The imaginary part of the complex number $\frac { 1 } { 1 - 3 \mathrm { i } }$ is
A. $- \frac { 3 } { 10 }$
B. $- \frac { 1 } { 10 }$
C. $\frac { 1 } { 10 }$
D. $\frac { 3 } { 10 }$

Given that the complex number $z$ satisfies $z = 1 - 2 i$ ($i$ is the imaginary unit), find $| z | =$ $\_\_\_\_$

Let complex numbers $z _ { 1 } , z _ { 2 }$ satisfy $\left| z _ { 1 } \right| = \left| z _ { 2 } \right| = 2 , z _ { 1 } + z _ { 2 } = \sqrt { 3 } + \mathrm { i }$ , then $\left| z _ { 1 } - z _ { 2 } \right| = $ $\_\_\_\_$.

1. The point corresponding to the complex number $\frac { 2 - \mathrm { i } } { 1 - 3 \mathrm { i } }$ in the complex plane is located in which quadrant?
A. First quadrant
B. Second quadrant
C. Third quadrant
D. Fourth quadrant 【Answer】A 【Solution】 【Analysis】Use complex division to simplify $\frac { 2 - \mathrm { i } } { 1 - 3 \mathrm { i } }$, and thus determine the location of the corresponding point. 【Detailed Solution】 $\frac { 2 - \mathrm { i } } { 1 - 3 \mathrm { i } } = \frac { ( 2 - \mathrm { i } ) ( 1 + 3 \mathrm { i } ) } { 10 } = \frac { 5 + 5 \mathrm { i } } { 10 } = \frac { 1 + \mathrm { i } } { 2 }$, so the point corresponding to this complex number is $\left( \frac { 1 } { 2 } , \frac { 1 } { 2 } \right)$, which is located in the first quadrant. Therefore, the answer is: A.

3. Given $( 1 - i ) ^ { 2 } z = 3 + 2 i$, then $z =$
A. $- 1 - \frac { 3 } { 2 } i$
B. $- 1 + \frac { 3 } { 2 } i$
C. $- \frac { 3 } { 2 } + i$
D. $- \frac { 3 } { 2 } - i$

3. Given $(1 - i)^2 z = 3 + 2i$, then $z =$
A. $-1 - \frac{3}{2}i$
B. $-1 + \frac{3}{2}i$
C. $-\frac{3}{2} + i$
D. $-\frac{3}{2} - i$

If $z = - 1 + \sqrt { 3 } \mathrm { i }$ , then $\frac { z } { z \bar { z } - 1 } =$
A. $- 1 + \sqrt { 3 } \mathrm { i }$
B. $- 1 - \sqrt { 3 } \mathrm { i }$
C. $- \frac { 1 } { 3 } + \frac { \sqrt { 3 } } { 3 } \mathrm { i }$
D. $- \frac { 1 } { 3 } - \frac { \sqrt { 3 } } { 3 } \mathrm { i }$

The point corresponding to the complex number $\frac { 2 - \mathrm { i } } { 1 - 3 \mathrm { i } }$ in the complex plane is located in which quadrant?
A. First quadrant
B. Second quadrant
C. Third quadrant
D. Fourth quadrant

Let $( 1 + 2 \mathrm { i } ) a + b = 2 \mathrm { i }$ , where $a , b$ are real numbers, then
A. $a = 1 , b = - 1$
B. $a = 1 , b = 1$
C. $a = - 1 , b = 1$
D. $a = - 1 , b = - 1$

Given $z = 1 - 2i$, and $z + a\bar{z} + b = 0$, where $a, b$ are real numbers, then
A. $a = 1, b = -2$
B. $a = -1, b = 2$
C. $a = 1, b = 2$
D. $a = -1, b = -2$

2. If $\mathrm { i } ( 1 - z ) = 1$, then $z + \bar { z } =$
A. $- 2$
B. $- 1$
C. $1$
D. $2$

Let $z = \frac { 2 + i } { 1 + i ^ { 2 } + i ^ { 5 } }$, then $\bar { z } =$
A. $1 - 2 i$
B. $1 + 2 i$
C. $2 - i$
D. $2 + i$

In the complex plane, the point corresponding to $(1+3i)(3-i)$ is located in
A. the first quadrant
B. the second quadrant
C. the third quadrant

If the complex number $( a + i )( 1 - ai ) = 2$ , then $a =$
A. $-1$
B. $0$
C. $1$
D. $2$

Given $z = - 1 - \mathrm { i }$, then $| z | =$
A. 0
B. 1
C. $\sqrt { 2 }$
D. 2

Given $\frac { Z } { \mathrm { i } } = \mathrm { i } - 1$, then $Z =$ \_\_\_\_

If $\frac { z } { z - 1 } = 1 + \mathrm { i }$ , then $z =$
A. $- 1 - \mathrm { i }$
B. $- 1 + \mathrm { i }$
C. $1 - \mathrm{i}$
D. $1 + \mathrm { i }$

The imaginary part of $(1 + 5\mathrm{i})\mathrm{i}$ is
A. $-1$
B. $0$
C. $1$
D. $6$

The imaginary part of $(1 + 5i)i$ is
A. $-1$
B. $0$
C. $1$
D. $6$

Given $z = 1 + \mathrm{i}$, then $\frac{1}{z-1} = $ ( )
A. $-i$
B. $i$
C. $-1$
D. $1$

Let $z \in \mathbb{C}$. We set $z = a + ib$, where $a, b \in \mathbb{R}$.
Let $n \in \mathbb{N}^*$. Determine the modulus and an argument of $\left(1 + \frac{z}{n}\right)^n$ as a function of $a$, $b$ and $n$.

Let $z \in \mathbb{C}$. We set $z = a + ib$, where $a, b \in \mathbb{R}$.
Deduce that $$\lim_{n \rightarrow \infty} \left(1 + \frac{z}{n}\right)^n = e^z$$

Throughout the rest of the problem, we fix an odd integer $\ell \geq 3$ and $q$ a primitive $\ell$-th root of unity. Show that $q^2$ is a primitive $\ell$-th root of unity.

Let $z$ be a complex number, with real part $x$ and imaginary part $y$, such that $(x,y) \notin \mathbb{R}^{-} \times \{0\}$. We denote $$\theta(z) = 2\arctan\left(\frac{y}{x + \sqrt{x^2 + y^2}}\right) \quad \text{and} \quad R(z) = \frac{z + |z|}{\sqrt{2(\operatorname{Re}(z) + |z|)}}$$ Justify that $\theta$ and $R$ are well defined.

Let $z$ be a complex number, with real part $x$ and imaginary part $y$, such that $(x,y) \notin \mathbb{R}^{-} \times \{0\}$. We denote $$\theta(z) = 2\arctan\left(\frac{y}{x + \sqrt{x^2 + y^2}}\right) \quad \text{and} \quad R(z) = \frac{z + |z|}{\sqrt{2(\operatorname{Re}(z) + |z|)}}$$ When $z$ takes successively the values $z_1 = 4$, $z_2 = 2\mathrm{i}$ and $z_3 = 1 - \mathrm{i}\sqrt{3}$, calculate $R(z)$, $\theta(z)$ and $(R(z))^2$.