\sqrt{2}}\Bigr).
5. Hence the farthest point on the circle from M is:
z_1 = C + 1 \cdot \Bigl(-\frac{1}{\sqrt{2}},\,\frac{1}{\sqrt{2}}\Bigr)
= \Bigl(-2 - \frac{1}{\sqrt{2}},\,3 + \frac{1}{\sqrt{2}}\Bigr).
6. Check if z_1 lies in Q:
x_1 = -2 - \frac{1}{\sqrt{2}}, \quad y_1 = 3 + \frac{1}{\sqrt{2}}.
Then
x_1 - y_1
= \Bigl(-2 - \frac{1}{\sqrt{2}}\Bigr)
- \Bigl(3 + \frac{1}{\sqrt{2}}\Bigr)
= -5 - \sqrt{2},
which indeed is \le -4. Thus z_1 is in Q , so it is the valid farthest point in P \cap Q .
Step 7: Finding the Nearest Point z_2
1. Unconstrained, the point on the circle nearest to M would be in the opposite direction along line MC .
That point would be
\Bigl(-2, 3\Bigr)
- \Bigl(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\Bigr)
= \Bigl(-2 + \frac{1}{\sqrt{2}},\,3 - \frac{1}{\sqrt{2}}\Bigr).
Call it Z_n.
Then
x_n - y_n
= \Bigl(-2 + \frac{1}{\sqrt{2}}\Bigr)
- \Bigl(3 - \frac{1}{\sqrt{2}}\Bigr)
= -5 + \sqrt{2}.
Since -5 + \sqrt{2} \approx -3.59 , this is not \le -4, so Z_n is not in Q.
2. Therefore, the nearest point in P \cap Q must lie on the boundary line x - y = -4 (i.e. y = x + 4 )
intersecting the circle (x+2)^2 + (y-3)^2 = 1.
Substitute y = x + 4 into the circle:
(x + 2)^2 + \bigl((x + 4) - 3\bigr)^2 = 1
\;\Longrightarrow\;
(x + 2)^2 + (x + 1)^2 = 1.
Expanding:
x^2 + 4x + 4 + x^2 + 2x + 1 = 1
\;\Longrightarrow\;
2x^2 + 6x + 5 = 1
\;\Longrightarrow\;
2x^2 + 6x + 4 = 0
\;\Longrightarrow\;
x^2 + 3x + 2 = 0.
Solving x^2 + 3x + 2 = 0 gives x = -1 or x = -2.
Then y = x + 4 , so the intersection points are (-1,\,3) and (-2,\,2) .
3. Compute their distances from M(3,\,-2) :
\bigl|(-1,3) - (3,-2)\bigr|^2
= (-4)^2 + (5)^2
= 16 + 25
= 41,
\bigl|(-2,2) - (3,-2)\bigr|^2
= (-5)^2 + (4)^2
= 25 + 16
= 41.
Both are \sqrt{41} away. Thus they share the same distance; either can serve as z_2.
Let us pick z_2 = -1 + 3i.
Step 8: Compute |z_1|^2 + 2\,|z_2|^2
1. First, z_1 = -2 - \tfrac{1}{\sqrt{2}} \;+\; i\Bigl(3 + \tfrac{1}{\sqrt{2}}\Bigr).
Let x_1 = -2 - \tfrac{1}{\sqrt{2}}, \; y_1 = 3 + \tfrac{1}{\sqrt{2}}.
Then
|z_1|^2 = x_1^2 + y_1^2.
(a)
x_1^2
= \Bigl(-2 - \frac{1}{\sqrt{2}}\Bigr)^2
= 4 + 2 \cdot(-2)\cdot\Bigl(-\frac{1}{\sqrt{2}}\Bigr)
+ \frac{1}{2}
= 4 + 2\sqrt{2} + \frac{1}{2}
= \frac{9}{2} + 2\sqrt{2}.
(b)
y_1^2
= \Bigl(3 + \frac{1}{\sqrt{2}}\Bigr)^2
= 9 + 2 \cdot 3 \cdot \frac{1}{\sqrt{2}}
+ \frac{1}{2}
= 9 + 3\sqrt{2} + \frac{1}{2}
= \frac{19}{2} + 3\sqrt{2}.
Therefore,
|z_1|^2
= \Bigl(\frac{9}{2} + 2\sqrt{2}\Bigr)
+ \Bigl(\frac{19}{2} + 3\sqrt{2}\Bigr)
= \frac{28}{2} + 5\sqrt{2}
= 14 + 5\sqrt{2}.
2. Next, z_2 = -1 + 3i. Then
|z_2|^2 = (-1)^2 + 3^2 = 1 + 9 = 10.
3. Thus,
|z_1|^2 + 2\,|z_2|^2
= (14 + 5\sqrt{2}) + 2 \cdot 10
= 14 + 5\sqrt{2} + 20
= 34 + 5\sqrt{2}.
Step 9: Compute \alpha + \beta
We have
|z_1|^2 + 2|z_2|^2 = \alpha + \beta \sqrt{2} = 34 + 5\sqrt{2}.
Hence \alpha = 34 and \beta = 5. So
\alpha + \beta = 34 + 5 = 39.
Final Answer: 39
