Complex Numbers

 

Throughout the history of mathematics, there have been times when existing number systems were insufficient to solve certain types of equations. In such cases, these number systems were expanded to obtain new number systems capable of solving them. For instance,

When an equation like

\[ x + 7 = 0 \]

could not be solved within the set of natural numbers (\(\mathbb{N}\)), this set was expanded to construct the set of integers (\(\mathbb{Z}\)).

When an equation like

\[ 7x + 19 = 0 \]

could not be solved within the set of integers, this set was expanded to construct the set of rational numbers (\(\mathbb{Q}\)).

When an equation like

\[ x^2 \ – \ 5 = 0 \]

could not be solved within the set of rational numbers, this set was expanded to construct the set of real numbers (\(\mathbb{R}\)).

Now, let us try to solve the following equation:

\[ x^4 + 7 = 0 \]

which implies:

\[ x^4 = \ – 7 \]

There is no real number that satisfies this equation, because any real number raised to an even power is non-negative. Therefore, there is a clear need to expand the set of real numbers to create a larger number system where such equations can be solved.

To achieve this, consider the equation:

\[ x^2 + 1 = 0 \Rightarrow x^2 = \ – 1 \]
\[ \Rightarrow \sqrt{x^2} = \sqrt{\ – 1} \]
\[ \Rightarrow |x| = \sqrt{\ – 1} \]
\[ \Rightarrow \pm x = \sqrt{\ – 1} \]
\[ \Rightarrow x = \pm \sqrt{\ – 1} \]

Accordingly, by utilizing the element \(\sqrt{-1}\), we can expand the set of real numbers to solve equations of this type. The number \(\sqrt{-1}\) is called the imaginary unit and is denoted as:

\[ i = \sqrt{\ – 1} \ \Rightarrow \ i^2 = \ – 1 \]

For \( a, b \in \mathbb{R} \) and \( i^2 = \ – 1 \), numbers of the form \( a + bi \) are called complex numbers. A complex number is typically denoted by \( z \), and the set of all complex numbers is denoted by \( \mathbb{C} \).

\[ \mathbb{C} = \{ a + bi : a,b \in \mathbb{R} \text{ and } i^2 = \ – 1 \} \]

In the complex number \( z = a + bi \), \( a \) is called the real part and is denoted by \( a = \mathrm{Re}(z) \), while \( b \) is called the imaginary part and is denoted by \( b = \mathrm{Im}(z) \).

 

Examples:

 

• In the complex number \( z = 2 \ – \ i \), we have \( \mathrm{Re}(z) = 2 \) and \( \mathrm{Im}(z) = \ – 1 \).

 

• In the complex number \( z = \ – 3 + 2i \), we have \( \mathrm{Re}(z) = \ – 3 \) and \( \mathrm{Im}(z) = 2 \).

 

• In the complex number \( z = \ – 1 + \sqrt{-2} \ = \ – \ 1 + \sqrt{2} \sqrt{-1 } = -1 + \sqrt{ 2} i \), we have \( \mathrm{Re}(z) = \ – 1 \) and \( \mathrm{Im}(z) = \sqrt{2} \).

 

• In the complex number \( z = \sqrt{\ – 4} = 0 + \sqrt{4} \sqrt{ -1} = 0+2i \), we have \( \mathrm{Re}(z) = 0 \) and \( \mathrm{Im}(z) = 2 \).

 

Example:

 

Given \( a < 0 < b \), let us find the real and imaginary parts of the complex number:

\[ z = \sqrt{(b \ – \ a)a^2} + \sqrt{a \ – \ b} \]

Since \( (b \ – \ a)^2 > 0 \), it follows that:

\[ \sqrt{(b \ – \ a)^2} \in \mathbb{R} \]

Since \( a \ – \ b < 0 \), it follows that:

\[ \sqrt{a \ – \ b} \notin \mathbb{R} \]

Thus, rewriting \( z \):

\[ z = \sqrt{(b \ – \ a)a^2} + \sqrt{a \ – \ b} \]
\[ = \sqrt{b \ – \ a}\,.\,|a| + \sqrt{\ – (b \ – \ a)} \]

Since \( a < 0 \), we know that \( |a| = \ – a \). Therefore:

\[ \Rightarrow z = \ – a\sqrt{b \ – \ a} + \sqrt{\ – 1}\sqrt{b \ – \ a} \]
\[ \Rightarrow z = \ – a\sqrt{b \ – \ a} + i\sqrt{b \ – \ a} \]

Consequently, the real and imaginary parts are:

\[ \mathrm{Re}(z) = \ – a\sqrt{b \ – \ a} \]

and

\[ \mathrm{Im}(z) = \sqrt{b \ – \ a} \]

 

Example:

 

Find the solution set of the equation:

\[ 3x^2 \ – \ \sqrt{3}\,x + 1 = 0 \]

The discriminant \(\Delta\) is calculated as:

\[ \Delta = (\ – \sqrt{3})^2 \ – \ 4 \cdot 3 \cdot 1 = \ – 9 < 0 \]

Since \(\Delta < 0\), the equation has two distinct complex roots:

\[ x_{1,2} = \frac{\sqrt{3} \pm \sqrt{\ – 9}}{2 \cdot 3} \]
\[ = \frac{\sqrt{3} \pm \sqrt{9}\sqrt{\ – 1}}{6} \]
\[ = \frac{\sqrt{3} \pm 3i}{6} \]

Thus, the solution set \(\text{S}\) is:

\[ \text{S} = \left\{\frac{\displaystyle \sqrt{3}}{\displaystyle 6} + \frac{\displaystyle 1}{\displaystyle 2}i \,;\, \frac{\displaystyle \sqrt{3}}{\displaystyle 6} \ – \ \frac{\displaystyle 1}{\displaystyle 2}i \right\} \]

 

Example:

 

Find the roots of the equation:

\[ z^2 \ – \ \sqrt{5}\,iz \ – \ 3i = 0 \]

The discriminant \(\Delta\) is:

\[ \Delta = (\ – \sqrt{5}\,i)^2 \ – \ 4 \cdot 1 \cdot ( \ – \ 3i ) \]
\[ = 5i^2 + 12i = \ – 5 + 12i = (2 + 3i)^2 \]

Using the quadratic formula:

\[ z_{1,2}= \frac{\displaystyle \sqrt{5}\,i \pm \sqrt{(2 + 3i)^2}} {\displaystyle 2 \cdot 1} = \frac{\displaystyle \sqrt{5}\,i \pm (2 + 3i)} {\displaystyle 2} \]
\[ z_1 = \frac{\displaystyle \sqrt{5}\,i + 2 + 3i} {\displaystyle 2} = 1 + \frac{\displaystyle \sqrt{5} + 3}{\displaystyle 2}i \]
\[ z_2 = \frac{\displaystyle \sqrt{5}\,i \ – \ 2 \ – \ 3i} {\displaystyle 2} = – 1 + \frac{\displaystyle \sqrt{5} \ – \ 3}{\displaystyle 2}i \]

 

Example:

 

Find the solution set of the equation:

\[ x^3 + 8 = 0 \]

Factoring using the sum of cubes:

\[ x^3 + 8 = 0 \Rightarrow (x + 2)(x^2 \ – \ 2x + 4) = 0 \]
\[ \Rightarrow x_1 = \ – 2 \]

or from the quadratic part:

\[ x^2 \ – \ 2x + 4 = 0 \]
\[ \Delta = (\ – 2)^2 \ – \ 4 \cdot 1 \cdot 4 = – 12 \]

Solving for the remaining roots:

\[ \Rightarrow x_{2,3} = \frac{\displaystyle 2 \pm \sqrt{\ – 12}} {\displaystyle 2 \cdot 1} \]
\[ = \frac{\displaystyle 2 \pm \sqrt{12}\,\sqrt{\ – 1}} {\displaystyle 2} \]
\[ \Rightarrow x_{2,3} = 1 \pm \sqrt{3}\,i \]

Therefore, the solution set in the complex field is:

\[ \text{S} = \{ – 2, \ 1 + \sqrt{3}\,i, \ 1 \ – \ \sqrt{3}\,i \} \]

 

QUESTION 1

 

Given \(x < y < 0\), what is the sum of the real part and the imaginary part of the complex number

\[ z = \sqrt{\ – x^2 + 2xy \ – \ y^2} \ – \ \sqrt[4]{x^4} \ – \ \sqrt[3]{y^3} \,? \]

\[ A) \ 0 \quad B) \ 2x \quad C) \ – \ 2x \quad D) \ 2y \quad E) \ – \ 2y \]

 

Solution:

 

Given expression:

\[ z = \sqrt{\ – x^2 + 2xy \ – \ y^2} \ – \ \sqrt[4]{x^4} \ – \ \sqrt[3]{y^3} \]

Simplifying the terms under the radicals:

\[ \Rightarrow z = \sqrt{\ – (x \ – \ y)^2} \ – \ |x| \ – \ y \]

Since \(x < 0\), we have \(|x| = \ – \ x\). Thus:

\[ \Rightarrow z = |x \ – \ y| \cdot \sqrt{\ – 1} \ – \ (\ – \ x) \ – \ y \]

Since \(x < y \Rightarrow x \ – \ y < 0\), the absolute value yields \(|x \ – \ y| = \ – \ (x \ – \ y)\). Substituting this back:

\[ \Rightarrow z = \ – \ (x \ – \ y)i + x \ – \ y = x \ – \ y + (\ – \ x + y)i \]

From this standard form, we identify:

\[ \mathrm{Re}(z) = x \ – \ y \quad \text{and} \quad \mathrm{Im}(z) = \ – \ x + y \]

Therefore, the sum of the real and imaginary parts is:

\[ \mathrm{Re}(z) + \mathrm{Im}(z) = x \ – \ y \ – \ x + y = 0 \]

\( \textbf{Answer: A} \)