AOS3 Topic 4: Complex Roots

Definition:

Complex roots are the imaginary roots of quadratic equations which have been represented as complex numbers. The square root of a negative number is not possible and hence we transform it into a complex number. The quadratic equations having discriminant values lesser than zero \( b^2 - 4ac < 0 \), is transformed using \( i^2 = -1 \), to obtain the complex roots.

Complex roots are expressed as complex numbers \(a \pm ib\). The complex root is made up of a real part and an imaginary part. The complex root is often represented as \(Z = a + ib\). Here 'a' is the real part of the complex number, which is denoted by Re(Z), and 'b' is the imaginary part, which is represented as Im(Z). Here \(ib\) is the imaginary number.

In the imaginary part of the complex number, the alphabet 'i' is referred to as iota. The iota - \(i\) is very useful to find the square root of any negative number. Here \(i^2 = -1\), and the negative number \(-N\) is represented as \(i^2N\), and it has now transformed into a positive number.


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Properties Of Complex Roots

Magnitude of Complex Roots:

The complex root \(α = a \pm ib\) is represented as a point \((a, \pm b)\) in the Argand plane, and the distance of this point from the origin \((0, 0)\) is called the modulus of the complex number. The distance is the simple linear distance, which is measured as \( r = |\sqrt{a^2 + b^2}| \). This can be easily understood with the use of Pythagoras theorem, and here the modulus of the complex root is represented by the hypotenuse of the right triangle, the base is the real part, and the height is the imaginary part.

Argument of Complex Roots

The complex root can be represented in the Argand plane as a point, and the line joining this point with the origin makes an angle \( \theta \) with the positive x-axis in the Argand plane, which is referred to as the argument of the complex number. The argument of the complex root is obtained from the inverse of the trigonometric tangent of the imaginary part divided by the real part, which is equal to \( \text{Arg}_z (\theta) = \tan^{-1} \frac{b}{a} \).

Polar Representation of Complex Roots

The complex root can be represented in the polar form with the help of the modulus and the argument of the complex number in the Argand plane. The complex root \( \alpha = a + ib \) can be represented in polar form as \( \alpha = r(\cos \theta + i\sin \theta) \). Here \( r \) is the modulus of the complex root and \( \theta \) is the argument of the complex root. The modulus of the complex root is computed as \( r = \sqrt{a^2 + n^2} \), and the argument of the complex root is computed using the formula \( \theta = \tan^{-1} \frac{b}{a} \).

Reciprocal of Complex Roots

The division of one complex root with another complex root is possible with the help of the reciprocal of a complex root. The division of one complex root with another complex root is equal to the product of one complex root with the reciprocal of another complex root. The reciprocal of a complex root \( \alpha = a + ib \) is \( \alpha^{-1} = \frac{1}{a+ib} = \frac{a-ib}{a^2 + b^2} = \frac{a}{a^2 + b^2} + \frac{-b}{a^2 + b^2}i \). Further, we can also understand that \( \alpha \neq \alpha^{-1} \).

Operations On Complex Roots

The complex roots of simple algebraic expressions can also be added, subtracted, multiplied, or divided similar to the normal numbers. Let us check each of the operations in detail:

  1. Addition: When adding complex roots, you simply add the real parts together and the imaginary parts together separately. For example, if you have two complex roots, \( z_1 = a_1 + ib_1 \) and \( z_2 = a_2 + ib_2 \), their sum would be \( z_1 + z_2 = (a_1 + a_2) + i(b_1 + b_2) \).
  2. Subtraction: Subtraction of complex roots follows a similar pattern to addition. You subtract the real parts and the imaginary parts separately. For example, if you have two complex roots, \( z_1 = a_1 + ib_1 \) and \( z_2 = a_2 + ib_2 \), their difference would be \( z_1 - z_2 = (a_1 - a_2) + i(b_1 - b_2) \).
  3. Multiplication: To multiply complex roots, you can use the distributive property and then combine like terms. For example, if you have two complex roots, \( z_1 = a_1 + ib_1 \) and \( z_2 = a_2 + ib_2 \), their product would be \( z_1 \cdot z_2 = (a_1a_2 - b_1b_2) + i(a_1b_2 + a_2b_1) \).
  4. Division: Dividing complex roots is similar to dividing polynomials. You multiply the numerator and denominator by the complex conjugate of the denominator to rationalize the denominator. For example, if you have two complex roots, \( z_1 = a_1 + ib_1 \) and \( z_2 = a_2 + ib_2 \), their quotient would be \( \frac{z_1}{z_2} = \frac{a_1a_2 + b_1b_2}{a_2^2 + b_2^2} + i \frac{a_2b_1 - a_1b_2}{a_2^2 + b_2^2} \).

Applications of Complex Roots

  • Engineering: Complex roots are used in solving electrical circuits, control systems, and signal analysis problems.
  • Physics: They play a crucial role in quantum mechanics, electromagnetism, and wave phenomena.
  • Signal Processing: Complex roots are employed in analyzing and processing signals in communication systems and digital signal processing.
  • Mathematical Modeling: They are utilized in modeling physical systems, such as oscillatory motion and fluid dynamics.
Example 1

Finding complex roots

Find the complex roots of the quadratic equation \(x^2 + 3x + 4 = 0\).

Solution:

To find the complex roots of the quadratic equation \(x^2 + 3x + 4 = 0\), we can use the quadratic formula:

\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]

Here, \(a = 1\), \(b = 3\), and \(c = 4\).

Substitute the values into the formula:

\[x = \frac{{-3 \pm \sqrt{{3^2 - 4 \cdot 1 \cdot 4}}}}{{2 \cdot 1}}\]

\[x = \frac{{-3 \pm \sqrt{{9 - 16}}}}{2}\]

\[x = \frac{{-3 \pm \sqrt{{-7}}}}{2}\]

The discriminant (\(b^2 - 4ac\)) is negative, indicating that the roots will be complex.

So, the roots are:

\[x = \frac{{-3 \pm i\sqrt{7}}}{2}\]

These are the complex roots of the given quadratic equation.

Example 2

Finding cube roots

Find the three cube roots of \( i \). In other words, find all \( z \) such that \( z^3 = i \).

Solution:

First, convert each number to polar form: \( z = re^{i\theta} \) and \( i = 1e^{i\frac{\pi}{2}} \). The equation now becomes:

\( (r^3e^{i\theta})^3 = r^3e^{3i\theta} = 1e^{i\frac{\pi}{2}} \)

Therefore, the two equations that we need to solve are \( r^3 = 1 \) and \( 3i\theta = i\frac{\pi}{2} \).

Given that \( r \in \mathbb{R} \) and \( r^3 = 1 \), it follows that \( r = 1 \).

Solving the second equation is as follows. First divide by \( i \). Then, since the argument of \( i \) is not unique we write \( 3\theta = \frac{\pi}{2} + 2\pi\ell \) for \( \ell = 0,1,2 \).

\( 3\theta = \frac{\pi}{2} + \frac{2}{3}\pi\ell \) for \( \ell = 0,1,2 \)
For \( \ell = 0 \): \( \theta = \frac{\pi}{6} + \frac{2}{3}\pi(0) = \frac{\pi}{6} \)
For \( \ell = 1 \): \( \theta = \frac{\pi}{6} + \frac{2}{3}\pi(1) = \frac{5\pi}{6} \)
For \( \ell = 2 \): \( \theta = \frac{\pi}{6} + \frac{2}{3}\pi(2) = \frac{3\pi}{2} \)

Therefore, the three roots are given by \( 1e^{i\frac{\pi}{6}}, 1e^{i\frac{5\pi}{6}}, 1e^{i\frac{3\pi}{2}} \).

Written in standard form, these roots are, respectively,
\( \frac{\sqrt{3}-i}{2}, -\frac{\sqrt{3}+i}{2}, -i \).

Example 3

Finding quadratic equation from complex roots

Find the quadratic equation having its complex roots as \(3 \pm 2i\).

Solution:

The given roots of the quadratic equation are \(3 \pm 2i\).

The values of \(x\) are \(x = 3 + 2i\), and \(x = 3 - 2i\).

This can be written as \(x - (3 + 2i) = 0\), and \(x - (3 - 2i) = 0\)

We can multiply these two to obtain the required equation. \[ (x - (3 + 2i))(x - (3 - 2i)) = 0 \] \[ x^2 - x[(3 + 2i) + (3 - 2i)] + ((3 + 2i)(3 - 2i)) = 0 \]

\[ x^2 - x[3 + 2i + 3 - 2i] + (3^2 - 2^2i^2) = 0 \] \[ x^2 - x[3 + 3] + (9 + 4) = 0 \] \[ x^2 - 6x + 13 = 0 \]

Therefore, the required quadratic equation is \(x^2 - 6x + 13 = 0\).

Example 4

Divide the complex roots of previous example.

Solution:

Suppose we have two complex roots: \( z_1 = 3 + 2i \) and \( z_2 = 3 - 2i \).

To find their quotient, we use the formula:

\[ \frac{z_1}{z_2} = \frac{a_1a_2 + b_1b_2}{a_2^2 + b_2^2} + i \frac{a_2b_1 - a_1b_2}{a_2^2 + b_2^2} \]

where \( a_1 = 3 \), \( b_1 = 2 \), \( a_2 = 3 \), and \( b_2 = -2 \).

Substituting these values into the formula, we get:

\[ \frac{3 \cdot 3 + 2 \cdot (-2)}{3^2 + (-2)^2} + i \frac{3 \cdot (-2) - 3 \cdot 2}{3^2 + (-2)^2} \]

Solving each term:

\[ \frac{9 - 4}{9 + 4} + i \frac{-6 - 6}{9 + 4} \] \[ \frac{5}{13} + i \frac{-12}{13} \]

So, the quotient of \( z_1 \) and \( z_2 \) is \( \frac{5}{13} - i \frac{12}{13} \).

Example 5

Finding product of complex roots

Find the product of the complex roots \( z_1 = 2 + 3i \) and \( z_2 = 4 - 5i \).

Solution:

To find the product of two complex roots, we multiply them together using the distributive property. Let's compute \( z_1 \times z_2 \):

\[ z_1 \times z_2 = (2 \times 4 - 3 \times (-5)) + i(2 \times (-5) + 4 \times 3) \]

\[ = (8 + 15) + i(-10 + 12) \]

\[ = 23 + 2i \]

So, the product of \( z_1 \) and \( z_2 \) is \( 23 + 2i \).

Exercise &&1&& (&&1&& Question)

What is the discriminant value of a quadratic equation having complex roots?

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Exercise &&2&& (&&1&& Question)

What is the value of \(i^{2}\)?

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Exercise &&3&& (&&1&& Question)

Find the quadratic equation with complex roots \(5 + 2i\) and \(5 - 2i\).

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Exercise &&4&& (&&1&& Question)

Let \( z_1 = 3 + 2i \) and \( z_2 = 5 - 4i \). The difference \( z_1 - z_2 \) is:

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Exercise &&5&& (&&1&& Question)

What does \(Arg(z)\) represent in complex numbers?

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