The Conjugate of a Complex Number

 

The Conjugate of a Complex Number

 

The complex number obtained by changing the sign of the imaginary part of a given complex number is called the complex conjugate of that number.

\[ a + bi \quad \text{and} \quad a \ – \ bi \quad \text{are given.} \]

Each of these numbers is the conjugate of the other. The conjugate of a complex number \( z \) is denoted by \( \overline{z} \).

\[ z = a + bi \Leftrightarrow \overline{z} = a \ – \ bi \]

In the complex plane, the geometric representation of a complex number and its conjugate are symmetric with respect to the real axis (x-axis).

 

Examples:

 

  • If \( z = 2 \ – \ i \), then \( \overline{z} = 2 + i \)

 

  • If \( z = -3 + i \), then \( \overline{z} = -3 \ – \ i \)

 

  • If \( z = 2 \ – \ \sqrt{-7} \), then \( z = 2 \ – \ \sqrt{7} \ i \) and \( \overline{z} = 2 + \sqrt{7} \ i \)

 

  • If \( z = -\sqrt{-9} \), then \( z = -3i \) and \( \overline{z} = 3i \)

 

  • If \( z = i \ – \ 5 \), then \( \overline{z} = -i \ – \ 5 \)

 

Properties:

 

1. \( \overline{(\overline{z})} = z \)

2. \( \overline{z_1 \pm z_2} = \overline{z_1} \pm \overline{z_2} \)

3. \( \overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2} \)

4. \( \overline{(z_1 : z_2)} = \overline{z_1} : \overline{z_2} \quad (z_2 \neq 0) \)

5. \( \overline{(z^n)} = (\overline{z})^n \quad (n \in \mathbb{Z}) \)

 

Examples:

 

  • \( \overline{z_1 : (z_2 + z_3)} = \overline{z_1} : \overline{(z_2 + z_3)} = \overline{z_1} : (\overline{z_2} + \overline{z_3}) \)

 

  • \( \overline{z_1 \cdot (z_2 – z_3)} = \overline{z_1} \cdot \overline{(z_2 \ – \ z_3)} = \overline{z_1} \cdot (\overline{z_2} \ – \ \overline{z_3}) \)

 

Product of a Complex Number and its Conjugate

 

Let \( z = a + bi \quad \text{and } \quad \overline{z} = a \ – \ bi \).

\[ z \cdot \overline{z} = (a + bi)(a \ – \ bi) = a^2 \ – \ (bi)^2 \]
\[ = a^2 \ – \ b^2 i^2 \]
\[ = a^2 + b^2 \]

Therefore, for any complex number \( z = a + bi \):

\[ z \cdot \overline{z} = a^2 + b^2 \]

 

Examples:

 

  • If \( z = -2 + \sqrt{2}i \), then:
    \[ z \cdot \overline{z} = (-2)^2 + (\sqrt{2})^2 = 6 \]

 

  • Evaluate \( (\sqrt{3} \ – \ \sqrt{2}i)(\sqrt{3} + \sqrt{2}i) \):
    \[ = (\sqrt{3})^2 + (-\sqrt{2})^2 = 5 \]

 

  • If \( z = \sqrt{2} + 1 + i \), then:
    \[ z \cdot \overline{z} = (\sqrt{2} + 1 + i)(\sqrt{2} + 1 \ – \ i) \]
    \[ = (\sqrt{2} + 1)^2 + 1^2 = 2 + 2\sqrt{2} + 1 + 1 = 4 + 2\sqrt{2} \]

 

Note:

 

If the discriminant (\( \Delta \)) of a quadratic equation with real coefficients is negative, the roots of this equation are complex conjugates of each other. That is, for \( a, b, c \in \mathbb{R} \), if one root of the equation \( ax^2 + bx + c = 0 \) is \( x + yi \), the other root must be \( x \ – \ yi \).

 

Example:

 

Show that the roots of the equation \( x^2 \ – \ 2x + 2 = 0 \) are complex conjugates of each other.

\[ \Delta = (-2)^2 \ – \ 4 \cdot 1 \cdot 2 = -4 < 0 \]
\[ x_{1,2} = \displaystyle \frac{2 \pm \sqrt{-4}}{2 \cdot 1} = \displaystyle \frac{2 \pm 2i}{2} = 1 \pm i \]
\[ x_1 = 1 + i, \quad x_2 = \overline{x_1} = 1 \ – \ i \]

 

QUESTION 6

 

If one of the roots of a quadratic equation with real coefficients is

\[ \sqrt{2} \ – \ \sqrt{3} \ – \ \sqrt[4]{24}i \]

what is the product of the roots of this equation?

\[ A) \ 1 \quad B) \ 2 \quad C) \ 3 \quad D) \ 4 \quad E) \ 5 \]

 

Solution:

 

Since the coefficients are real, the roots must be complex conjugates:

\[ (\sqrt{2} – \sqrt{3} – \sqrt[4]{24}i)(\sqrt{2} – \sqrt{3} + \sqrt[4]{24}i) \]
\[ \quad = (\sqrt{2} – \sqrt{3})^2 + (\sqrt[4]{24})^2 \]
\[ \quad = 2 + 3 – 2\sqrt{6} + \sqrt{24}\]
\[ \quad = 5 – 2\sqrt{6} + 2\sqrt{6} = 5 \]

\[ \textbf{Answer: E} \]

 

QUESTION 7

 

Given \( a, b \in \mathbb{R} \), if two of the roots of the equation \( x^3 – ax^2 + 7x – b = 0 \) are \(1\) and \( 1 – 2i \), what is the value of the sum \( a + b \)?

\[ A) \ 5 \quad B) \ 6 \quad C) \ 7 \quad D) \ 8 \quad E) \ 9 \]

 

Solution:

 

Since the coefficients of the polynomial equation are real, complex roots must occur in conjugate pairs. Thus, the third root is \( \overline{1 – 2i} = 1 + 2i \). The three roots are:

\[ x_1 = 1 , \quad x_2 = 1 – 2i , \quad x_3 = 1 + 2i \]

Using Vieta’s formulas:

\[ \begin{array}{l@{\quad \quad \quad \quad}l}
x_1 + x_2 + x_3 = a & x_1 \cdot x_2 \cdot x_3 = b \\
\Rightarrow 1 + 1 – 2i + 1 + 2i = a & \Rightarrow 1 \cdot (1 – 2i)(1 + 2i) = b \\
\Rightarrow a = 3 & \Rightarrow b = 1^2 + (-2)^2 = 5
\end{array} \]

Therefore, the sum is:

\[ a + b = 3 + 5 = 8 \]

\[ \textbf{Answer: D} \]