Let $z \in \mathbb{C}$. We denote $C_z$ (respectively $\Omega_z$) the set of points in the plane with complex affixe $Z$ such that $|Z(Z-2z)| = 1$ (respectively $|Z(Z-2z)| < 1$). In this question we assume that $z$ is a real number denoted $a$. We work in the orthonormal frame $\mathcal{R}'$ with center $O'$ with affixe $a$, obtained from $\mathcal{R}$ by translation. Show that an equation of the curve $C_a$ in ``polar coordinates $(\rho, \theta)$'' in the frame $\mathcal{R}'$ is $$\left(\rho^2 + a^2\right)^2 - 4a^2 \rho^2 \cos^2\theta = 1$$

Let $z \in \mathbb{C}$. We denote $\Omega_z$ the set of points in the plane with complex affixe $Z$ such that $|Z(Z-2z)| < 1$. Justify that $\Omega_z$ is a bounded subset of the plane. Is it open? closed? compact?

Let $z \in \mathbb{C}$. We denote $\Omega_z$ the set of points in the plane with complex affixe $Z$ such that $|Z(Z-2z)| < 1$. Justify that the origin $O$ is an interior point of $\Omega_z$.

Let $z \in \mathbb{C}$. Determine the domain of definition $D_z$ of the function $Z \mapsto \dfrac{1}{1 - 2zZ + Z^2}$.

Let $w \in \mathbb{C}$ be a number that is neither real nor purely imaginary.
(a) Show that the equation $$\left|\frac{z-1}{z+1}\right| = \left|\frac{w-1}{w+1}\right|$$ defines a circle in the complex plane, which passes through $w$. Verify that the interval $]-1,1[$ intersects this circle at a unique point; we denote this point by $y$. We will express $y$ in terms of the number $$\lambda = \left|\frac{w-1}{w+1}\right|.$$
(b) Show the inequality $$\left|\frac{1-w}{1-y}\right| > 1.$$
(c) Show that the equation $$\left|\frac{z-w}{z-y}\right| = \left|\frac{1-w}{1-y}\right|$$ defines a circle in the complex plane, which passes through $1$ and through $-1$.
Deduce that, for all $x \in [-1,1] \setminus \{y\}$, we have $$\left|\frac{w-x}{y-x}\right| \geqslant \left|\frac{w-1}{y-1}\right| = \left|\frac{w+1}{y+1}\right|$$

Consider the following two subsets of $\mathbb { C }$ : $$A = \left\{ \frac { 1 } { z } : | z | = 2 \right\} \text { and } B = \left\{ \frac { 1 } { z } : | z - 1 | = 2 \right\} .$$ Then
(A) $A$ is a circle, but $B$ is not a circle.
(B) $B$ is a circle, but $A$ is not a circle.
(C) $A$ and $B$ are both circles.
(D) Neither $A$ nor $B$ is a circle.

Let complex numbers $\alpha$ and $\frac { 1 } { \bar { \alpha } }$ lie on circles $\left( x - x _ { 0 } \right) ^ { 2 } + \left( y - y _ { 0 } \right) ^ { 2 } = r ^ { 2 }$ and $\left( x - x _ { 0 } \right) ^ { 2 } + \left( y - y _ { 0 } \right) ^ { 2 } = 4 r ^ { 2 }$, respectively. If $z _ { 0 } = x _ { 0 } + i y _ { 0 }$ satisfies the equation $2 \left| z _ { 0 } \right| ^ { 2 } = r ^ { 2 } + 2$, then $| \alpha | =$
(A) $\frac { 1 } { \sqrt { 2 } }$
(B) $\frac { 1 } { 2 }$
(C) $\frac { 1 } { \sqrt { 7 } }$
(D) $\frac { 1 } { 3 }$

Let $a , b \in \mathbb { R }$ and $a ^ { 2 } + b ^ { 2 } \neq 0$. Suppose $S = \left\{ z \in \mathbb { C } : z = \frac { 1 } { a + i b t } , t \in \mathbb { R } , t \neq 0 \right\}$, where $i = \sqrt { - 1 }$.
If $z = x + i y$ and $z \in S$, then $( x , y )$ lies on
(A) the circle with radius $\frac { 1 } { 2 a }$ and centre $\left( \frac { 1 } { 2 a } , 0 \right)$ for $a > 0 , b \neq 0$
(B) the circle with radius $- \frac { 1 } { 2 a }$ and centre $\left( - \frac { 1 } { 2 a } , 0 \right)$ for $a < 0 , b \neq 0$
(C) the $x$-axis for $a \neq 0 , b = 0$
(D) the $y$-axis for $a = 0 , b \neq 0$

A complex number $z$ is said to be unimodular if $|z| = 1$. Suppose $z_1$ and $z_2$ are complex numbers such that $\frac{z_1 - 2z_2}{2 - z_1\bar{z}_2}$ is unimodular and $z_2$ is not unimodular. Then the point $z_1$ lies on a:
(1) straight line parallel to $x$-axis
(2) straight line parallel to $y$-axis
(3) circle of radius 2
(4) circle of radius $\sqrt{2}$

A complex number $z$ is said to be unimodular if $| z | = 1$. Let $z _ { 1 }$ and $z _ { 2 }$ are complex numbers such that $\frac { z _ { 1 } - 2 z _ { 2 } } { 2 - z _ { 1 } \bar { z } _ { 2 } }$ is unimodular and $z _ { 2 }$ is not unimodular, then the point $z _ { 1 }$ lies on a
(1) circle of radius $\sqrt { 2 }$
(2) straight line parallel to $x$-axis
(3) straight line parallel to $y$-axis
(4) circle of radius 2

Let $z \in C$, the set of complex numbers. Then the equation, $2 | z + 3 i | - | z - i | = 0$ represents:
(1) A circle with radius $\frac { 8 } { 3 }$
(2) An ellipse with length of minor axis $\frac { 16 } { 9 }$
(3) An ellipse with length of major axis $\frac { 16 } { 3 }$
(4) A circle with diameter $\frac { 10 } { 3 }$

All the points in the set $S = \left\{ \frac { \alpha + i } { \alpha - i } , \alpha \in R \right\} , i = \sqrt { - 1 }$ lie on a
(1) straight line whose slope is - 1
(2) circle whose radius is $\sqrt { 2 }$
(3) circle whose radius is 1
(4) straight line whose slope is 1

If $\operatorname { Re } \left( \frac { z - 1 } { 2 z + i } \right) = 1$, where $z = x + i y$, then the point $(x, y)$ lies on a
(1) circle whose centre is at $\left( - \frac { 1 } { 2 } , - \frac { 3 } { 2 } \right)$
(2) straight line whose slope is $- \frac { 2 } { 3 }$
(3) straight line whose slope is $\frac { 3 } { 2 }$
(4) circle whose diameter is $\frac { \sqrt { 5 } } { 2 }$

Let $u = \frac { 2 z + i } { z - k i } , z = x + i y$ and $k > 0$. If the curve represented by Re $( u ) + \operatorname { Im } ( u ) = 1$ intersects the $y$-axis at points P and Q where $\mathrm { PQ } = 5$ then the value of k is
(1) $\frac { 3 } { 2 }$
(2) $\frac { 1 } { 2 }$
(3) 4
(4) 2

If the equation $a | z | ^ { 2 } + \overline { \bar { \alpha } z + \alpha \bar { z } } + d = 0$ represents a circle where $a , d$ are real constants then which of the following condition is correct?
(1) $| \alpha | ^ { 2 } - a d \neq 0$
(2) $| \alpha | ^ { 2 } - a d > 0$ and $a \in R - \{ 0 \}$
(3) $| \alpha | ^ { 2 } - a d \geq 0$ and $a \in R$
(4) $\alpha = 0 , a , d \in R ^ { + }$

Let a circle $C$ in complex plane pass through the points $z _ { 1 } = 3 + 4 i , z _ { 2 } = 4 + 3 i$ and $z _ { 3 } = 5 i$. If $z \neq z _ { 1 }$ is a point on $C$ such that the line through $z$ and $z _ { 1 }$ is perpendicular to the line through $z _ { 2 }$ and $z _ { 3 }$, then $\arg z$ is equal to
(1) $\tan ^ { - 1 } \frac { 24 } { 7 } - \pi$
(2) $\tan ^ { - 1 } \frac { 2 } { \sqrt { 5 } } - \pi$
(3) $\tan ^ { - 1 } 3 - \pi$
(4) $\tan ^ { - 1 } \frac { 3 } { 4 } - \pi$

For $\alpha, \beta, z \in \mathbb{C}$ and $\lambda > 1$, if $\sqrt{\lambda - 1}$ is the radius of the circle $|z - \alpha|^{2} + |z - \beta|^{2} = 2\lambda$, then $|\alpha - \beta|$ is equal to $\_\_\_\_$.

If the center and radius of the circle $\left|\frac{z - 2}{z - 3}\right| = 2$ are respectively $(\alpha, \beta)$ and $\gamma$, then $3\alpha + \beta + \gamma$ is equal to
(1) 11
(2) 9
(3) 10
(4) 12

Let z be a complex number such that $\left| \frac { z - 2 i } { z + i } \right| = 2 , z \neq - i$. Then $z$ lies on the circle of radius 2 and centre
(1) $( 2,0 )$
(2) $( 0,2 )$
(3) $( 0,0 )$
(4) $( 0 , - 2 )$

Let $z$ be a complex number such that $| z + 2 | = 1$ and $\operatorname { Im } \left( \frac { z + 1 } { z + 2 } \right) = \frac { 1 } { 5 }$. Then the value of $| \operatorname { Re } ( \overline { z + 2 } ) |$ is
(1) $\frac { 2 \sqrt { 6 } } { 5 }$
(2) $\frac { 24 } { 5 }$
(3) $\frac { 1 + \sqrt { 6 } } { 5 }$
(4) $\frac { \sqrt { 6 } } { 5 }$

If $z _ { 1 } , z _ { 2 }$ are two distinct complex number such that $\left| \frac { z _ { 1 } - 2 z _ { 2 } } { \frac { 1 } { 2 } - z _ { 1 } \bar { z } _ { 2 } } \right| = 2$, then
(1) $z _ { 1 }$ lies on a circle of radius $\frac { 1 } { 2 }$ and $z _ { 2 }$ lies on a circle of radius 1.
(2) both $z _ { 1 }$ and $z _ { 2 }$ lie on the same circle.
(3) either $z _ { 1 }$ lies on a circle of radius $\frac { 1 } { 2 }$ or $z _ { 2 }$ lies on a circle of radius 1.
(4) either $z _ { 1 }$ lies on a circle of radius 1 or $z _ { 2 }$ lies on a circle of radius $\frac { 1 } { 2 }$.

Let $\left| \frac { \bar { z } - i } { 2 \bar { z } + i } \right| = \frac { 1 } { 3 } , z \in C$, be the equation of a circle with center at $C$. If the area of the triangle, whose vertices are at the points $( 0,0 ) , \mathrm { C }$ and $( \alpha , 0 )$ is 11 square units, then $\alpha ^ { 2 }$ equals:
(1) 50
(2) 100
(3) $\frac { 81 } { 25 }$
(4) $\frac { 121 } { 25 }$