Exercise 4 (3 points)
In the complex plane equipped with a direct orthonormal coordinate system $(\mathrm{O}; \vec{u}, \vec{v})$, we consider the points A and B with complex numbers respectively $z_{\mathrm{A}} = 2\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}$ and $z_{\mathrm{B}} = 2\mathrm{e}^{\mathrm{i}\frac{3\pi}{4}}$.

1. Show that OAB is a right isosceles triangle.
2. We consider the equation $$(E): z^2 - \sqrt{6}\, z + 2 = 0$$ Show that one of the solutions of $(E)$ is the complex number of a point located on the circumscribed circle of triangle OAB.

Let $z_1$ be a purely imaginary complex number and $z_2$ be any complex number such that $|z_1 + z_2| = |z_1 - z_2|$. Find the circumcentre of the triangle with vertices $0, z_1, z_2$.
(A) $z_1/2$ (B) $z_2/2$ (C) $(z_1 + z_2)/2$ (D) $(z_1 - z_2)/2$

Let $\alpha , \beta , \gamma$ be complex numbers which are the vertices of an equilateral triangle. Then, we must have:
(A) $\alpha + \beta + \gamma = 0$
(B) $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } = 0$
(C) $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \alpha \beta + \beta \gamma + \gamma \alpha = 0$
(D) $( \alpha - \beta ) ^ { 2 } + ( \beta - \gamma ) ^ { 2 } + ( \gamma - \alpha ) ^ { 2 } = 0$

The area of the triangle whose vertices are complex numbers $z, iz, z+iz$ in the Argand diagram is
(1) $2|z|^{2}$
(2) $\frac{1}{2}|z|^{2}$
(3) $4|z|^{2}$
(4) $|z|^{2}$

If the four complex numbers $z , \bar { z } , \bar { z } - 2 \operatorname { Re } ( \bar { z } )$ and $z - 2 \operatorname { Re } ( z )$ represent the vertices of a square of side 4 units in the Argand plane, then $| z |$ is equal to :
(1) $4 \sqrt { 2 }$
(2) 4
(3) $2 \sqrt { 2 }$
(4) 2

Let a complex number be $w = 1 - \sqrt { 3 } i$. Let another complex number $z$ be such that $| z w | = 1$ and $\arg ( z ) - \arg ( w ) = \frac { \pi } { 2 }$. Then the area of the triangle (in sq. units) with vertices origin, $z$ and $w$ is equal to
(1) 4
(2) $\frac { 1 } { 2 }$
(3) $\frac { 1 } { 4 }$
(4) 2

Let $O$ be the origin and $A$ be the point $z _ { 1 } = 1 + 2i$. If $B$ is the point $z _ { 2 } , \operatorname { Re } \left( z _ { 2 } \right) < 0$, such that $OAB$ is a right angled isosceles triangle with $OB$ as hypotenuse, then which of the following is NOT true?
(1) $\arg z _ { 2 } = \pi - \tan ^ { - 1 } 3$
(2) $\arg \left( z _ { 1 } - 2 z _ { 2 } \right) = - \tan ^ { - 1 } \frac { 4 } { 3 }$
(3) $\left| z _ { 2 } \right| = \sqrt { 10 }$
(4) $\left| 2 z _ { 1 } - z _ { 2 } \right| = 5$

Let $O$ be the origin, the point $A$ be $z _ { 1 } = \sqrt { 3 } + 2 \sqrt { 2 } i$, the point $B \left( z _ { 2 } \right)$ be such that $\sqrt { 3 } \left| z _ { 2 } \right| = \left| z _ { 1 } \right|$ and $\arg \left( z _ { 2 } \right) = \arg \left( z _ { 1 } \right) + \frac { \pi } { 6 }$. Then
(1) area of triangle ABO is $\frac { 11 } { \sqrt { 3 } }$
(2) ABO is an obtuse angled isosceles triangle
(3) area of triangle ABO is $\frac { 11 } { 4 }$
(4) ABO is a scalene triangle

Let $z$ be a complex number satisfying $| z | = 2$. In the complex number plane with the origin O, let A and B be the points representing $1 + z$ and $1 - \frac { 1 } { 2 } z$, respectively.
First of all, we can express the complex number $z$ as
$$z = \mathbf { M } ( \cos \theta + i \sin \theta ) \quad ( - \pi \leqq \theta < \pi ) .$$
(1) If $z$ is not a real number, then the area $S$ of the triangle OAB is $S = \mathbf { N }$. For $\mathbf{N}$, choose the correct answer from among (0) $\sim$ (8) below.
Hence, when $\theta = \pm \frac { \mathbf { O } } { \mathbf { P } } \pi$, $S$ is maximized.
(0) $\frac { 1 } { 2 } \left| \sin \left( \theta + \frac { 1 } { 3 } \pi \right) \right|$ (1) $\frac { 1 } { 2 } | \sin \theta |$ (2) $\frac { 1 } { 2 } \left| \sin \left( \theta - \frac { 1 } { 3 } \pi \right) \right|$ (3) $\left| \sin \left( \theta + \frac { 1 } { 3 } \pi \right) \right|$ (4) $| \sin \theta |$ (5) $\left| \sin \left( \theta - \frac { 1 } { 3 } \pi \right) \right|$ (6) $\frac { 3 } { 2 } \left| \sin \left( \theta + \frac { 1 } { 3 } \pi \right) \right|$ (7) $\frac { 3 } { 2 } | \sin \theta |$ (8) $\frac { 3 } { 2 } \left| \sin \left( \theta - \frac { 1 } { 3 } \pi \right) \right|$
(2) When the triangle OAB is an isosceles triangle where $\mathrm { OA } = \mathrm { OB }$, we see that
$$| 1 + z | = \left| 1 - \frac { 1 } { 2 } z \right| = \sqrt { \mathbf { Q } }$$
and
$$\arg ( 1 + z ) = \pm \frac { \mathbf { R } } { \mathbf { S } } \pi , \quad \arg \left( 1 - \frac { 1 } { 2 } z \right) = \mp \frac { \mathbf { T } } { \mathbf{U} } \pi ,$$
where the right-hand sides of the equations are of opposite signs, and where $- \pi \leqq \arg ( 1 + z ) < \pi$ and $- \pi \leqq \arg \left( 1 - \frac { 1 } { 2 } z \right) < \pi$.

In the complex plane, let $O$ be the origin, and let $A$ and $B$ represent points with coordinates corresponding to complex numbers $z$ and $z + 1$ respectively. Given that both points $A$ and $B$ lie on the unit circle centered at $O$, select the correct options.
(1) Line $AB$ is parallel to the real axis
(2) $\triangle OAB$ is a right triangle
(3) Point $A$ is in the second quadrant
(4) $z^{3} = 1$
(5) The point with coordinate $1 + \frac{1}{z}$ also lies on the same unit circle