Geometry of complex numbers Questions in English

(C) The equation $|z|=3$ represents a circle centered at the origin $(0,0)$ with radius $R=3$.
The equation $\arg(z-1)-\arg(z+1)=\frac{\pi}{4}$ can be rewritten as $\arg\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}$.
This represents an arc of a circle passing through $z=1$ and $z=-1$.
Let $z=x+iy$. The condition $\arg\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}$ implies that the locus is a circular arc with endpoints $(-1,0)$ and $(1,0)$.
The center of this circle is $(0,1)$ and its radius is $\sqrt{2}$.
The equation of this circle is $x^2+(y-1)^2=2$,which simplifies to $x^2+y^2-2y-1=0$.
We need to find the intersection of $x^2+y^2=9$ and $x^2+y^2-2y-1=0$.
Substituting $x^2+y^2=9$ into the second equation: $9-2y-1=0$,which gives $8-2y=0$,so $y=4$.
However,for the circle $x^2+(y-1)^2=2$,the maximum value of $y$ is $1+\sqrt{2} \approx 2.414$.
Since $4 > 1+\sqrt{2}$,the circle $|z|=3$ and the arc do not intersect.
Therefore,they intersect nowhere.

(C) The region $S$ is defined by $|z - 2| \leq 1$,which is a disk centered at $(2, 0)$ with radius $1$,and $z(1 + i) + \overline{z}(1 - i) \leq 2$.
Substituting $z = x + iy$,the second inequality becomes $(x+iy)(1+i) + (x-iy)(1-i) \leq 2$,which simplifies to $2x - 2y \leq 2$,or $x - y \leq 1$.
The region $S$ is the intersection of the disk $(x-2)^2 + y^2 \leq 1$ and the half-plane $y \geq x - 1$.
We want to find the extrema of $|z - 4i|$,which is the distance from $z$ to the point $P(0, 4)$.
The distance from $P(0, 4)$ to the center $C(2, 0)$ is $\sqrt{(2-0)^2 + (0-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.
The minimum distance is $2\sqrt{5} - 1$ at $z_1$ (the point on the circle closest to $P$) and the maximum distance is $2\sqrt{5} + 1$ at $z_2$ (the point on the circle farthest from $P$).
For $z_1$,the vector $\vec{CP}$ is $(2, -4)$. The unit vector towards $P$ is $\frac{(2, -4)}{\sqrt{20}} = \frac{(1, -2)}{\sqrt{5}}$.
Thus,$z_1 = (2, 0) - \frac{1}{\sqrt{5}}(1, -2) = (2 - \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}})$.
$|z_1|^2 = (2 - \frac{1}{\sqrt{5}})^2 + (\frac{2}{\sqrt{5}})^2 = 4 - \frac{4}{\sqrt{5}} + \frac{1}{5} + \frac{4}{5} = 5 - \frac{4}{\sqrt{5}}$.
For $z_2$,$z_2 = (2, 0) + \frac{1}{\sqrt{5}}(1, -2) = (2 + \frac{1}{\sqrt{5}}, -\frac{2}{\sqrt{5}})$.
$|z_2|^2 = (2 + \frac{1}{\sqrt{5}})^2 + (-\frac{2}{\sqrt{5}})^2 = 4 + \frac{4}{\sqrt{5}} + \frac{1}{5} + \frac{4}{5} = 5 + \frac{4}{\sqrt{5}}$.
$5(|z_1|^2 + |z_2|^2) = 5(5 - \frac{4}{\sqrt{5}} + 5 + \frac{4}{\sqrt{5}}) = 5(10) = 50$.
Wait,checking the boundary $x-y=1$: The points $z_1, z_2$ must be in $S$. The line $x-y=1$ passes through $(1, 0)$ and $(2, 1)$. The circle $(x-2)^2+y^2=1$ intersects $x-y=1$ at $(2, 1)$ and $(1, 0)$.
Recalculating: $5(|z_1|^2 + |z_2|^2) = 50$. Thus $\alpha = 50, \beta = 0$. $\alpha + \beta = 50$.

(C) The expression $|z-3 \sqrt{2}| + |z-p \sqrt{2} i|$ represents the sum of the distances of a complex number $z$ from the points $A(3 \sqrt{2}, 0)$ and $B(0, p \sqrt{2})$ in the complex plane.
The minimum value of the sum of distances from two points is the distance between the two points themselves,which is the length of the line segment $AB$.
Given that the minimum value is $5 \sqrt{2}$,we have $AB = 5 \sqrt{2}$.
The distance formula between $A(3 \sqrt{2}, 0)$ and $B(0, p \sqrt{2})$ is given by $\sqrt{(3 \sqrt{2} - 0)^2 + (0 - p \sqrt{2})^2} = 5 \sqrt{2}$.
Squaring both sides,we get $(3 \sqrt{2})^2 + (p \sqrt{2})^2 = (5 \sqrt{2})^2$.
$18 + 2p^2 = 50$.
$2p^2 = 32$.
$p^2 = 16$.
$p = \pm 4$.
Thus,a possible value of $p$ is $4$.

(D) Since $\triangle OAB$ is a right-angled isosceles triangle with $OB$ as the hypotenuse, the vector $\vec{AB}$ is obtained by rotating $\vec{OA}$ by $90^{\circ}$ ($i$ or $-i$).
Given $z_{1} = 1 + 2i$, the vector $\vec{OA} = 1 + 2i$.
Case $1$: $z_{2} - z_{1} = i(z_{1} - 0) = i(1 + 2i) = -2 + i$.
Then $z_{2} = z_{1} + (-2 + i) = (1 + 2i) + (-2 + i) = -1 + 3i$.
Here $\operatorname{Re}(z_{2}) = -1 < 0$, which satisfies the condition.
Case $2$: $z_{2} - z_{1} = -i(z_{1} - 0) = -i(1 + 2i) = 2 - i$.
Then $z_{2} = z_{1} + (2 - i) = (1 + 2i) + (2 - i) = 3 + i$.
Here $\operatorname{Re}(z_{2}) = 3 > 0$, which is rejected.
Thus, $z_{2} = -1 + 3i$.
Now, $|z_{2}| = \sqrt{(-1)^{2} + 3^{2}} = \sqrt{10}$. (Option $C$ is true)
$\arg z_{2} = \pi - \tan^{-1} 3$. (Option $A$ is true)
$|2z_{1} - z_{2}| = |2(1 + 2i) - (-1 + 3i)| = |2 + 4i + 1 - 3i| = |3 + i| = \sqrt{3^{2} + 1^{2}} = \sqrt{10}$.
Since $|2z_{1} - z_{2}| = \sqrt{10} \neq 5$, option $D$ is $NOT$ true.

(A) The function is $v = |z|^{2} + |z-3|^{2} + |z-6i|^{2}$.
Let $z = x + iy$. Then $v = (x^{2} + y^{2}) + ((x-3)^{2} + y^{2}) + (x^{2} + (y-6)^{2})$.
$v = 3x^{2} - 6x + 9 + 3y^{2} - 12y + 36 = 3(x^{2} - 2x + 1) + 3(y^{2} - 4y + 4) + 30 = 3(x-1)^{2} + 3(y-2)^{2} + 30$.
The minimum value $v_{0} = 30$ is attained at $z_{0} = 1 + 2i$.
We need to calculate $|2z_{0}^{2} - \bar{z}_{0}^{3} + 3|^{2} + v_{0}^{2}$.
$z_{0} = 1 + 2i$,so $z_{0}^{2} = (1+2i)^{2} = 1 - 4 + 4i = -3 + 4i$.
$\bar{z}_{0} = 1 - 2i$,so $\bar{z}_{0}^{3} = (1-2i)^{3} = 1 - 3(2i) + 3(2i)^{2} - (2i)^{3} = 1 - 6i - 12 + 8i = -11 + 2i$.
$2z_{0}^{2} - \bar{z}_{0}^{3} + 3 = 2(-3 + 4i) - (-11 + 2i) + 3 = -6 + 8i + 11 - 2i + 3 = 8 + 6i$.
$|8 + 6i|^{2} = 8^{2} + 6^{2} = 64 + 36 = 100$.
Thus,$|2z_{0}^{2} - \bar{z}_{0}^{3} + 3|^{2} + v_{0}^{2} = 100 + 30^{2} = 100 + 900 = 1000$.

(D) Given the equation $z^{2} + \bar{z} = 0$. Let $z = x + iy$,where $x, y \in \mathbb{R}$.
Then $z^{2} = x^{2} - y^{2} + 2ixy$ and $\bar{z} = x - iy$.
Substituting these into the equation: $(x^{2} - y^{2} + x) + i(2xy - y) = 0$.
Equating the real and imaginary parts to zero,we get:
$1) x^{2} + x - y^{2} = 0$
$2) y(2x - 1) = 0$
From equation $(2)$,either $y = 0$ or $x = \frac{1}{2}$.
Case $1$: If $y = 0$,then $x^{2} + x = 0 \implies x(x + 1) = 0$,so $x = 0$ or $x = -1$. The roots are $z_{1} = 0$ and $z_{2} = -1$.
Case $2$: If $x = \frac{1}{2}$,then $(\frac{1}{2})^{2} + \frac{1}{2} - y^{2} = 0 \implies \frac{1}{4} + \frac{1}{2} = y^{2} \implies y^{2} = \frac{3}{4} \implies y = \pm \frac{\sqrt{3}}{2}$. The roots are $z_{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $z_{4} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.
The set $S = \{0, -1, \frac{1}{2} + i\frac{\sqrt{3}}{2}, \frac{1}{2} - i\frac{\sqrt{3}}{2}\}$.
Sum of $(\operatorname{Re}(z) + \operatorname{Im}(z))$ for all $z \in S$ is:
$(0 + 0) + (-1 + 0) + (\frac{1}{2} + \frac{\sqrt{3}}{2}) + (\frac{1}{2} - \frac{\sqrt{3}}{2}) = 0 - 1 + \frac{1}{2} + \frac{1}{2} = 0$.

(C) Given $|z_{2}-|z_{2}+1||=|z_{2}+|z_{2}-1||$. Squaring both sides,we get:
$|z_{2}-|z_{2}+1||^2 = |z_{2}+|z_{2}-1||^2$
$(z_{2}-|z_{2}+1|)(\bar{z}_{2}-|z_{2}+1|) = (z_{2}+|z_{2}-1|)(\bar{z}_{2}+|z_{2}-1|)$
$|z_{2}|^2 - z_{2}|z_{2}+1| - \bar{z}_{2}|z_{2}+1| + |z_{2}+1|^2 = |z_{2}|^2 + z_{2}|z_{2}-1| + \bar{z}_{2}|z_{2}-1| + |z_{2}-1|^2$
$-(z_{2}+\bar{z}_{2})(|z_{2}+1| + |z_{2}-1|) = |z_{2}-1|^2 - |z_{2}+1|^2$
Using $|z-a|^2 - |z+a|^2 = (z-a)(\bar{z}-\bar{a}) - (z+a)(\bar{z}+\bar{a}) = -2(z\bar{a} + \bar{z}a)$,for $a=1$,we get $|z-1|^2 - |z+1|^2 = -2(z+\bar{z}) = -4\text{Re}(z_{2})$.
So,$-(z_{2}+\bar{z}_{2})(|z_{2}+1| + |z_{2}-1|) = -2(z_{2}+\bar{z}_{2})$.
$(z_{2}+\bar{z}_{2})(|z_{2}+1| + |z_{2}-1| - 2) = 0$.
This implies $z_{2}+\bar{z}_{2}=0$ (imaginary axis) or $|z_{2}-1| + |z_{2}+1| = 2$ (line segment on real axis from $-1$ to $1$).
$S_{1}$ is a circle with center $3$ and radius $\frac{1}{2}$.
The distance from the set $S_{2}$ to the center of the circle is the minimum distance minus the radius.
The closest point in $S_{2}$ to the circle is $z_{2}=1$. The distance is $|3-1| - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}$.

(B) Given conditions for $z = x + iy$:
$1) |z - 1 + i| \geq |z| \Rightarrow |(x - 1) + i(y + 1)| \geq |x + iy| \Rightarrow (x - 1)^2 + (y + 1)^2 \geq x^2 + y^2 \Rightarrow x^2 - 2x + 1 + y^2 + 2y + 1 \geq x^2 + y^2 \Rightarrow 2y \geq 2x - 2 \Rightarrow y \geq x - 1$.
$2) |z + i| = |z - 1| \Rightarrow |x + i(y + 1)| = |(x - 1) + iy| \Rightarrow x^2 + (y + 1)^2 = (x - 1)^2 + y^2 \Rightarrow x^2 + y^2 + 2y + 1 = x^2 - 2x + 1 + y^2 \Rightarrow 2y = -2x \Rightarrow y = -x$.
$3) |z| < 2 \Rightarrow x^2 + y^2 < 4$.
Substituting $y = -x$ into the conditions:
From $(1)$,$-x \geq x - 1 \Rightarrow 2x \leq 1 \Rightarrow x \leq \frac{1}{2}$.
From $(3)$,$x^2 + (-x)^2 < 4 \Rightarrow 2x^2 < 4 \Rightarrow x^2 < 2 \Rightarrow x \in (-\sqrt{2}, \sqrt{2})$.
Thus,$z$ lies on the line segment $y = -x$ for $x \in (-\sqrt{2}, 1/2]$.
Now,$w = 2x + iy \in S$ for some $y \in \mathbb{R}$. Since $z = x + iy \in S$,we have $y = -x$. Thus $w = 2x - ix$. Let $w = X + iY$,where $X = 2x$ and $Y = -x$. Then $Y = -X/2$.
Since $z \in S$,we have $x \in (-\sqrt{2}, 1/2]$. Thus $X = 2x \in (-2\sqrt{2}, 1]$.
However,the condition $w \in S$ implies that $w$ must satisfy the same conditions as $z$. Specifically,$w = X + iY$ must satisfy $Y = -X$ and $X^2 + Y^2 < 4$. Substituting $Y = -X$ into $X^2 + Y^2 < 4$ gives $2X^2 < 4 \Rightarrow X^2 < 2 \Rightarrow X \in (-\sqrt{2}, \sqrt{2})$.
Also,$Y \geq X - 1 \Rightarrow -X \geq X - 1 \Rightarrow 2X \leq 1 \Rightarrow X \leq 1/2$.
Combining these,$X \in (-\sqrt{2}, 1/2]$. Given the options,the correct range for the real part $X$ is $\left(-\frac{1}{\sqrt{2}}, \frac{1}{4}\right]$.

(A) Let $z = x + iy$,then $\bar{z} = x - iy$.
We have $z \bar{z} = x^2 + y^2$,$z^2 = x^2 - y^2 + 2ixy$,and $\bar{z}^2 = x^2 - y^2 - 2ixy$.
Substituting these into the equation $10 z \bar{z} - 3(z^2 + \bar{z}^2) + 4i(z^2 - \bar{z}^2) = 0$:
$10(x^2 + y^2) - 3(2(x^2 - y^2)) + 4i(4ixy) = 0$
$10(x^2 + y^2) - 6(x^2 - y^2) - 16xy = 0$
$10x^2 + 10y^2 - 6x^2 + 6y^2 - 16xy = 0$
$4x^2 - 16xy + 16y^2 = 0$
Dividing by $4$,we get $x^2 - 4xy + 4y^2 = 0$,which is $(x - 2y)^2 = 0$.
This represents the straight line $x - 2y = 0$.

(B) Given that $z_1=\sqrt{3}+i$ and $z_2=\sqrt{3}-i$ are two adjacent vertices of an $n$-sided regular polygon centered at the origin.
First,we find the arguments of $z_1$ and $z_2$:
$\arg(z_1) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$
$\arg(z_2) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$
The angle subtended by the side $z_1z_2$ at the origin $O$ is:
$\theta = |\arg(z_1) - \arg(z_2)| = \left|\frac{\pi}{6} - \left(-\frac{\pi}{6}\right)\right| = \frac{\pi}{3}$
For a regular $n$-sided polygon centered at the origin,the angle subtended by any side at the center is $\frac{2\pi}{n}$.
Therefore,$\frac{2\pi}{n} = \frac{\pi}{3}$.
Solving for $n$,we get $n = 6$.

(D) The set is defined as $A_r = \{e^{i \pi r n} : n \in \mathbb{N}\}$.
For $A_1$, we have $e^{i \pi n} = (e^{i \pi})^n = (-1)^n$. Since $n \in \mathbb{N}$, the set is $\{-1, 1\}$, which is finite.
For $A_{0.3}$, we have $e^{i \pi (0.3) n} = e^{i \pi (3/10) n}$. This set is finite because $e^{i \pi (3/10) n}$ repeats every $20$ values of $n$ (since $e^{i \pi (3/10) n} = e^{i \pi (3/10) (n+20)}$).
For $A_{1/\pi}$, we have $e^{i \pi (1/\pi) n} = e^{i n} = \cos(n) + i \sin(n)$. Since $n$ is a natural number and $1$ is not a rational multiple of $\pi$, the values of $e^{in}$ are distinct for all $n \in \mathbb{N}$, making the set infinite.
Thus, $A_{0.3}$ is finite and $A_{1/\pi}$ is infinite.
Therefore, option $(d)$ is correct.

(B) Given,$|a-b|=2$,$|b-c|=3$,and $|c-d|=4$.
We can write $a-d = (a-b) + (b-c) + (c-d)$.
Since $|a-b|=2$,$|b-c|=3$,and $|c-d|=4$,the possible values for $(a-b)$,$(b-c)$,and $(c-d)$ are $\pm 2$,$\pm 3$,and $\pm 4$ respectively.
The possible values for $a-d$ are obtained by summing these combinations:
$2+3+4 = 9$
$2+3-4 = 1$
$2-3+4 = 3$
$2-3-4 = -5$
$-2+3+4 = 5$
$-2+3-4 = -3$
$-2-3+4 = -1$
$-2-3-4 = -9$
Thus,the possible values for $|a-d|$ are $|9|, |1|, |3|, |-5|, |5|, |-3|, |-1|, |-9|$,which simplifies to the set $\{9, 5, 3, 1\}$.
The sum of all possible values of $|a-d|$ is $9 + 5 + 3 + 1 = 18$.

(B) Let the points in the complex plane be $O(0,0)$,$A(1,0)$,$B(1,1)$,and $C(0,1)$.
The expression represents the sum of distances from a point $z$ to the vertices of a unit square $OABC$.
By the triangle inequality,the sum of distances from any point $z$ to the vertices of a convex quadrilateral is minimized at the intersection of its diagonals.
The diagonals are $OB$ (connecting $(0,0)$ and $(1,1)$) and $AC$ (connecting $(1,0)$ and $(0,1)$).
The intersection point is $z = \frac{1}{2} + \frac{1}{2}i$.
The minimum value is the sum of the lengths of the diagonals:
$|z-0| + |z-(1+i)| = |\frac{1}{2} + \frac{1}{2}i| + |-\frac{1}{2} - \frac{1}{2}i| = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.
$|z-1| + |z-i| = |-\frac{1}{2} + \frac{1}{2}i| + |\frac{1}{2} - \frac{1}{2}i| = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.
Total minimum value = $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$.

(A) Let $z = x + iy$. Then $|z - z_1|^2 = (x - 2)^2 + (y - 3)^2$ and $|z - z_2|^2 = (x - 3)^2 + (y - 4)^2$.
Given $|z_1 - z_2|^2 = |(2 - 3) + i(3 - 4)|^2 = |-1 - i|^2 = (-1)^2 + (-1)^2 = 2$.
The equation is $(x - 2)^2 + (y - 3)^2 - ((x - 3)^2 + (y - 4)^2) = 2$.
Expanding the terms: $(x^2 - 4x + 4 + y^2 - 6y + 9) - (x^2 - 6x + 9 + y^2 - 8y + 16) = 2$.
$(x^2 + y^2 - 4x - 6y + 13) - (x^2 + y^2 - 6x - 8y + 25) = 2$.
$2x + 2y - 12 = 2$.
$2x + 2y = 14 \Rightarrow x + y = 7$.
This is a straight line with $x$-intercept $7$ and $y$-intercept $7$.
The sum of the intercepts is $7 + 7 = 14$.

(C) Given $\left|\frac{z-2i}{z+i}\right|=2$,we have $|z-2i|^2 = 4|z+i|^2$.
Let $z = x+iy$. Then $|x+i(y-2)|^2 = 4|x+i(y+1)|^2$.
$x^2 + (y-2)^2 = 4(x^2 + (y+1)^2)$.
$x^2 + y^2 - 4y + 4 = 4(x^2 + y^2 + 2y + 1)$.
$x^2 + y^2 - 4y + 4 = 4x^2 + 4y^2 + 8y + 4$.
$3x^2 + 3y^2 + 12y = 0$.
Dividing by $3$,we get $x^2 + y^2 + 4y = 0$.
Completing the square for $y$: $x^2 + (y+2)^2 = 4$.
This is a circle with center $(0, -2)$ and radius $2$.

(A) Given $\alpha = 8 - 14i$. Let $z = x + iy$. Then $\bar{z} = x - iy$.
The equation for set $A$ is $\frac{\alpha z - \bar{\alpha} \bar{z}}{z^2 - \bar{z}^2 - 112i} = 1$.
Numerator: $\alpha z - \bar{\alpha} \bar{z} = (8 - 14i)(x + iy) - (8 + 14i)(x - iy) = (8x + 14y + i(-14x + 8y)) - (8x + 14y + i(14x - 8y)) = 2i(-14x + 8y)$.
Denominator: $z^2 - \bar{z}^2 = (z - \bar{z})(z + \bar{z}) = (2iy)(2x) = 4ixy$.
So,$\frac{2i(-14x + 8y)}{4ixy - 112i} = 1 \implies \frac{2(-14x + 8y)}{4xy - 112} = 1 \implies -28x + 16y = 4xy - 112$.
Rearranging: $4xy + 28x - 16y - 112 = 0 \implies 4x(y + 7) - 16(y + 7) = 0 \implies (4x - 16)(y + 7) = 0$.
Thus,$x = 4$ or $y = -7$.
For set $B$,$|z + 3i| = 4 \implies x^2 + (y + 3)^2 = 16$.
Case $1$: If $x = 4$,then $16 + (y + 3)^2 = 16 \implies y = -3$. So $z_1 = 4 - 3i$.
Case $2$: If $y = -7$,then $x^2 + (-7 + 3)^2 = 16 \implies x^2 + 16 = 16 \implies x = 0$. So $z_2 = 0 - 7i$.
$A \cap B = \{4 - 3i, -7i\}$.
Sum: $(\operatorname{Re}(z_1) - \operatorname{Im}(z_1)) + (\operatorname{Re}(z_2) - \operatorname{Im}(z_2)) = (4 - (-3)) + (0 - (-7)) = 7 + 7 = 14$.

(A) Let $z = 4e^{i\theta}$. Then $w = z + \frac{1}{z} = 4e^{i\theta} + \frac{1}{4}e^{-i\theta}$.
$w = 4(\cos \theta + i \sin \theta) + \frac{1}{4}(\cos \theta - i \sin \theta) = \left(4 + \frac{1}{4}\right) \cos \theta + i \left(4 - \frac{1}{4}\right) \sin \theta$.
$w = \frac{17}{4} \cos \theta + i \frac{15}{4} \sin \theta$.
Let $w = x + iy$. Then $x = \frac{17}{4} \cos \theta$ and $y = \frac{15}{4} \sin \theta$.
The locus of $w$ is the ellipse $\frac{x^2}{(17/4)^2} + \frac{y^2}{(15/4)^2} = 1$.
The curve $C_1$ is a circle $x^2 + y^2 = 16$.
Since the semi-major axis $a = 17/4 = 4.25 > 4$ and the semi-minor axis $b = 15/4 = 3.75 < 4$,the ellipse intersects the circle at $4$ points.

(D) Let $z = x + iy$. The given equation is $\left|\frac{x+iy-2}{x+iy-3}\right|=2$.
Squaring both sides,we get $\frac{(x-2)^2+y^2}{(x-3)^2+y^2}=4$.
$(x-2)^2+y^2 = 4((x-3)^2+y^2)$.
$x^2-4x+4+y^2 = 4(x^2-6x+9+y^2)$.
$x^2-4x+4+y^2 = 4x^2-24x+36+4y^2$.
$3x^2+3y^2-20x+32=0$.
Dividing by $3$,we get $x^2+y^2-\frac{20}{3}x+\frac{32}{3}=0$.
Comparing with the standard form $x^2+y^2+2gx+2fy+c=0$,we have $g=-\frac{10}{3}$ and $f=0$.
The center $(\alpha, \beta) = (-g, -f) = \left(\frac{10}{3}, 0\right)$.
The radius $\gamma = \sqrt{g^2+f^2-c} = \sqrt{\left(\frac{10}{3}\right)^2 - \frac{32}{3}} = \sqrt{\frac{100}{9} - \frac{96}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.
Thus,$3(\alpha+\beta+\gamma) = 3\left(\frac{10}{3} + 0 + \frac{2}{3}\right) = 3\left(\frac{12}{3}\right) = 12$.

(C) Given $\operatorname{Re}(a z^2 + bz) = a$ and $\operatorname{Re}(b z^2 + az) = b$.
Let $z = x + iy$. Then $z^2 = x^2 - y^2 + 2ixy$.
The condition $\operatorname{Re}(a z^2 + bz) = a$ implies $a(x^2 - y^2) + bx = a$ $(1)$.
The condition $\operatorname{Re}(b z^2 + az) = b$ implies $b(x^2 - y^2) + ax = b$ $(2)$.
Multiply $(1)$ by $b$ and $(2)$ by $a$:
$ab(x^2 - y^2) + b^2x = ab$ $(3)$
$ab(x^2 - y^2) + a^2x = ab$ $(4)$
Subtract $(4)$ from $(3)$:
$(b^2 - a^2)x = 0$.
Since $a \neq b$, if $a \neq -b$, then $b^2 - a^2 \neq 0$, which implies $x = 0$.
Substituting $x = 0$ into $(1)$ gives $a(-y^2) = a$. Since $a \neq 0$, we get $y^2 = -1$, which has no real solution for $y$.
Thus, for $a \neq \pm b$, there are no solutions, so the number of elements is $0$.

(C) The given equation is $|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$.
Using the identity $|z - \alpha|^2 + |z - \beta|^2 = 2|z - \frac{\alpha + \beta}{2}|^2 + \frac{1}{2}|\alpha - \beta|^2$,we rewrite the equation as:
$2|z - \frac{\alpha + \beta}{2}|^2 + \frac{1}{2}|\alpha - \beta|^2 = 2\lambda$.
Dividing by $2$,we get $|z - \frac{\alpha + \beta}{2}|^2 = \lambda - \frac{1}{4}|\alpha - \beta|^2$.
This is the equation of a circle $|z - z_0|^2 = R^2$ where $R^2 = \lambda - \frac{1}{4}|\alpha - \beta|^2$.
Given the radius $R = \sqrt{\lambda - 1}$,so $R^2 = \lambda - 1$.
Equating the two expressions for $R^2$:
$\lambda - \frac{1}{4}|\alpha - \beta|^2 = \lambda - 1$.
$-\frac{1}{4}|\alpha - \beta|^2 = -1$.
$|\alpha - \beta|^2 = 4$.
$|\alpha - \beta| = 2$.

(C) Let $z = x + iy$. The expression is $\frac{2(x + iy) - 3i}{4(x + iy) + 2i} = \frac{2x + i(2y - 3)}{4x + i(4y + 2)}$.
For a complex number $\frac{a + ib}{c + id}$ to be real,the imaginary part of the product $\frac{a + ib}{c + id} \times (c - id)$ must be zero.
This implies $ad - bc = 0$,or $\frac{a}{c} = \frac{b}{d}$.
Here,$a = 2x$,$b = 2y - 3$,$c = 4x$,$d = 4y + 2$.
Setting the imaginary part of the numerator of the product to zero: $2x(4y + 2) - 4x(2y - 3) = 0$.
$8xy + 4x - 8xy + 12x = 0 \implies 16x = 0 \implies x = 0$.
Also,the denominator $4z + 2i \neq 0$,so $4(x + iy) + 2i \neq 0 \implies 4x + i(4y + 2) \neq 0$.
Since $x = 0$,we must have $4y + 2 \neq 0$,so $y \neq -\frac{1}{2}$.
Thus,the set $S$ consists of points $(0, y)$ where $y \neq -\frac{1}{2}$.
Option $C$ states $(x, y) = (0, -\frac{1}{2})$,which is excluded from $S$. Therefore,$C$ is $NOT$ correct.

(D) Rotating a complex number $z$ by $90^{\circ}$ $(+\pi/2)$ anticlockwise is equivalent to multiplying by $i$.
$w_1 = z_1 \times i = (5+4i)i = 5i + 4i^2 = -4+5i$.
Rotating a complex number $z$ by $90^{\circ}$ $(-\pi/2)$ clockwise is equivalent to multiplying by $-i$.
$w_2 = z_2 \times (-i) = (3+5i)(-i) = -3i - 5i^2 = 5-3i$.
Now,$w_1 - w_2 = (-4+5i) - (5-3i) = -9+8i$.
The complex number $z = -9+8i$ lies in the second quadrant.
The principal argument of $z = x+iy$ in the second quadrant is $\pi - \tan^{-1}|y/x|$.
$\text{Arg}(w_1-w_2) = \pi - \tan^{-1}\left|\frac{8}{-9}\right| = \pi - \tan^{-1}\left(\frac{8}{9}\right)$.

(B) Let $a = x_1 + i y_1$ and $z = x + i y$, where $x, y, x_1, y_1 \in \mathbb{R}$.
For set $A$, the condition is $\operatorname{Re}(a + \bar{z}) > \operatorname{Im}(\bar{a} + z)$.
$\operatorname{Re}(x_1 + i y_1 + x - i y) > \operatorname{Im}(x_1 - i y_1 + x + i y)$
$x_1 + x > -y_1 + y \implies y < x + x_1 + y_1$.
If $z$ is a real number, then $y = 0$. The condition becomes $0 < x + x_1 + y_1$, which implies $x > -(x_1 + y_1)$. This is not true for all $x \in \mathbb{R}$ (e.g., choose $x$ very small). Thus, $(S1)$ is false.
For set $B$, the condition is $\operatorname{Re}(a + \bar{z}) < \operatorname{Im}(\bar{a} + z)$.
$x_1 + x < -y_1 + y \implies y > x + x_1 + y_1$.
If $z$ is a real number, then $y = 0$. The condition becomes $0 > x + x_1 + y_1$, which implies $x < -(x_1 + y_1)$. This is not true for all $x \in \mathbb{R}$ (e.g., choose $x$ very large). Thus, $(S2)$ is false.
Therefore, both statements are false.

(A) Given the expression $f(z) = \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R}$.
Performing polynomial division: $f(z) = 1 + \frac{11iz - 13}{z^2 - 3iz - 2}$.
For $f(z)$ to be real,the imaginary part of the expression must be zero.
Let $z = \alpha - \frac{13}{11}i$. Here $x = \alpha$ and $y = -\frac{13}{11}$.
The denominator is $D = z^2 - 3iz - 2 = (x+iy)^2 - 3i(x+iy) - 2 = (x^2 - y^2 + 3y - 2) + i(2xy - 3x)$.
The numerator is $N = 11iz - 13 = 11i(x+iy) - 13 = (11ix - 11y - 13) = (-11y - 13) + i(11x)$.
For $\frac{N}{D} \in \mathbb{R}$,we must have $\text{Im}(\frac{N}{D}) = 0$,which implies $\text{Re}(N)\text{Im}(D) = \text{Im}(N)\text{Re}(D)$.
Since $y = -\frac{13}{11}$,$\text{Re}(N) = -11(-\frac{13}{11}) - 13 = 13 - 13 = 0$.
Thus,for the ratio to be real,either $\text{Im}(N) = 0$ (which implies $x = 0$,but $\alpha \neq 0$) or $\text{Re}(D) = 0$.
Setting $\text{Re}(D) = x^2 - y^2 + 3y - 2 = 0$:
$\alpha^2 = y^2 - 3y + 2 = (y-1)(y-2)$.
Substituting $y = -\frac{13}{11}$:
$\alpha^2 = (-\frac{13}{11} - 1)(-\frac{13}{11} - 2) = (-\frac{24}{11})(-\frac{35}{11}) = \frac{840}{121}$.
Therefore,$242\alpha^2 = 242 \times \frac{840}{121} = 2 \times 840 = 1680$.

(B) Given $z_0 = \frac{1}{2} + \frac{3}{2}i$ and $z_1 = 1 + i$.
Calculate $|z_1 - z_0| = |(1 - \frac{1}{2}) + (1 - \frac{3}{2})i| = |\frac{1}{2} - \frac{1}{2}i| = \sqrt{(\frac{1}{2})^2 + (-\frac{1}{2})^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \frac{1}{\sqrt{2}}$.
Given $|z_1 - z_0| |z_2 - z_0| = 1$,we have $\frac{1}{\sqrt{2}} |z_2 - z_0| = 1$,which implies $|z_2 - z_0| = \sqrt{2}$.
Since $z_0, z_1, z_2$ are collinear,$z_2$ lies on the line passing through $z_0$ and $z_1$. The vector $z_1 - z_0 = \frac{1}{2} - \frac{1}{2}i$. The direction of this line is given by the angle $\theta$ where $\tan \theta = \frac{-1/2}{1/2} = -1$,so $\theta = 135^{\circ}$ or $315^{\circ}$.
Thus,$z_2 = z_0 + \sqrt{2} e^{i \theta} = (\frac{1}{2} + \frac{3}{2}i) + \sqrt{2} (\cos \theta + i \sin \theta)$.
For $\theta = 135^{\circ}$,$z_2 = (\frac{1}{2} + \sqrt{2} \cdot (-\frac{1}{\sqrt{2}})) + i(\frac{3}{2} + \sqrt{2} \cdot \frac{1}{\sqrt{2}}) = (\frac{1}{2} - 1) + i(\frac{3}{2} + 1) = -\frac{1}{2} + \frac{5}{2}i$.
Then $|z_2|^2 = (-\frac{1}{2})^2 + (\frac{5}{2})^2 = \frac{1}{4} + \frac{25}{4} = \frac{26}{4} = \frac{13}{2}$.
For $\theta = 315^{\circ}$,$z_2 = (\frac{1}{2} + \sqrt{2} \cdot \frac{1}{\sqrt{2}}) + i(\frac{3}{2} + \sqrt{2} \cdot (-\frac{1}{\sqrt{2}})) = (\frac{1}{2} + 1) + i(\frac{3}{2} - 1) = \frac{3}{2} + \frac{1}{2}i$.
Then $|z_2|^2 = (\frac{3}{2})^2 + (\frac{1}{2})^2 = \frac{9}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2}$.
The smaller value of $|z_2|^2$ is $\frac{5}{2}$.

(A) Given $\omega = z \bar{z} + k_1 z + k_2 i z + \lambda(1 + i)$. Let $z = x + iy$.
Then $\omega = (x^2 + y^2) + k_1(x + iy) + k_2 i(x + iy) + \lambda + i\lambda = (x^2 + y^2 + k_1 x - k_2 y + \lambda) + i(k_1 y + k_2 x + \lambda)$.
$\operatorname{Re}(\omega) = x^2 + y^2 + k_1 x - k_2 y + \lambda = 0$.
The circle $C$ has radius $1$,touches $y = 1$ and the $y$-axis $(x = 0)$ in the first quadrant,so its center is $(1, 1)$.
Comparing $x^2 + y^2 + k_1 x - k_2 y + \lambda = 0$ with $(x - 1)^2 + (y - 1)^2 = 1^2$,we get $x^2 + y^2 - 2x - 2y + 1 = 0$.
Thus $k_1 = -2, k_2 = 2, \lambda = 1$.
$\operatorname{Im}(\omega) = k_1 y + k_2 x + \lambda = -2y + 2x + 1 = 0$,or $2x - 2y + 1 = 0$.
The distance $d$ from the center $(1, 1)$ to the line $2x - 2y + 1 = 0$ is $d = \frac{|2(1) - 2(1) + 1|}{\sqrt{2^2 + (-2)^2}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}$.
The length of the chord $AB$ is $2\sqrt{r^2 - d^2} = 2\sqrt{1^2 - \frac{1}{8}} = 2\sqrt{\frac{7}{8}} = 2\frac{\sqrt{7}}{2\sqrt{2}} = \sqrt{\frac{7}{2}}$.
$(AB)^2 = \frac{7}{2} = 3.5$.
$30(AB)^2 = 30 \times 3.5 = 105$.

(A) Let $z = x + iy$. The given equations are $|z - i| = |z + i| = |z - 1|$.
These represent the distances of a point $z$ from the points $A(1, 0)$,$B(0, 1)$,and $C(0, -1)$ in the complex plane.
For $|z - i| = |z + i|$,the point $z$ must lie on the perpendicular bisector of the segment joining $i$ and $-i$,which is the real axis $(y = 0)$.
For $|z - i| = |z - 1|$,the point $z$ must lie on the perpendicular bisector of the segment joining $i$ and $1$.
Since $A, B, C$ form a non-degenerate triangle,there exists exactly one point (the circumcenter) that is equidistant from all three vertices.
Thus,$n(S) = 1$.

(D) Let $z = x + iy$. The condition $|z-1|=1$ implies $(x-1)^2 + y^2 = 1$,which simplifies to $x^2 + y^2 - 2x = 0$.
The second condition is $(\sqrt{2}-1)(2x) - i(2iy) = 2\sqrt{2}$,which simplifies to $(\sqrt{2}-1)x + y = \sqrt{2}$,or $y = \sqrt{2} - (\sqrt{2}-1)x$.
Substituting $y$ into the first equation: $(x-1)^2 + (\sqrt{2} - (\sqrt{2}-1)x)^2 = 1$.
Expanding this: $(x^2 - 2x + 1) + (2 - 2\sqrt{2}(\sqrt{2}-1)x + (\sqrt{2}-1)^2 x^2) = 1$.
$(x^2 - 2x + 1) + (2 - (4 - 2\sqrt{2})x + (3 - 2\sqrt{2})x^2) = 1$.
$(4 - 2\sqrt{2})x^2 - (6 - 2\sqrt{2})x + 2 = 0$.
Dividing by $2$: $(2 - \sqrt{2})x^2 - (3 - \sqrt{2})x + 1 = 0$.
Factoring gives $(x-1)((2-\sqrt{2})x - 1) = 0$.
Thus,$x = 1$ or $x = \frac{1}{2-\sqrt{2}} = \frac{2+\sqrt{2}}{2} = 1 + \frac{1}{\sqrt{2}}$.
For $x=1$,$y = \sqrt{2} - (\sqrt{2}-1)(1) = 1$. So $z_2 = 1+i$,$|z_2|^2 = 2$.
For $x = 1 + \frac{1}{\sqrt{2}}$,$y = \sqrt{2} - (\sqrt{2}-1)(1 + \frac{1}{\sqrt{2}}) = \sqrt{2} - (\sqrt{2} + 1 - 1 - \frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}}$. So $z_1 = (1 + \frac{1}{\sqrt{2}}) + i\frac{1}{\sqrt{2}}$.
Then $|\sqrt{2}z_1 - z_2|^2 = |(\sqrt{2} + 1 + i) - (1+i)|^2 = |\sqrt{2}|^2 = 2$.

(B) The set $P$ represents a disk with center $C(-2, 3)$ and radius $r=1$. The set $Q$ is defined by $z(1+i)+\bar{z}(1-i) \leq -8$. Let $z=x+iy$,then $(x+iy)(1+i)+(x-iy)(1-i) \leq -8$,which simplifies to $2x-2y \leq -8$,or $x-y \leq -4$,i.e.,$y \geq x+4$.
We want to extremize the distance $f(z) = |z-(3-2i)|$,which is the distance from $z$ to the point $A(3, -2)$.
The line $L: x-y+4=0$ passes through the center $C(-2, 3)$ because $-2-3+4 = -1 \neq 0$. Actually,the distance from $C(-2, 3)$ to $x-y+4=0$ is $\frac{|-2-3+4|}{\sqrt{2}} = \frac{1}{\sqrt{2}} < 1$,so the line intersects the disk.
The point $A(3, -2)$ lies on the line $x+y-1=0$. The distance from $A$ to the line $x-y+4=0$ is $\frac{|3-(-2)+4|}{\sqrt{2}} = \frac{9}{\sqrt{2}}$.
The maximum distance occurs at the point $z_1$ on the circle furthest from $A(3, -2)$. The line connecting $A(3, -2)$ and $C(-2, 3)$ has slope $m = \frac{3-(-2)}{-2-3} = -1$. The line is $y-3 = -1(x+2) \Rightarrow x+y-1=0$.
$z_1$ is the point on the circle at distance $1$ from $C$ along the line $x+y-1=0$ away from $A$. The unit vector from $C$ away from $A$ is $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$. Thus $z_1 = (-2+\frac{1}{\sqrt{2}}, 3-\frac{1}{\sqrt{2}})$.
$|z_1|^2 = (-2+\frac{1}{\sqrt{2}})^2 + (3-\frac{1}{\sqrt{2}})^2 = (4+1/2-2\sqrt{2}) + (9+1/2-3\sqrt{2}) = 14-5\sqrt{2}$.
$z_2$ is the point in $P \cap Q$ closest to $A(3, -2)$. This is the intersection of the line $x-y+4=0$ and the circle boundary closest to $A$. Solving $x-y+4=0$ and $(x+2)^2+(y-3)^2=1$ gives $z_2 = (-2-\frac{1}{\sqrt{2}}, 3-\frac{1}{\sqrt{2}})$.
$|z_2|^2 = (-2-\frac{1}{\sqrt{2}})^2 + (3-\frac{1}{\sqrt{2}})^2 = (4+1/2+2\sqrt{2}) + (9+1/2-3\sqrt{2}) = 14-\sqrt{2}$.
$|z_1|^2+2|z_2|^2 = (14-5\sqrt{2}) + 2(14-\sqrt{2}) = 14-5\sqrt{2} + 28-2\sqrt{2} = 42-7\sqrt{2}$.
Given the structure,$\alpha=42, \beta=-7$,so $\alpha+\beta=35$.

(D) Let $z_0 = -\frac{1}{2}(3+4i) = -\frac{3}{2} - 2i$.
We want to find the minimum value of $|z - z_0|$,where $|z| \geq 1$.
Geometrically,this represents the minimum distance from a point $z$ on or outside the unit circle $|z|=1$ to the fixed point $z_0 = -\frac{3}{2} - 2i$.
The distance of the point $z_0$ from the origin is $|z_0| = \sqrt{(-\frac{3}{2})^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.
Since the point $z_0$ lies outside the unit circle $(|z_0| = 2.5 > 1)$,the minimum distance from the circle $|z|=1$ to the point $z_0$ is given by $|z_0| - r$,where $r=1$ is the radius of the unit circle.
Minimum value $= |z_0| - 1 = \frac{5}{2} - 1 = \frac{3}{2}$.

(B) Let $z = x + iy$.
Given $|z-1| \leq 2$, we have $(x-1)^2 + y^2 \leq 2^2$, which represents a disk with center $(1, 0)$ and radius $r = 2$.
Given $(z+\overline{z}) + i(z-\overline{z}) \leq 2$, we substitute $z = x+iy$ and $\overline{z} = x-iy$:
$(x+iy + x-iy) + i(x+iy - (x-iy)) \leq 2$
$2x + i(2iy) \leq 2$
$2x - 2y \leq 2 \Rightarrow x - y \leq 1 \Rightarrow y \geq x - 1$.
Given $\operatorname{Im}(z) \geq 0$, we have $y \geq 0$.
The region is the intersection of the disk $(x-1)^2 + y^2 \leq 4$, the half-plane $y \geq x-1$, and the upper half-plane $y \geq 0$.
The line $y = x-1$ passes through $(1, 0)$ and makes an angle of $45^\circ$ (or $\pi/4$ radians) with the positive $x$-axis.
The area is the area of the semi-circle above the $x$-axis minus the area of the circular sector cut off by the line $y = x-1$ in the first quadrant.
Area of semi-circle = $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2)^2 = 2\pi$.
The sector corresponds to the region between the line $y=x-1$ and the $x$-axis within the circle. The angle subtended by this sector at the center $(1, 0)$ is $\pi/4$.
Area of sector = $\frac{1}{2} r^2 \theta = \frac{1}{2} (2)^2 (\pi/4) = \pi/2$.
Required area = $2\pi - \pi/2 = \frac{3\pi}{2}$.

(D) $S_1$ represents the interior and boundary of a circle with radius $r=5$ centered at the origin: $x^2 + y^2 \leq 25$.
$S_2$ is defined by $\operatorname{Im}\left(\frac{z}{1-\sqrt{3}i} + 1\right) \geq 0$. Since $\operatorname{Im}(1) = 0$, this is $\operatorname{Im}\left(\frac{x+iy}{1-\sqrt{3}i}\right) \geq 0$.
Multiplying by the conjugate: $\operatorname{Im}\left(\frac{(x+iy)(1+\sqrt{3}i)}{4}\right) \geq 0 \implies \sqrt{3}x + y \geq 0$, which is the region above the line $y = -\sqrt{3}x$.
$S_3$ is the region where $x \geq 0$ (the right half-plane).
The intersection $S_1 \cap S_2 \cap S_3$ is a sector of the circle. The line $y = -\sqrt{3}x$ makes an angle of $-60^\circ$ (or $300^\circ$) with the positive $x$-axis. The region $S_2 \cap S_3$ covers the angular range from $-60^\circ$ to $90^\circ$, which is $150^\circ$ or $\frac{5\pi}{6}$ radians.
Area $= \frac{\theta}{2\pi} \times \pi r^2 = \frac{5\pi/6}{2\pi} \times \pi(5)^2 = \frac{5}{12} \times 25\pi = \frac{125\pi}{12}$.

(D) Let $P(x) = 4x^4 + 8x^3 - 17x^2 - 12x + 9 = 4(x-x_1)(x-x_2)(x-x_3)(x-x_4)$.
We want to evaluate the product $S = (4+x_1^2)(4+x_2^2)(4+x_3^2)(4+x_4^2)$.
Note that $4+x_k^2 = (2i+x_k)(-2i+x_k) = (x_k - 2i)(x_k + 2i)$.
Thus,$S = \prod_{k=1}^4 (x_k - 2i) \prod_{k=1}^4 (x_k + 2i) = \prod_{k=1}^4 (2i - x_k) \prod_{k=1}^4 (-2i - x_k)$.
From $P(x) = 4(x-x_1)(x-x_2)(x-x_3)(x-x_4)$,we have $\prod_{k=1}^4 (x-x_k) = \frac{P(x)}{4}$.
So,$\prod_{k=1}^4 (2i - x_k) = \frac{P(2i)}{4}$ and $\prod_{k=1}^4 (-2i - x_k) = \frac{P(-2i)}{4}$.
$P(2i) = 4(2i)^4 + 8(2i)^3 - 17(2i)^2 - 12(2i) + 9 = 4(16) + 8(-8i) - 17(-4) - 24i + 9 = 64 - 64i + 68 - 24i + 9 = 141 - 88i$.
$P(-2i) = 4(-2i)^4 + 8(-2i)^3 - 17(-2i)^2 - 12(-2i) + 9 = 4(16) + 8(8i) - 17(-4) + 24i + 9 = 64 + 64i + 68 + 24i + 9 = 141 + 88i$.
$S = \frac{P(2i)}{4} \times \frac{P(-2i)}{4} = \frac{(141-88i)(141+88i)}{16} = \frac{141^2 + 88^2}{16} = \frac{19881 + 7744}{16} = \frac{27625}{16}$.
Given $S = \frac{125}{16}m$,we have $\frac{27625}{16} = \frac{125}{16}m$.
$m = \frac{27625}{125} = 221$.

(A) Given $\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2$.
Squaring both sides,we get $\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|^2=4$.
Using the property $|z|^2 = z \bar{z}$,we have $\frac{(z_1-2 z_2)(\bar{z}_1-2 \bar{z}_2)}{(\frac{1}{2}-z_1 \bar{z}_2)(\frac{1}{2}-\bar{z}_1 z_2)}=4$.
Expanding the numerator: $z_1 \bar{z}_1 - 2 z_1 \bar{z}_2 - 2 z_2 \bar{z}_1 + 4 z_2 \bar{z}_2 = |z_1|^2 - 2 z_1 \bar{z}_2 - 2 \bar{z}_1 z_2 + 4 |z_2|^2$.
Expanding the denominator: $\frac{1}{4} - \frac{1}{2} \bar{z}_1 z_2 - \frac{1}{2} z_1 \bar{z}_2 + |z_1|^2 |z_2|^2$.
So,$|z_1|^2 - 2 z_1 \bar{z}_2 - 2 \bar{z}_1 z_2 + 4 |z_2|^2 = 4 (\frac{1}{4} - \frac{1}{2} \bar{z}_1 z_2 - \frac{1}{2} z_1 \bar{z}_2 + |z_1|^2 |z_2|^2)$.
$|z_1|^2 - 2 z_1 \bar{z}_2 - 2 \bar{z}_1 z_2 + 4 |z_2|^2 = 1 - 2 \bar{z}_1 z_2 - 2 z_1 \bar{z}_2 + 4 |z_1|^2 |z_2|^2$.
Canceling $-2 z_1 \bar{z}_2 - 2 \bar{z}_1 z_2$ from both sides: $|z_1|^2 + 4 |z_2|^2 = 1 + 4 |z_1|^2 |z_2|^2$.
Rearranging: $|z_1|^2 - 1 - 4 |z_2|^2 + 4 |z_1|^2 |z_2|^2 = 0$.
$(|z_1|^2 - 1) - 4 |z_2|^2 (1 - |z_1|^2) = 0$.
$(|z_1|^2 - 1) + 4 |z_2|^2 (|z_1|^2 - 1) = 0$.
$(|z_1|^2 - 1)(1 + 4 |z_2|^2) = 0$.
Since $1 + 4 |z_2|^2$ cannot be $0$,we must have $|z_1|^2 - 1 = 0$,which means $|z_1| = 1$.
Wait,checking the original equation again,if we write $4|z_2|^2$ as $|2z_2|^2$,the expression factors as $(|z_1|^2 - 1)(1 - |2z_2|^2) = 0$.
Thus,$|z_1| = 1$ or $|2z_2| = 1$,which means $|z_1| = 1$ or $|z_2| = \frac{1}{2}$.

(B) Given $|z-1| \leq 1$,where $z=x+iy$.
$(x-1)^2 + y^2 \leq 1$. This represents a disk centered at $(1,0)$ with radius $1$.
Also,$|z-5| \leq |z-5i|$.
$(x-5)^2 + y^2 \leq x^2 + (y-5)^2$.
$x^2 - 10x + 25 + y^2 \leq x^2 + y^2 - 10y + 25$.
$-10x \leq -10y \Rightarrow x \geq y$.
We need to find $z=x+iy$ such that $x, y \in \mathbb{Z}$ satisfying $(x-1)^2 + y^2 \leq 1$ and $x \geq y$.
Possible integer points $(x,y)$ satisfying $(x-1)^2 + y^2 \leq 1$:
If $x=0$,$(0-1)^2 + y^2 \leq 1$ $\Rightarrow 1 + y^2 \leq 1$ $\Rightarrow y^2 \leq 0$ $\Rightarrow y=0$. Point: $(0,0)$. Check $x \geq y$: $0 \geq 0$ (True).
If $x=1$,$(1-1)^2 + y^2 \leq 1$ $\Rightarrow y^2 \leq 1$ $\Rightarrow y \in \{-1, 0, 1\}$. Points: $(1,-1), (1,0), (1,1)$. Check $x \geq y$: $1 \geq -1$ (True),$1 \geq 0$ (True),$1 \geq 1$ (True).
If $x=2$,$(2-1)^2 + y^2 \leq 1$ $\Rightarrow 1 + y^2 \leq 1$ $\Rightarrow y^2 \leq 0$ $\Rightarrow y=0$. Point: $(2,0)$. Check $x \geq y$: $2 \geq 0$ (True).
The set of elements is $z \in \{0, 1-i, 1, 1+i, 2\}$.
The sum of the squares of the modulus is:
$|0|^2 + |1-i|^2 + |1|^2 + |1+i|^2 + |2|^2 = 0 + (1^2+(-1)^2) + 1^2 + (1^2+1^2) + 2^2 = 0 + 2 + 1 + 2 + 4 = 9$.

(A) Let $w = \frac{z-2i}{z+2i}$. Given that $\text{Re}(w) = 0$,we have $w + \bar{w} = 0$.
$\frac{z-2i}{z+2i} + \frac{\bar{z}+2i}{\bar{z}-2i} = 0$
$(z-2i)(\bar{z}-2i) + (\bar{z}+2i)(z+2i) = 0$
$z\bar{z} - 2iz - 2i\bar{z} - 4 + z\bar{z} + 2iz + 2i\bar{z} - 4 = 0$
$2|z|^2 - 8 = 0$ $\Rightarrow |z|^2 = 4$ $\Rightarrow |z| = 2$.
This represents a circle centered at the origin with radius $r = 2$.
We want to find the maximum value of $|z - (6+8i)|$,which is the distance from a point $z$ on the circle to the point $P = 6+8i$.
The distance from the origin $O(0,0)$ to $P(6,8)$ is $OP = \sqrt{6^2 + 8^2} = \sqrt{36+64} = 10$.
The maximum distance from a point on the circle to $P$ is $OP + r = 10 + 2 = 12$.

(D) Let the origin be $O(0, 0)$. The man walks $3$ units in the direction of $N 45^{\circ} E$,which corresponds to an angle of $\pi / 4$ with the positive $x$-axis. The position of point $A$ is $z_A = 3 e^{i \pi / 4}$.
From $A$,he walks $4$ units in the direction of $N 45^{\circ} W$. The direction $N 45^{\circ} W$ makes an angle of $45^{\circ} + 90^{\circ} = 135^{\circ}$ or $3\pi / 4$ with the positive $x$-axis.
The displacement vector from $A$ to $P$ is $4 e^{i 3\pi / 4}$.
Thus,the position of $P$ is $z_P = z_A + 4 e^{i 3\pi / 4} = 3 e^{i \pi / 4} + 4 e^{i 3\pi / 4}$.
Since $e^{i 3\pi / 4} = e^{i \pi / 4} \cdot e^{i \pi / 2} = i e^{i \pi / 4}$,we have:
$z_P = 3 e^{i \pi / 4} + 4 i e^{i \pi / 4} = (3 + 4 i) e^{i \pi / 4}$.

(B, C, D) $1.$ $A$ represents the region $y \geq 1$. $B$ is a circle with center $(2, 1)$ and radius $3$. $C$ is the line $\operatorname{Re}((1-i)(x+iy)) = x+y = \sqrt{2}$.
Substituting $y = \sqrt{2}-x$ into the circle equation $(x-2)^2 + (y-1)^2 = 9$ gives $(x-2)^2 + (\sqrt{2}-x-1)^2 = 9$.
Expanding this: $(x^2 - 4x + 4) + (x^2 + 2x(1-\sqrt{2}) + (1-\sqrt{2})^2) = 9$.
$2x^2 + x(2-2\sqrt{2}-4) + 4 + 1 - 2\sqrt{2} + 2 = 9 \implies 2x^2 - (2+2\sqrt{2})x - 2 - 2\sqrt{2} = 0$.
Solving this quadratic,we find one point with $y \geq 1$. Thus,the number of elements is $1$.
$2.$ Let $z = x+iy$. The expression is $|(x+1)+i(y-1)|^2 + |(x-5)+i(y-1)|^2 = (x+1)^2 + (y-1)^2 + (x-5)^2 + (y-1)^2$.
Since $z$ is on the circle $(x-2)^2 + (y-1)^2 = 9$,we have $(y-1)^2 = 9 - (x-2)^2$.
Substituting this,the expression becomes $(x+1)^2 + (x-5)^2 + 2(9 - (x-2)^2) = x^2+2x+1 + x^2-10x+25 + 18 - 2(x^2-4x+4) = 2x^2-8x+26 + 18 - 2x^2+8x-8 = 36$.
Since $36$ is between $35$ and $39$,the answer is $(C)$.
$3.$ Since $|z-2-i|=3$ and $|w-2-i| < 3$,by the triangle inequality $|z-w| < |z-(2+i)| + |w-(2+i)| < 3+3 = 6$.
Also $|z-w| > ||z-(2+i)| - |w-(2+i)|| = |3 - |w-(2+i)|| > 0$.
Thus $0 < |z-w| < 6$. Using $||z|-|w|| \leq |z-w|$,we have $-6 < |z|-|w| < 6$.
Adding $3$,we get $-3 < |z|-|w|+3 < 9$.

(D) The initial position is $z_0 = 1 + 2i$.
Moving horizontally by $5$ units: $z = (1 + 5) + 2i = 6 + 2i$.
Moving vertically by $3$ units: $z_1 = 6 + (2 + 3)i = 6 + 5i$.
From $z_1$,moving $\sqrt{2}$ units in the direction of $\hat{i} + \hat{j}$ (which is the direction of $1+i$): The unit vector is $\frac{1+i}{\sqrt{2}}$. The displacement is $\sqrt{2} \times \frac{1+i}{\sqrt{2}} = 1 + i$.
So,the position becomes $z' = (6 + 5i) + (1 + i) = 7 + 6i$.
Finally,rotating $z'$ by $\frac{\pi}{2}$ anticlockwise about the origin is equivalent to multiplying by $e^{i\pi/2} = i$.
$z_2 = (7 + 6i) \times i = 7i + 6i^2 = 7i - 6 = -6 + 7i$.

(D) Given $z = \frac{1}{a+ibt}$.
$x+iy = \frac{a-ibt}{a^2+b^2t^2}$.
Equating real and imaginary parts,$x = \frac{a}{a^2+b^2t^2}$ and $y = \frac{-bt}{a^2+b^2t^2}$.
If $a \neq 0$ and $b \neq 0$,then $a^2+b^2t^2 = \frac{a}{x}$,so $b^2t^2 = \frac{a}{x} - a^2 = \frac{a(1-ax)}{x}$.
Also $y^2 = \frac{b^2t^2}{(a^2+b^2t^2)^2} = \frac{a(1-ax)/x}{(a/x)^2} = \frac{x(1-ax)}{a} = \frac{x}{a} - x^2$.
Rearranging gives $x^2 - \frac{x}{a} + y^2 = 0$,which is $(x - \frac{1}{2a})^2 + y^2 = (\frac{1}{2a})^2$. This represents a circle with radius $|\frac{1}{2a}|$ and center $(\frac{1}{2a}, 0)$.
If $b=0$,then $z = \frac{1}{a}$,so $y=0$,which is the $x$-axis.
If $a=0$,then $z = \frac{1}{ibt} = -i(\frac{1}{bt})$,so $x=0$,which is the $y$-axis.
Thus,options $A, C, D$ are correct.

(C) Let the vertices of the regular octagon be represented by complex numbers $z_k = 2e^{i(\theta_0 + \frac{2\pi(k-1)}{8})}$ for $k=1, 2, \ldots, 8$. Without loss of generality,let $\theta_0 = 0$. The vertices are the roots of the equation $z^8 - 2^8 = 0$.
Thus,$z^8 - 2^8 = \prod_{k=1}^8 (z - A_k)$.
Let $P$ be represented by the complex number $z = 2e^{i\theta}$. The distance $PA_k = |z - A_k|$.
The product is $\prod_{k=1}^8 PA_k = |\prod_{k=1}^8 (z - A_k)| = |z^8 - 2^8|$.
Substituting $z = 2e^{i\theta}$,we get $|(2e^{i\theta})^8 - 2^8| = |2^8 e^{i8\theta} - 2^8| = 2^8 |e^{i8\theta} - 1|$.
Using the identity $|e^{i\phi} - 1| = |\cos \phi + i \sin \phi - 1| = \sqrt{(\cos \phi - 1)^2 + \sin^2 \phi} = \sqrt{2 - 2\cos \phi} = 2|\sin(\frac{\phi}{2})|$.
Here,$\phi = 8\theta$,so the product is $2^8 \cdot 2|\sin(4\theta)| = 2^9 |\sin(4\theta)| = 512 |\sin(4\theta)|$.
The maximum value of $|\sin(4\theta)|$ is $1$.
Therefore,the maximum value of the product is $512 \times 1 = 512$.

(A) Given $z = (1-t)z_1 + tz_2$,where $0 < t < 1$. This implies that $z$ lies on the line segment joining $z_1$ and $z_2$.
$1$. Since $z$ lies on the segment $AB$,the sum of distances $|z-z_1| + |z-z_2|$ is equal to the total distance $|z_1-z_2|$. Thus,$(A)$ is true.
$2$. The vector $z-z_1$ is in the same direction as $z_2-z_1$ because $z-z_1 = t(z_2-z_1)$ and $t > 0$. Therefore,$\operatorname{Arg}(z-z_1) = \operatorname{Arg}(z_2-z_1)$. Thus,$(D)$ is true.
$3$. The condition $\frac{z-z_1}{z_2-z_1} = t$ (where $t$ is real) implies that the ratio is purely real. This is equivalent to $\frac{z-z_1}{z_2-z_1} = \overline{\left(\frac{z-z_1}{z_2-z_1}\right)} = \frac{\bar{z}-\bar{z}_1}{\bar{z}_2-\bar{z}_1}$. Cross-multiplying gives $(z-z_1)(\bar{z}_2-\bar{z}_1) - (\bar{z}-\bar{z}_1)(z_2-z_1) = 0$,which is the determinant form $\left|\begin{array}{cc} z-z_1 & \bar{z}-\bar{z}_1 \\ z_2-z_1 & \bar{z}_2-\bar{z}_1 \end{array}\right| = 0$. Thus,$(C)$ is true.
$4$. $\operatorname{Arg}(z-z_1)$ and $\operatorname{Arg}(z-z_2)$ represent the angles of vectors $P-A$ and $P-B$. Since $P$ is between $A$ and $B$,these vectors point in opposite directions,so $\operatorname{Arg}(z-z_1) \neq \operatorname{Arg}(z-z_2)$. Thus,$(B)$ is false.
Therefore,the correct options are $(A), (C), (D)$.

(C) $(A)-(q)$: $|z-i|z||=|z+i|z|| \Rightarrow |\frac{z}{|z|}-i|=|\frac{z}{|z|}+i|$,for $z \neq 0$. The expression $\frac{z}{|z|}$ represents a point on the unit circle. The equation implies that the point $\frac{z}{|z|}$ is equidistant from $i$ and $-i$. The locus of such points is the real axis,where $\operatorname{Im}(z)=0$.
$(B)-(p)$: $|z+4|+|z-4|=10$ represents an ellipse with foci at $(\pm 4, 0)$ and major axis length $2a=10$. Here $2ae=8$ and $2a=10$,so $e=4/5$. Thus,it is an ellipse with eccentricity $4/5$.
$(C)-(p), (t)$: Let $\omega=2(\cos \theta+i \sin \theta)$. Then $z = 2(\cos \theta+i \sin \theta) - \frac{1}{2}(\cos \theta-i \sin \theta) = \frac{3}{2} \cos \theta + i \frac{5}{2} \sin \theta$. This is an ellipse $\frac{x^2}{(3/2)^2} + \frac{y^2}{(5/2)^2} = 1$ with $e^2 = 1 - \frac{9/4}{25/4} = 16/25$,so $e=4/5$. Since the semi-major axis is $2.5 < 3$,it is contained in $|z| \leq 3$.
$(D)-(q), (t)$: Let $\omega = \cos \theta + i \sin \theta$. Then $z = (\cos \theta + i \sin \theta) + (\cos \theta - i \sin \theta) = 2 \cos \theta$. Since $z$ is purely real,$\operatorname{Im}(z)=0$ and $|z| = |2 \cos \theta| \leq 2 < 3$.

(A) Let $w = z - (3 + 2i)$. Given $|w| \leq 2$.
We need to find the minimum value of $|2z - 6 + 5i|$.
$|2z - 6 + 5i| = |2(z - 3) + 5i| = |2(z - 3 - 2i + 2i) + 5i| = |2(z - 3 - 2i) + 4i + 5i| = |2(z - 3 - 2i) + 9i|$.
Let $w = z - 3 - 2i$,where $|w| \leq 2$.
The expression becomes $|2w + 9i| = 2|w + \frac{9}{2}i|$.
By the triangle inequality,$|w + \frac{9}{2}i| \geq ||\frac{9}{2}i| - |w||$.
Since $|w| \leq 2$,the minimum value of $|w + \frac{9}{2}i|$ occurs when $w$ is in the direction of $\frac{9}{2}i$,giving $|\frac{9}{2} - |w|| = |4.5 - 2| = 2.5$.
Thus,the minimum value is $2 \times 2.5 = 5$.

(A) Let $z = x + iy$,$s = s_1 + is_2$,$t = t_1 + it_2$,and $r = r_1 + ir_2$.
The equation $sz + t\bar{z} + r = 0$ becomes:
$(s_1 + is_2)(x + iy) + (t_1 + it_2)(x - iy) + (r_1 + ir_2) = 0$
$(s_1x - s_2y + t_1x + t_2y + r_1) + i(s_2x + s_1y - t_2x + t_1y + r_2) = 0$
Equating real and imaginary parts to zero:
$(s_1 + t_1)x + (t_2 - s_2)y + r_1 = 0$
$(s_2 - t_2)x + (s_1 + t_1)y + r_2 = 0$
This is a system of two linear equations in $x$ and $y$. The determinant of the coefficient matrix is $D = (s_1 + t_1)^2 - (t_2 - s_2)(s_2 - t_2) = (s_1 + t_1)^2 + (t_2 - s_2)^2 = |s + t|^2$.
If $|s| \neq |t|$,then $D \neq 0$ (unless $s = -t$),leading to a unique solution (a point).
If $|s| = |t|$,the lines are either parallel or coincident. Specifically,if $s \neq -t$,the system represents a line or is inconsistent.
Thus,if $L$ has more than one element,it must be a line (infinitely many elements).
$L$ is either a point or a line. The intersection of a line and a circle is at most $2$ points. If $L$ is a point,the intersection is at most $1$ point. Thus,$C$ is true.
Therefore,statements $A, B, C, D$ are all true.

(B) The condition $|z-(2-i)| \geq \sqrt{5}$ represents the region outside or on the circle with center $C(2, -1)$ and radius $r = \sqrt{5}$.
To maximize $\frac{1}{|z-1|}$,we must minimize $|z-1|$,which is the distance of $z$ from the point $A(1, 0)$.
The point $z_0$ on the circle closest to $A(1, 0)$ lies on the line segment connecting $A(1, 0)$ and $C(2, -1)$.
The line $AC$ has the equation $y - 0 = \frac{-1-0}{2-1}(x-1)$,so $y = -x+1$,or $x+y-1=0$.
The intersection of this line with the circle $(x-2)^2 + (y+1)^2 = 5$ gives $z_0$. Since $z_0$ is on the circle and on the line $y = 1-x$,we have $(x-2)^2 + (1-x+1)^2 = 5 \implies (x-2)^2 + (2-x)^2 = 5 \implies 2(x-2)^2 = 5 \implies x-2 = \pm \sqrt{2.5}$.
Since $z_0$ is closer to $A(1, 0)$,$x_0 = 2 - \sqrt{2.5} \approx 0.42$,so $y_0 = 1 - x_0 = \sqrt{2.5} - 1 \approx 0.58$.
Let $z_0 = x_0 + iy_0$. Then $z_0 - \bar{z}_0 = 2iy_0$ and $z_0 + \bar{z}_0 = 2x_0$.
The expression becomes $\frac{4-2x_0}{2iy_0+2i} = \frac{2-x_0}{i(y_0+1)}$.
Since $y_0 = 1-x_0$,$y_0+1 = 2-x_0$.
Thus,the expression is $\frac{2-x_0}{i(2-x_0)} = \frac{1}{i} = -i$.
The principal argument of $-i$ is $-\frac{\pi}{2}$.

(C) Given $|z_1| = |z_2| = \ldots = |z_{10}| = 1$.
The distance between two points $z_a$ and $z_b$ on the unit circle is $|z_a - z_b| = 2 \sin(\frac{\Delta \theta}{2})$,where $\Delta \theta$ is the angle between them.
Since $\sin(x) \leq x$ for $x \geq 0$,we have $|z_k - z_{k-1}| = 2 \sin(\frac{\theta_k}{2}) \leq 2(\frac{\theta_k}{2}) = \theta_k$.
Summing these,$\sum_{k=1}^{10} |z_{k+1} - z_k| \leq \sum_{k=1}^{10} \theta_k = 2 \pi$ (where $z_{11} = z_1$). Thus,$P$ is $TRUE$.
For $Q$,let $w_k = z_k^2 = e^{i 2 \phi_k}$,where $\phi_k = \sum_{j=1}^k \theta_j$. The angle between $w_k$ and $w_{k-1}$ is $2 \theta_k$.
Similarly,$|w_k - w_{k-1}| = |z_k^2 - z_{k-1}^2| = 2 \sin(\frac{2 \theta_k}{2}) = 2 \sin(\theta_k) \leq 2 \theta_k$.
Summing these,$\sum |z_k^2 - z_{k-1}^2| \leq \sum 2 \theta_k = 2(2 \pi) = 4 \pi$. Thus,$Q$ is $TRUE$.

(D) The condition $\arg \left(\frac{z+\alpha}{z+\beta}\right)=\frac{\pi}{4}$ represents an arc of a circle passing through the points $(-\alpha, 0)$ and $(-\beta, 0)$,where the angle subtended by the chord joining these points at any point $z$ on the arc is $\frac{\pi}{4}$.
Given the circle $x^2+y^2+5x-3y+4=0$,we find its intersection with the $x$-axis by setting $y=0$:
$x^2+5x+4=0 \Rightarrow (x+1)(x+4)=0 \Rightarrow x=-1, x=-4$.
Thus,the points $(-\alpha, 0)$ and $(-\beta, 0)$ are $(-1, 0)$ and $(-4, 0)$.
This implies ${-\alpha, -\beta} = {-1, -4}$,so ${\alpha, \beta} = {1, 4}$.
For $\arg \left(\frac{z+\alpha}{z+\beta}\right)=\frac{\pi}{4}$ to be positive,the order of points must be such that the rotation from the vector $(z+\beta)$ to $(z+\alpha)$ is counter-clockwise. Comparing with the standard form $\arg \left(\frac{z-z_1}{z-z_2}\right) = \theta$,we identify $\alpha=1$ and $\beta=4$ (or vice versa depending on the arc orientation). Checking the options,$\beta=4$ and $\alpha\beta = 1 \times 4 = 4$ are consistent with $(B)$ and $(D)$.

(C) The given circles are $|z-z_0|=r$ and $|z-z_0|=2r$.
Since $\alpha$ lies on the first circle,$|\alpha-z_0|=r$,which implies $|\alpha-z_0|^2=r^2$.
Expanding this,$|\alpha|^2 - z_0\bar{\alpha} - \bar{z}_0\alpha + |z_0|^2 = r^2$ $(1)$.
Since $\frac{1}{\bar{\alpha}}$ lies on the second circle,$|\frac{1}{\bar{\alpha}}-z_0|=2r$,which implies $|\frac{1}{\bar{\alpha}}-z_0|^2=4r^2$.
Using $\bar{\alpha}\alpha = |\alpha|^2$,we have $|\frac{\alpha}{|\alpha|^2}-z_0|^2=4r^2$.
Expanding this,$\frac{1}{|\alpha|^2} - \frac{z_0\bar{\alpha}}{|\alpha|^2} - \frac{\bar{z}_0\alpha}{|\alpha|^2} + |z_0|^2 = 4r^2$.
Multiplying by $|\alpha|^2$,$1 - z_0\bar{\alpha} - \bar{z}_0\alpha + |z_0|^2|\alpha|^2 = 4r^2|\alpha|^2$ $(2)$.
Subtracting $(1)$ from $(2)$,$(|z_0|^2|\alpha|^2 - |z_0|^2) + (1 - |\alpha|^2) = 4r^2|\alpha|^2 - r^2$.
$|z_0|^2(|\alpha|^2-1) - (|\alpha|^2-1) = r^2(4|\alpha|^2-1)$.
$(|z_0|^2-1)(|\alpha|^2-1) = r^2(4|\alpha|^2-1)$.
Given $2|z_0|^2 = r^2+2$,so $|z_0|^2 = \frac{r^2+2}{2}$.
Substituting this,$(\frac{r^2+2}{2}-1)(|\alpha|^2-1) = r^2(4|\alpha|^2-1)$.
$\frac{r^2}{2}(|\alpha|^2-1) = r^2(4|\alpha|^2-1)$.
$\frac{1}{2}|\alpha|^2 - \frac{1}{2} = 4|\alpha|^2 - 1$.
$\frac{1}{2} = \frac{7}{2}|\alpha|^2$.
$|\alpha|^2 = \frac{1}{7} \Rightarrow |\alpha| = \frac{1}{\sqrt{7}}$.