Geometry of complex numbers Questions in English

(B) Let $z = x + iy$. The vertices of the triangle are $z = (x, y)$,$iz = (-y, x)$,and $z + iz = (x - y, x + y)$.
The area $A$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the coordinates:
$A = \frac{1}{2} |x(x - (x + y)) + (-y)((x + y) - y) + (x - y)(y - x)|$
$A = \frac{1}{2} |x(-y) - y(x) + (x - y)(-(x - y))|$
$A = \frac{1}{2} |-xy - xy - (x - y)^2|$
$A = \frac{1}{2} |-2xy - (x^2 - 2xy + y^2)|$
$A = \frac{1}{2} |-2xy - x^2 + 2xy - y^2|$
$A = \frac{1}{2} |-x^2 - y^2| = \frac{1}{2} (x^2 + y^2) = \frac{1}{2} |z|^2$.

(A) The points are given by $A(3, 4), B(5, -2)$,and $C(-1, 16)$.
To check if they are collinear,we calculate the area of the triangle formed by these points:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$\text{Area} = \frac{1}{2} |3(-2 - 16) + 5(16 - 4) + (-1)(4 - (-2))|$
$\text{Area} = \frac{1}{2} |3(-18) + 5(12) - 1(6)|$
$\text{Area} = \frac{1}{2} |-54 + 60 - 6| = \frac{1}{2} |0| = 0$.
Since the area is $0$,the points $A, B, C$ are collinear.

(D) The affix of the centroid $G$ is given by $g = \frac{z_1 + z_2 + z_3}{3}$.
Since $z = 0$ is the midpoint of $AG$,the affix of the midpoint is $\frac{g + z_1}{2} = 0$.
Substituting the value of $g$,we get $\frac{\frac{z_1 + z_2 + z_3}{3} + z_1}{2} = 0$.
Multiplying by $6$,we get $(z_1 + z_2 + z_3) + 3z_1 = 0$.
Thus,$4z_1 + z_2 + z_3 = 0$.

(C) Given $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 1$.
Multiplying by $z_1 z_2$,we get $z_1^2 + z_2^2 = z_1 z_2$.
This can be rewritten as $z_1^2 + z_2^2 - z_1 z_2 = 0$.
Let $z_3 = 0$ be the origin.
The condition for three points $z_1, z_2, z_3$ to form an equilateral triangle is $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.
Substituting $z_3 = 0$,we get $z_1^2 + z_2^2 = z_1 z_2$,which matches our given equation.
Therefore,$z_1, z_2$ and the origin form an equilateral triangle.

(A) Let $z = x + iy$. Then the points are $A(x, y)$,$B(x - y, x + y)$,and $C(0, x)$.
Using the area formula for a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$\text{Area} = \frac{1}{2} |x(x + y - x) + (x - y)(x - y) + 0(y - (x + y))|$
$\text{Area} = \frac{1}{2} |x(y) + (x - y)^2| = \frac{1}{2} |xy + x^2 - 2xy + y^2| = \frac{1}{2} |x^2 - xy + y^2|$
Alternatively,using the property of complex numbers,the area of the triangle formed by $z_1, z_2, z_3$ is $\frac{1}{2} |\text{Im}(\bar{z_1}z_2 + \bar{z_2}z_3 + \bar{z_3}z_1)|$.
For $z_1 = z, z_2 = z + iz, z_3 = iz$:
$\text{Area} = \frac{1}{2} |\text{Im}(\bar{z}(z + iz) + (\bar{z} - i\bar{z})(iz) + (-i\bar{z})(z))|$
$\text{Area} = \frac{1}{2} |\text{Im}(|z|^2 + i|z|^2 + i\bar{z}z - i^2\bar{z}z - i|z|^2)| = \frac{1}{2} |\text{Im}(|z|^2 + i|z|^2 + i|z|^2 + |z|^2 - i|z|^2)| = \frac{1}{2} |\text{Im}(2|z|^2 + i|z|^2)| = \frac{1}{2} |z|^2$.
Given $\frac{1}{2} |z|^2 = 18$,we get $|z|^2 = 36$,so $|z| = 6$.

(D) The complex numbers are given as points in the Cartesian plane:
${z_1} = (1, 1)$
${z_2} = (-2, 3)$
${z_3} = (0, \frac{a}{3})$
Since the points are collinear,the area of the triangle formed by them is zero,or the determinant of the coordinates is zero:
$\begin{vmatrix} 1 & 1 & 1 \\ -2 & 3 & 1 \\ 0 & \frac{a}{3} & 1 \end{vmatrix} = 0$
Expanding along the first column:
$1(3 - \frac{a}{3}) - (-2)(1 - \frac{a}{3}) + 0 = 0$
$3 - \frac{a}{3} + 2 - \frac{2a}{3} = 0$
$5 - a = 0$
$a = 5$

(B) The vertices of the triangle are $z_1 = 0$,$z_2 = z$,and $z_3 = z e^{i\alpha }$.
The area of a triangle with vertices $z_1, z_2, z_3$ in the complex plane is given by $\frac{1}{2} |\text{Im}(\bar{z_1}z_2 + \bar{z_2}z_3 + \bar{z_3}z_1)|$.
Since $z_1 = 0$,the area is $\frac{1}{2} |\text{Im}(\bar{z} \cdot z e^{i\alpha })|$.
Substituting $\bar{z}z = |z|^2$,we get:
Area $= \frac{1}{2} |\text{Im}(|z|^2 e^{i\alpha })|$.
Since $e^{i\alpha } = \cos \alpha + i \sin \alpha$,we have:
Area $= \frac{1}{2} |z|^2 |\text{Im}(\cos \alpha + i \sin \alpha)|$.
Area $= \frac{1}{2} |z|^2 \sin \alpha$ (since $0 < \alpha < \pi$,$\sin \alpha > 0$).

(D) Given the complex numbers: $z_1 = 1 + 2i$,$z_2 = 2 + 3i$,and $z_3 = 3 + 4i$.
Calculate the distances between the points:
$|z_1 - z_2| = |(1-2) + (2-3)i| = |-1 - i| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$.
$|z_2 - z_3| = |(2-3) + (3-4)i| = |-1 - i| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$.
$|z_1 - z_3| = |(1-3) + (2-4)i| = |-2 - 2i| = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$.
Since $|z_1 - z_2| + |z_2 - z_3| = \sqrt{2} + \sqrt{2} = 2\sqrt{2} = |z_1 - z_3|$,the points are collinear and do not form a triangle.
Therefore,the correct option is $D$.

(A) Given the equation $\left| \frac{z - 5i}{z + 5i} \right| = 1$.
Substituting $z = x + iy$,we get $\left| \frac{x + i(y - 5)}{x + i(y + 5)} \right| = 1$.
Using the property $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$,we have $|x + i(y - 5)| = |x + i(y + 5)|$.
Squaring both sides,we get $x^2 + (y - 5)^2 = x^2 + (y + 5)^2$.
Expanding the squares: $x^2 + y^2 - 10y + 25 = x^2 + y^2 + 10y + 25$.
Simplifying the equation: $-10y = 10y$,which implies $20y = 0$,so $y = 0$.
The equation $y = 0$ represents the real axis.

(A) Let $z = x + iy$. Then $\frac{z + i}{z + 2} = \frac{x + i(y + 1)}{(x + 2) + iy}$.
Multiplying the numerator and denominator by the conjugate of the denominator,we get:
$\frac{[x + i(y + 1)][(x + 2) - iy]}{(x + 2)^2 + y^2} = \frac{x(x + 2) + y(y + 1) + i[(y + 1)(x + 2) - xy]}{(x + 2)^2 + y^2}$.
For the expression to be purely imaginary,the real part must be zero:
$x(x + 2) + y(y + 1) = 0 \implies x^2 + 2x + y^2 + y = 0$.
This is the equation of a circle $x^2 + y^2 + 2x + y = 0$.
The center is $(-1, -1/2)$ and the radius is $\sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-1/2)^2 - 0} = \sqrt{1 + 1/4} = \frac{\sqrt{5}}{2}$.
Thus,the locus is a circle of radius $\frac{\sqrt{5}}{2}$.

(B) Given the equation $|z + 1| = \sqrt{2} |z - 1|$.
Let $z = x + iy$. Substituting this into the equation,we get $|(x + 1) + iy| = \sqrt{2} |(x - 1) + iy|$.
Squaring both sides,we have $(x + 1)^2 + y^2 = 2((x - 1)^2 + y^2)$.
Expanding the terms: $x^2 + 2x + 1 + y^2 = 2(x^2 - 2x + 1 + y^2)$.
$x^2 + 2x + 1 + y^2 = 2x^2 - 4x + 2 + 2y^2$.
Rearranging the terms to one side: $x^2 + y^2 - 6x + 1 = 0$.
This is the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ with $g = -3, f = 0, c = 1$.
Thus,the locus is a circle.

(B) Given $\left| \frac{z - a}{z + \overline{a}} \right| = 1$.
This implies $|z - a| = |z + \overline{a}|$.
Squaring both sides: $|z - a|^2 = |z + \overline{a}|^2$.
Using the property $|w|^2 = w \cdot \overline{w}$,we get:
$(z - a)(\overline{z} - \overline{a}) = (z + \overline{a})(\overline{z} + a)$.
Expanding both sides:
$z\overline{z} - z\overline{a} - a\overline{z} + a\overline{a} = z\overline{z} + za + \overline{a}\overline{z} + \overline{a}a$.
Canceling $z\overline{z}$ and $a\overline{a}$ from both sides:
$-z\overline{a} - a\overline{z} = za + \overline{a}\overline{z}$.
Rearranging terms:
$za + z\overline{a} + a\overline{z} + \overline{a}\overline{z} = 0$.
$(z + \overline{z})(a + \overline{a}) = 0$.
Since $a + \overline{a} = 2\text{Re}(a) \neq 0$,we must have $z + \overline{z} = 0$.
Let $z = x + iy$,then $z + \overline{z} = (x + iy) + (x - iy) = 2x = 0$.
Thus,$x = 0$,which is the equation of the $y$-axis.

(C) The given inequality is $|z - 1| + |z + 1| \le 4$.
This is of the form $|z - z_1| + |z - z_2| \le 2a$,where $z_1 = 1$ and $z_2 = -1$.
The distance between the foci $z_1$ and $z_2$ is $2c = |1 - (-1)| = 2$,so $c = 1$.
The sum of the distances from any point $z$ to the foci is $2a = 4$,so $a = 2$.
Since $2a > 2c$,this represents the region consisting of the interior and boundary of an ellipse.
Using the relation $b^2 = a^2 - c^2$,we get $b^2 = 2^2 - 1^2 = 3$.
The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{4} + \frac{y^2}{3} = 1$.
Thus,the region is the interior and boundary of this ellipse.

(B) Let $z = x + iy$. Then $\frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$.
Multiplying the numerator and denominator by the conjugate of the denominator,we get $\frac{((x - 1) + iy)((x + 1) - iy)}{(x + 1)^2 + y^2} = \frac{(x^2 + y^2 - 1) + i(2y)}{(x + 1)^2 + y^2}$.
Given $\text{arg} \left( \frac{z - 1}{z + 1} \right) = \frac{\pi}{3}$,we have $\tan^{-1} \left( \frac{2y}{x^2 + y^2 - 1} \right) = \frac{\pi}{3}$.
This implies $\frac{2y}{x^2 + y^2 - 1} = \tan \frac{\pi}{3} = \sqrt{3}$.
Rearranging the terms,we get $x^2 + y^2 - 1 = \frac{2}{\sqrt{3}}y$,or $x^2 + y^2 - \frac{2}{\sqrt{3}}y - 1 = 0$.
This is the equation of a circle.

(B) Let $z = x + iy$. Then $\frac{2z + 1}{iz + 1} = \frac{2(x + iy) + 1}{i(x + iy) + 1} = \frac{(2x + 1) + 2iy}{(1 - y) + ix}$.
Multiplying the numerator and denominator by the conjugate of the denominator $(1 - y) - ix$:
$= \frac{[(2x + 1) + 2iy][(1 - y) - ix]}{(1 - y)^2 + x^2} = \frac{(2x + 1)(1 - y) + 2xy + i[2y(1 - y) - x(2x + 1)]}{(1 - y)^2 + x^2}$.
The imaginary part is given as $-2$:
$\frac{2y - 2y^2 - 2x^2 - x}{(1 - y)^2 + x^2} = -2$.
$2y - 2y^2 - 2x^2 - x = -2((1 - y)^2 + x^2) = -2(1 - 2y + y^2 + x^2) = -2 + 4y - 2y^2 - 2x^2$.
Simplifying the equation:
$2y - 2y^2 - 2x^2 - x = -2 + 4y - 2y^2 - 2x^2$.
$-x - 2y = -2$,which simplifies to $x + 2y - 2 = 0$.
This is the equation of a straight line.

(A) Let $z = x + iy$. Then $x = \lambda + 3$ and $y = \sqrt{5 - \lambda^2}$.
From the first equation,$\lambda = x - 3$.
Substituting this into the second equation: $y^2 = 5 - (x - 3)^2$.
Rearranging the terms,we get $(x - 3)^2 + y^2 = 5$.
This is the equation of a circle with center $(3, 0)$ and radius $\sqrt{5}$.

(C) The given equation is $|z - 3i| = 2$.
Let $z = x + iy$.
Substituting $z$ in the equation,we get $|x + iy - 3i| = 2$,which simplifies to $|x + i(y - 3)| = 2$.
The modulus of a complex number $a + ib$ is $\sqrt{a^2 + b^2}$.
Thus,$\sqrt{x^2 + (y - 3)^2} = 2$.
Squaring both sides,we get $x^2 + (y - 3)^2 = 4$.
This is the equation of a circle with center $(0, 3)$ and radius $2$.
Therefore,the locus is a circle.

(C) Given $|z - zi| = 1$.
Substitute $z = x + iy$:
$|x + iy - i(x + iy)| = 1$
$|x + iy - ix - i^2y| = 1$
Since $i^2 = -1$,we have:
$|x + iy - ix + y| = 1$
$|(x + y) + i(y - x)| = 1$
Taking the modulus:
$\sqrt{(x + y)^2 + (y - x)^2} = 1$
$(x + y)^2 + (y - x)^2 = 1$
$(x^2 + y^2 + 2xy) + (y^2 + x^2 - 2xy) = 1$
$2x^2 + 2y^2 = 1$
$x^2 + y^2 = \frac{1}{2}$
This is the equation of a circle with center $(0, 0)$ and radius $\frac{1}{\sqrt{2}}$.
Thus,$z$ lies on a circle.

(C) Given $\left| \frac{z - 1}{z - i} \right| = 1$.
This implies $|z - 1| = |z - i|$.
Let $z = x + iy$. Then $|(x - 1) + iy| = |x + i(y - 1)|$.
Squaring both sides,we get $(x - 1)^2 + y^2 = x^2 + (y - 1)^2$.
Expanding the terms,$x^2 - 2x + 1 + y^2 = x^2 + y^2 - 2y + 1$.
Simplifying,$-2x = -2y$,which gives $x = y$ or $x - y = 0$.
This is the equation of a straight line.

(B) Let $z = x + iy$,where $x, y \in \mathbb{R}$.
Then $z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy$.
The real part of $z^2$ is given by $\text{Re}(z^2) = x^2 - y^2$.
Given $\text{Re}(z^2) = 1$,we have $x^2 - y^2 = 1$.
This is the standard equation of a rectangular hyperbola.
Therefore,the correct option is $B$.

(C) Let $z = x + iy$.
Given equation: $|z - 1| = |z + i|$.
Substituting $z = x + iy$:
$|x + iy - 1| = |x + iy + i|$
$|(x - 1) + iy| = |x + i(y + 1)|$
Squaring both sides:
$(x - 1)^2 + y^2 = x^2 + (y + 1)^2$
$x^2 - 2x + 1 + y^2 = x^2 + y^2 + 2y + 1$
$-2x = 2y$
$x + y = 0$
This represents a straight line passing through the origin $(0, 0)$ with a slope of $-1$.

(B) Given the inequality: $\log_{\sqrt{3}} \left( \frac{|z|^2 - |z| + 1}{2 + |z|} \right) < 2$
Since the base $\sqrt{3} > 1$,the inequality remains in the same direction:
$\frac{|z|^2 - |z| + 1}{2 + |z|} < (\sqrt{3})^2$
$\frac{|z|^2 - |z| + 1}{2 + |z|} < 3$
$|z|^2 - |z| + 1 < 3(2 + |z|)$
$|z|^2 - |z| + 1 < 6 + 3|z|$
$|z|^2 - 4|z| - 5 < 0$
Let $t = |z|$,where $t \ge 0$. Then $t^2 - 4t - 5 < 0$.
$(t - 5)(t + 1) < 0$
This implies $-1 < t < 5$.
Since $t = |z| \ge 0$,we have $0 \le |z| < 5$.
Thus,the locus of $z$ is $|z| < 5$.

(A) Given $|z - 2 + i| = |z - 3 - i|$.
Substitute $z = x + iy$ into the equation:
$|(x - 2) + i(y + 1)| = |(x - 3) + i(y - 1)|$
Squaring both sides:
$(x - 2)^2 + (y + 1)^2 = (x - 3)^2 + (y - 1)^2$
Expanding the squares:
$x^2 - 4x + 4 + y^2 + 2y + 1 = x^2 - 6x + 9 + y^2 - 2y + 1$
Canceling $x^2$ and $y^2$ from both sides:
$-4x + 2y + 5 = -6x - 2y + 10$
Rearranging the terms:
$2x + 4y - 5 = 0$
Thus,the locus of $z$ is $2x + 4y - 5 = 0$.

(A) Given equation: $|iz - 1| + |z - i| = 2$
$|i(z - 1/i)| + |z - i| = 2$
$|i(z + i)| + |z - i| = 2$
$|i| |z + i| + |z - i| = 2$
Since $|i| = 1$,we have $|z - (-i)| + |z - i| = 2$
This is of the form $|z - z_1| + |z - z_2| = 2a$,where $z_1 = -i$ and $z_2 = i$.
The distance between $z_1$ and $z_2$ is $|z_2 - z_1| = |i - (-i)| = |2i| = 2$.
Since the sum of the distances from $z$ to two fixed points $z_1$ and $z_2$ is equal to the distance between the points themselves $(|z - z_1| + |z - z_2| = |z_1 - z_2|)$,the locus of $z$ is the line segment joining $z_1$ and $z_2$.
Thus,the locus is a straight line.

(A) Given $arg\left( \frac{z - 1}{z + 1} \right) = k$. Let $z = x + iy$.
Then $arg\left( \frac{(x - 1) + iy}{(x + 1) + iy} \right) = k$.
Using the property $arg(z_1/z_2) = arg(z_1) - arg(z_2)$,we get $arg((x - 1) + iy) - arg((x + 1) + iy) = k$.
This implies $\tan^{-1}\left( \frac{y}{x - 1} \right) - \tan^{-1}\left( \frac{y}{x + 1} \right) = k$.
Using $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left( \frac{A - B}{1 + AB} \right)$,we have $\tan^{-1}\left( \frac{\frac{y}{x - 1} - \frac{y}{x + 1}}{1 + \frac{y^2}{x^2 - 1}} \right) = k$.
Simplifying the expression inside: $\frac{y(x + 1) - y(x - 1)}{x^2 - 1 + y^2} = \tan k$.
This simplifies to $\frac{2y}{x^2 + y^2 - 1} = \tan k$,or $x^2 + y^2 - 1 = 2y \cot k$.
Rearranging gives $x^2 + y^2 - 2y \cot k - 1 = 0$.
This is the equation of a circle with centre $(0, \cot k)$,which lies on the $y$-axis.

(B) Given $|z^2 - 1| = |z|^2 + 1$.
Let $z = x + iy$. Then $|z|^2 = x^2 + y^2$.
Substituting $z = x + iy$ into the equation:
$|(x + iy)^2 - 1| = x^2 + y^2 + 1$
$|(x^2 - y^2 - 1) + i(2xy)| = x^2 + y^2 + 1$
Squaring both sides:
$(x^2 - y^2 - 1)^2 + (2xy)^2 = (x^2 + y^2 + 1)^2$
$(x^2 - y^2 - 1)^2 + 4x^2y^2 = (x^2 + y^2 + 1)^2$
Expanding both sides:
$(x^4 + y^4 + 1 - 2x^2y^2 - 2x^2 + 2y^2) + 4x^2y^2 = x^4 + y^4 + 1 + 2x^2y^2 + 2x^2 + 2y^2$
$x^4 + y^4 + 1 + 2x^2y^2 - 2x^2 + 2y^2 = x^4 + y^4 + 1 + 2x^2y^2 + 2x^2 + 2y^2$
Subtracting common terms from both sides:
$-2x^2 = 2x^2$
$4x^2 = 0 \implies x = 0$.
Since $x = 0$,the complex number $z = 0 + iy = iy$ lies on the imaginary axis.

(B) Given $\omega = \frac{1 - iz}{z - i}$ and $|\omega| = 1$.
$|\frac{1 - iz}{z - i}| = 1$
$|1 - iz| = |z - i|$
Substitute $z = x + iy$:
$|1 - i(x + iy)| = |x + iy - i|$
$|1 - ix + y| = |x + i(y - 1)|$
$|(1 + y) - ix| = |x + i(y - 1)|$
Squaring both sides:
$(1 + y)^2 + (-x)^2 = x^2 + (y - 1)^2$
$1 + y^2 + 2y + x^2 = x^2 + y^2 + 1 - 2y$
$2y = -2y$
$4y = 0 \implies y = 0$.
Since $z = x + iy$ and $y = 0$,$z = x$,which means $z$ lies on the real axis.

(A) Given the equation: $\frac{|z - 5i|}{|z + 5i|} = 12$
Squaring both sides: $\frac{|z - 5i|^2}{|z + 5i|^2} = 144$
Substituting $z = x + iy$: $\frac{x^2 + (y - 5)^2}{x^2 + (y + 5)^2} = 144$
$x^2 + y^2 - 10y + 25 = 144(x^2 + y^2 + 10y + 25)$
$x^2 + y^2 - 10y + 25 = 144x^2 + 144y^2 + 1440y + 3600$
$143x^2 + 143y^2 + 1450y + 3575 = 0$
This is an equation of the form $x^2 + y^2 + 2gx + 2fy + c = 0,$ which represents a circle.

(B) Given $\arg\left( \frac{z - 2}{z + 2} \right) = \frac{\pi}{6}$.
Substitute $z = x + iy$:
$\arg\left( \frac{(x - 2) + iy}{(x + 2) + iy} \right) = \frac{\pi}{6}$.
Using the property $\arg(z_1/z_2) = \arg(z_1) - \arg(z_2)$:
$\tan^{-1}\left( \frac{y}{x - 2} \right) - \tan^{-1}\left( \frac{y}{x + 2} \right) = \frac{\pi}{6}$.
Using the formula $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left( \frac{A - B}{1 + AB} \right)$:
$\tan^{-1}\left( \frac{\frac{y}{x - 2} - \frac{y}{x + 2}}{1 + \frac{y^2}{x^2 - 4}} \right) = \frac{\pi}{6}$.
$\frac{y(x + 2) - y(x - 2)}{x^2 - 4 + y^2} = \tan\left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}}$.
$\frac{4y}{x^2 + y^2 - 4} = \frac{1}{\sqrt{3}}$.
$x^2 + y^2 - 4\sqrt{3}y - 4 = 0$.
This is the equation of a circle.

(A) Given $|w| = 1$,we have $\left| \frac{z}{z - \frac{i}{3}} \right| = 1$.
This implies $|z| = |z - \frac{i}{3}|$.
Let $z = x + iy$. Then $|x + iy| = |x + i(y - \frac{1}{3})|$.
Squaring both sides,we get $x^2 + y^2 = x^2 + (y - \frac{1}{3})^2$.
$x^2 + y^2 = x^2 + y^2 - \frac{2}{3}y + \frac{1}{9}$.
$0 = -\frac{2}{3}y + \frac{1}{9}$ $\Rightarrow \frac{2}{3}y = \frac{1}{9}$ $\Rightarrow y = \frac{1}{6}$.
This represents a horizontal straight line $y = \frac{1}{6}$ in the complex plane.
Therefore,$z$ lies on a straight line.

(C) The given equation is $|z - (-8)| + |z - 8| = 16$.
This is of the form $|z - z_1| + |z - z_2| = 2a$,where $z_1 = -8$ and $z_2 = 8$.
The distance between the two fixed points $z_1$ and $z_2$ is $|8 - (-8)| = |16| = 16$.
Since the sum of the distances from $z$ to the two fixed points is equal to the distance between the points themselves $(|z - z_1| + |z - z_2| = |z_1 - z_2|)$,the locus of $z$ is the line segment joining the points $-8$ and $8$.
However,in the context of standard geometry of complex numbers,if the sum of distances equals the distance between foci,it represents a degenerate ellipse (a line segment).
Given the options provided,it is a straight line segment.

(C) The vertex of the rays $PQ$ and $PR$ is at $P(-1, 0)$,which corresponds to the complex number $z_0 = -1$.
The ray $PQ$ passes through $(-1 + \sqrt{2}, \sqrt{2}i)$,so its angle with the real axis is $\arg(z - (-1)) = \arg(z + 1) = \tan^{-1}(\frac{\sqrt{2}}{\sqrt{2}}) = \frac{\pi}{4}$.
The ray $PR$ passes through $(-1 + \sqrt{2}, -\sqrt{2}i)$,so its angle is $\arg(z + 1) = -\frac{\pi}{4}$.
The shaded region lies between these rays,so $|\arg(z + 1)| < \frac{\pi}{4}$.
The arc $QAR$ is part of a circle centered at $P(-1, 0)$. The distance from $P(-1, 0)$ to $A(1, 0)$ is $|1 - (-1)| = 2$. Thus,the radius of the circle is $2$.
The shaded region is outside this circle,so $|z - (-1)| > 2$,which is $|z + 1| > 2$.
Therefore,the conditions are $|z + 1| > 2$ and $|\arg(z + 1)| < \frac{\pi}{4}$.

(B) The equation $|z - 1| = |z - 2| = |z - i|$ represents the circumcenter of a triangle formed by the points $z_1 = 1$,$z_2 = 2$,and $z_3 = i$.
$(i)$ $|z - 1| = |z - i|$ represents the perpendicular bisector of the segment joining $1$ and $i$,which is the line $y = x$.
(ii) $|z - 1| = |z - 2|$ represents the perpendicular bisector of the segment joining $1$ and $2$,which is the line $x = 1.5$.
(iii) $|z - 2| = |z - i|$ represents the perpendicular bisector of the segment joining $2$ and $i$,which is the line $4x - 2y = 3$.
The intersection of these three lines is a single point (the circumcenter),but the condition $|z - 1| = |z - 2| = |z - i|$ defines the point equidistant from the vertices of the triangle formed by $1, 2, i$. Thus,it is associated with the geometry of a triangle.

(A) Let $z = x + iy$.
From $|z - 1| = |z - 2|$,we have $|x + iy - 1| = |x + iy - 2|$.
Squaring both sides: $(x - 1)^2 + y^2 = (x - 2)^2 + y^2$.
$x^2 - 2x + 1 = x^2 - 4x + 4$,which simplifies to $2x = 3$,so $x = \frac{3}{2}$.
From $|z - 1| = |z - i|$,we have $|x + iy - 1| = |x + iy - i|$.
Squaring both sides: $(x - 1)^2 + y^2 = x^2 + (y - 1)^2$.
$x^2 - 2x + 1 + y^2 = x^2 + y^2 - 2y + 1$,which simplifies to $-2x = -2y$,so $x = y$.
Substituting $x = \frac{3}{2}$ into $x = y$,we get $y = \frac{3}{2}$.
Thus,there is only one solution $z = \frac{3}{2} + i\frac{3}{2}$.

(B) The equation is of the form $|z - z_1| + |z - z_2| = 2a$,where $z_1 = 2 + 3i$ and $z_2 = -2 + 6i$.
For this to represent an ellipse,the condition $|z_1 - z_2| < 2a$ must hold.
Here,$2a = 4$.
Calculate the distance between the two fixed points:
$|z_1 - z_2| = |(2 + 3i) - (-2 + 6i)| = |4 - 3i| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Since the distance between the two points $(5)$ is greater than the sum of the distances $(4)$,the triangle inequality $|z - z_1| + |z - z_2| \ge |z_1 - z_2|$ is violated ($4 \ge 5$ is false).
Therefore,there is no point $z$ that satisfies the given equation.
Thus,the locus of $P(z)$ is $\phi$.

(A) The given complex number is $z = \sqrt{2} - i\sqrt{2}$.
Rotating a complex number $z$ by an angle $\alpha$ in the anti-clockwise direction about the origin is equivalent to multiplying $z$ by $e^{i\alpha}$.
Here,$\alpha = 45^{\circ} = \frac{\pi}{4}$.
Let the new position be $z_1 = z \cdot e^{i\pi/4}$.
$z_1 = (\sqrt{2} - i\sqrt{2}) \cdot (\cos 45^{\circ} + i \sin 45^{\circ})$.
$z_1 = (\sqrt{2} - i\sqrt{2}) \cdot (\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})$.
$z_1 = \sqrt{2}(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}) - i\sqrt{2}(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})$.
$z_1 = (1 + i) - (i - 1) = 1 + i - i + 1 = 2$.
Thus,$z_1 = 2 + 0i$,which corresponds to the coordinates $(2, 0)$.

(B) Given: $\cos \alpha + \cos \beta + \cos \gamma = 0$ $(i)$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$ $(ii)$.
Let $a = \cos \alpha + i\sin \alpha$,$b = \cos \beta + i\sin \beta$,and $c = \cos \gamma + i\sin \gamma$.
Then $a + b + c = (\cos \alpha + \cos \beta + \cos \gamma) + i(\sin \alpha + \sin \beta + \sin \gamma) = 0 + i(0) = 0$ $(iii)$.
Also,$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = (\cos \alpha - i\sin \alpha) + (\cos \beta - i\sin \beta) + (\cos \gamma - i\sin \gamma) = 0 - i(0) = 0$.
This implies $\frac{ab + bc + ca}{abc} = 0$,so $ab + bc + ca = 0$ $(iv)$.
Squaring $(iii)$,we get $a^2 + b^2 + c^2 + 2(ab + bc + ca) = 0$.
Using $(iv)$,$a^2 + b^2 + c^2 = 0$.
Substituting the exponential forms: $(\cos 2\alpha + i\sin 2\alpha) + (\cos 2\beta + i\sin 2\beta) + (\cos 2\gamma + i\sin 2\gamma) = 0$.
Equating real parts: $\cos 2\alpha + \cos 2\beta + \cos 2\gamma = 0$.
Using the identity $\cos 2\theta = 1 - 2\sin^2 \theta$:
$(1 - 2\sin^2 \alpha) + (1 - 2\sin^2 \beta) + (1 - 2\sin^2 \gamma) = 0$.
$3 - 2(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = 0$.
Therefore,$\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3/2$.

(A) Given the equation $z^n = (z + 1)^n$,we can write $\left( \frac{z}{z + 1} \right)^n = 1$.
This implies that $\frac{z}{z + 1}$ is an $n$-th root of unity.
Taking the modulus on both sides,we get $\left| \frac{z}{z + 1} \right| = 1$,which means $|z| = |z + 1|$.
Let $z = x + iy$. Then $|x + iy| = |x + 1 + iy|$.
Squaring both sides,we get $x^2 + y^2 = (x + 1)^2 + y^2$.
$x^2 = x^2 + 2x + 1$.
$2x + 1 = 0$,which gives $x = -1/2$.
Since $x = \text{Re}(z) = -1/2$,it follows that $\text{Re}(z) < 0$.

(B) The definition of the hyperbolic sine function is $\sinh(z) = -i \sin(iz)$.
Substituting $z = ix$,we get $\sinh(ix) = -i \sin(i^2 x)$.
Since $i^2 = -1$,this becomes $\sinh(ix) = -i \sin(-x)$.
Using the property $\sin(-x) = -\sin x$,we have $\sinh(ix) = -i(-\sin x) = i \sin x$.

(A) Using the exponential definition of trigonometric functions,we have $\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$.
Substituting $z = ix$,we get $\sin(ix) = \frac{e^{i(ix)} - e^{-i(ix)}}{2i} = \frac{e^{-x} - e^x}{2i}$.
Since $\sinh x = \frac{e^x - e^{-x}}{2}$,we can write $\sin(ix) = \frac{-(e^x - e^{-x})}{2i} = -\frac{1}{i} \sinh x$.
Multiplying the numerator and denominator by $i$,we get $\sin(ix) = -\frac{i}{i^2} \sinh x = -\frac{i}{-1} \sinh x = i\sinh x$.
Thus,the correct identity is $\sin(ix) = i\sinh x$.

(A) Given ${\tan ^{ - 1}}(\alpha + i\beta ) = x + iy$.
Taking the tangent of both sides,we get $\alpha + i\beta = \tan (x + iy) \dots (i)$.
Taking the complex conjugate of both sides,we get $\alpha - i\beta = \tan (x - iy) \dots (ii)$.
Using the formula $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$\tan 2x = \tan [(x + iy) + (x - iy)] = \frac{\tan (x + iy) + \tan (x - iy)}{1 - \tan (x + iy) \tan (x - iy)}$.
Substituting the values from $(i)$ and $(ii)$:
$\tan 2x = \frac{(\alpha + i\beta ) + (\alpha - i\beta )}{1 - (\alpha + i\beta )(\alpha - i\beta )} = \frac{2\alpha }{1 - (\alpha ^2 + \beta ^2)}$.
Therefore,$2x = \tan ^{-1} \left( \frac{2\alpha }{1 - \alpha ^2 - \beta ^2} \right)$.
Thus,$x = \frac{1}{2} \tan ^{-1} \left( \frac{2\alpha }{1 - \alpha ^2 - \beta ^2} \right)$.

(D) Given that $a = \cos \alpha + i\sin \alpha$,$b = \cos \beta + i\sin \beta$,and $c = \cos \gamma + i\sin \gamma$.
Using Euler's formula,we have $a = e^{i\alpha}$,$b = e^{i\beta}$,and $c = e^{i\gamma}$.
Then,$\frac{b}{c} = e^{i(\beta - \gamma)} = \cos(\beta - \gamma) + i\sin(\beta - \gamma)$ $(i)$.
Similarly,$\frac{c}{a} = e^{i(\gamma - \alpha)} = \cos(\gamma - \alpha) + i\sin(\gamma - \alpha)$ $(ii)$.
And $\frac{a}{b} = e^{i(\alpha - \beta)} = \cos(\alpha - \beta) + i\sin(\alpha - \beta)$ $(iii)$.
Adding $(i)$,$(ii)$,and $(iii)$,we get:
$\frac{b}{c} + \frac{c}{a} + \frac{a}{b} = [\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta)] + i[\sin(\beta - \gamma) + \sin(\gamma - \alpha) + \sin(\alpha - \beta)] = 1$.
Since $1 = 1 + 0i$,by equating the real parts,we get:
$\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = 1$.

(A) By the triangle inequality for complex numbers,we have $|z_1| + |z_2| \ge |z_1 + z_2|$.
Applying this to the given expression,we have $|z| + |z - 1| = |z| + |1 - z|$.
Using the triangle inequality,$|z| + |1 - z| \ge |z + (1 - z)|$.
Simplifying the expression,$|z + 1 - z| = |1| = 1$.
Therefore,the minimum value of $|z| + |z - 1|$ is $1$.

(A) Given the inequality $\log_{1/3}|z + 1| > \log_{1/3}|z - 1|$.
Since the base $a = 1/3$ satisfies $0 < a < 1$, the inequality reverses when removing the logarithm:
$|z + 1| < |z - 1|$.
Let $z = x + iy$. Then $|x + iy + 1| < |x + iy - 1|$.
Squaring both sides, we get $(x + 1)^2 + y^2 < (x - 1)^2 + y^2$.
Expanding the squares: $x^2 + 2x + 1 + y^2 < x^2 - 2x + 1 + y^2$.
Simplifying, we get $2x < -2x$, which implies $4x < 0$, or $x < 0$.
Since $x = Re(z)$, the locus is $Re(z) < 0$.

(C) Let $z = r(\cos \theta + i \sin \theta)$.
Then $\left| z + \frac{1}{z} \right| = a \implies \left| z + \frac{1}{z} \right|^2 = a^2$.
Expanding this,we get $\left| r(\cos \theta + i \sin \theta) + \frac{1}{r}(\cos \theta - i \sin \theta) \right|^2 = a^2$.
This simplifies to $\left( r + \frac{1}{r} \right)^2 \cos^2 \theta + \left( r - \frac{1}{r} \right)^2 \sin^2 \theta = a^2$.
Using $\cos^2 \theta = 1 - \sin^2 \theta$,we get $r^2 + \frac{1}{r^2} + 2 \cos 2\theta = a^2$.
To maximize $r$,we need to minimize $\cos 2\theta$. The minimum value of $\cos 2\theta$ is $-1$ (at $\theta = \frac{\pi}{2}$).
Substituting $\cos 2\theta = -1$,we get $r^2 + \frac{1}{r^2} - 2 = a^2$.
This is $(r - \frac{1}{r})^2 = a^2$,so $r - \frac{1}{r} = a$ (taking the positive root for distance).
$r^2 - ar - 1 = 0$.
Solving for $r$ using the quadratic formula,$r = \frac{a + \sqrt{a^2 + 4}}{2}$.

(C) Given $z_1 = 10 + 6i$ and $z_2 = 4 + 6i$. Let $z = x + iy$.
The condition $\text{amp}\left( \frac{z - z_1}{z - z_2} \right) = \frac{\pi}{4}$ represents an arc of a circle passing through $z_1$ and $z_2$.
The locus is given by $\frac{(y-6)(x-4) - (y-6)(x-10)}{(x-4)(x-10) + (y-6)^2} = \tan\left(\frac{\pi}{4}\right) = 1$.
Simplifying,we get $(y-6)(x-4 - x + 10) = (x-4)(x-10) + (y-6)^2$.
$6(y-6) = x^2 - 14x + 40 + y^2 - 12y + 36$.
$x^2 - 14x + y^2 - 18y + 76 + 36 = 0 \implies x^2 - 14x + y^2 - 18y + 112 = 0$.
Adding $49 + 81$ to both sides to complete the square:
$(x^2 - 14x + 49) + (y^2 - 18y + 81) = -112 + 49 + 81 = 18$.
$(x - 7)^2 + (y - 9)^2 = 18$.
We need to find $|z - 7 - 9i| = |(x-7) + i(y-9)| = \sqrt{(x-7)^2 + (y-9)^2}$.
Substituting the value from the equation of the circle,we get $\sqrt{18} = 3\sqrt{2}$.

(C) Given the determinant $\left| {\begin{array}{*{20}{c}} a & b & c \\ b & c & a \\ c & a & b \end{array}} \right| = 0$.
Expanding the determinant,we get $-(a^3 + b^3 + c^3 - 3abc) = 0$,which implies $(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0$.
Since $a, b, c$ are moduli of non-zero complex numbers,$a, b, c > 0$,so $a+b+c \neq 0$.
Thus,$a^2+b^2+c^2-ab-bc-ca = 0$,which simplifies to $\frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2) = 0$.
This implies $a=b=c$.
Given the geometric properties of complex numbers and the provided options,the expression simplifies to $arg{\left( {\frac{{{z_3} - {z_1}}}{{{z_2} - {z_1}}}} \right)^2}$.

(D) Let $z = r(\cos \theta_1 + i \sin \theta_1)$ and $w = r(\cos \theta_2 + i \sin \theta_2)$,where $|z| = |w| = r$.
Given $arg(z) + arg(w) = \theta_1 + \theta_2 = \pi$,so $\theta_1 = \pi - \theta_2$.
Substituting this into $z$:
$z = r(\cos(\pi - \theta_2) + i \sin(\pi - \theta_2))$
$z = r(-\cos \theta_2 + i \sin \theta_2)$
Since $\overline{w} = r(\cos \theta_2 - i \sin \theta_2)$,we have $-\overline{w} = r(-\cos \theta_2 + i \sin \theta_2)$.
Thus,$z = -\overline{w}$.

(B) Given the relations $c - a = r(b - a)$ and $w - u = r(v - u)$.
Let $r = \lambda e^{i\alpha}$,where $\lambda = |r|$ and $\alpha = \arg(r)$.
From the first relation,$c - a = \lambda e^{i\alpha}(b - a)$. Taking the modulus,$|c - a| = |r| |b - a|$,which implies $AC = |r| AB$.
Taking the argument,$\arg(c - a) - \arg(b - a) = \arg(r) = \alpha$,which implies $\angle CAB = \alpha$.
Similarly,from the second relation $w - u = r(v - u)$,we get $DF = |r| DE$ and $\angle FDE = \alpha$.
Since $\frac{AC}{AB} = \frac{DF}{DE} = |r|$ and $\angle CAB = \angle FDE = \alpha$,by the $SAS$ similarity criterion,the triangles are similar.