Sachchidanand Prasad

28-01-2025

Problem: Draw the following set in the complex plane.
  1. $A = \left\{ z \in \mathbb{C} : \text{ Re } z \leq 1 \right\} $
  2. $B = \left\{ z \in \mathbb{C} : \text{ Re} \left( z^2 \right) = 2 \right\} $
  3. $C = \left\{ z \in \mathbb{C} : \text{ Im} \left( \frac{1}{z} \right) = \frac{1}{4} \right\} $
  4. $D = \left\{ z \in \mathbb{C} : \vert z \vert = \text{ Re } z + 1 \right\} $
  5. $E = \left\{ z \in \mathbb{C} : \vert z - 2 \vert = \vert 1 - 2 \bar{z} \vert \right\} $
complex-analysis
Solution: We recall that for a complex number $z = x + \iota y$, the real part is denoted by $\text{Re}(z) = x$ and the imaginary part is denoted by $\text{Im}(z) = y$. The modulus of a complex number is given by $\vert z \vert = \sqrt{x^2 + y^2}$ and the complex conjugate is denoted by $\bar{z} = x - \iota y$.
  1. The given set is $A = \left\{ z \in \mathbb{C} : \text{ Re } z \leq 1 \right\} $ which is the set of all complex numbers whose real part is less than or equal to $1$ that is \[ A = \left\{ x + \iota y \in \mathbb{C} : x \leq 1 \right\}. \]
    A = {z in C : Re z <= 1}
  2. The given set is $B = \left\{ z \in \mathbb{C} : \text{ Re} \left( z^2 \right) = 2 \right\} $. We have \begin{align*} B & = \left\{ x + \iota y \in \mathbb{C} : \text{ Re} \left( (x + \iota y)^2 \right) = 2 \right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : \text{ Re} \left( x^2 - y^2 + 2\iota xy \right) = 2 \right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : x^2 - y^2 = 2 \right\}. \end{align*} The above set is a hyperbola in the complex plane.
    B = {z in C : Re(z^2) = 2}
  3. The given set is $C = \left\{ z \in \mathbb{C} : \text{ Im} \left( \frac{1}{z} \right) = \frac{1}{4} \right\} $. We have \begin{align*} C & = \left\{ x + \iota y \in \mathbb{C} : \text{ Im} \left( \frac{1}{x + \iota y} \right) = \frac{1}{4} \right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : \text{ Im} \left( \frac{x - \iota y}{x^2 + y^2} \right) = \frac{1}{4} \right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : \frac{-y}{x^2 + y^2} = \frac{1}{4} \right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : -4y = x^2 + y^2 \text{ and } x^2 + y^2 \neq 0\right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : x^2 + (y + 2)^2 = 4 \right\} \setminus \{ 0 \} . \end{align*} The above set is a circle with center at $(0, -2)$ and radius $2$.
    C = {z in C : Im(1/z) = 1/4}
  4. The set $D = \left\{ z \in \mathbb{C} : \vert z \vert = \text{ Re } z + 1 \right\} $ is given by \begin{align*} D & = \left\{ x + \iota y \in \mathbb{C} : \sqrt{x^2 + y^2} = x + 1 \right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : x^2 + y^2 = x^2 + 2x + 1 \right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : y^2 = 2x + 1 \right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : x = \frac{y^2 - 1}{2} \right\}. \end{align*} The above set is a parabola in the complex plane.
    D = {z in C : |z| = Re z + 1}
  5. The set $E = \left\{ z \in \mathbb{C} : \vert z - 2 \vert = \vert 1 - 2 \bar{z} \vert \right\} $ is given by \begin{align*} E & = \left\{ x + \iota y \in \mathbb{C} : \sqrt{(x - 2)^2 + y^2} = \sqrt{(1 - 2x)^2 + 4y^2} \right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : (x - 2)^2 + y^2 = (1 - 2x)^2 + 4y^2 \right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : x^2 - 4x + 4 + y^2 = 1 - 4x + 4x^2 + 4y^2 \right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : 3x^2 + 3y^2 = 3 \right\} \\ & = \left\{ x + \iota y \in \mathbb{C} : x^2 + y^2 = 1 \right\}. \end{align*} The above set is a circle with center at $(0, 0)$ and radius $1$.
    E = {z in C : |z - 2| = |1 - 2 bar{z}|}