Complex Numbers | Electrical Energy Technology

1. Introduction to Complex Numbers

1.1 Definition

A complex number \(z\) is a number of the form:

\[ z = a + bi \]

where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit defined by \(i^2 = -1\). In this representation:

\(a\) is called the real part of \(z\), denoted by \(\text{Re}(z)\)
\(b\) is called the imaginary part of \(z\), denoted by \(\text{Im}(z)\)

Example:

For the complex number \(z = 3 + 4i\):

\(\text{Re}(z) = 3\) and \(\text{Im}(z) = 4\)

For the complex number \(z = -2i\):

\(\text{Re}(z) = 0\) and \(\text{Im}(z) = -2\)

1.2 The Complex Plane

Complex numbers can be visualized as points in a two-dimensional plane called the complex plane or Argand diagram. The horizontal axis represents the real part, and the vertical axis represents the imaginary part.

Complex Plane Illustration — The complex plane showing a complex number \(z = a + bi\)

1.3 Visualization Techniques

Beyond simple plotting, there are several visualization techniques that help understand complex number operations:

Vector representation: Complex numbers can be viewed as 2D vectors, with operations like addition corresponding to vector addition
Color mapping: Using color to represent magnitude and hue to represent phase
Dynamic plotting: Showing how complex numbers transform under various operations

Example: Visualizing Complex Addition

To add two complex numbers \(z_1 = 3 + 2i\) and \(z_2 = 1 + 4i\):

1. Plot both numbers as points in the complex plane

2. Construct vectors from the origin to each point

3. Apply the parallelogram rule from vector addition

4. The result \(z_1 + z_2 = 4 + 6i\) can be verified geometrically

2. Operations with Complex Numbers

2.1 Addition & Subtraction

Complex numbers are added or subtracted by combining their real and imaginary parts separately.

\[ (a + bi) + (c + di) = (a + c) + (b + d)i \] \[ (a + bi) - (c + di) = (a - c) + (b - d)i \]

Example:

\((3 + 2i) + (1 + 5i) = (3 + 1) + (2 + 5)i = 4 + 7i\)

\((3 + 2i) - (1 + 5i) = (3 - 1) + (2 - 5)i = 2 - 3i\)

\((5 - 7i) - i = 5 + (-7 - 1)i = 5 - 8i\)

2.2 Multiplication

Complex numbers are multiplied using algebraic distribution, remembering that \(i^2 = -1\).

\[ (a + bi)(c + di) = ac + adi + bci + bdi^2 \] \[ = (ac - bd) + (ad + bc)i \]

Example:

\((3 + 2i)(1 + 5i) = 3 \cdot 1 + 3 \cdot 5i + 2i \cdot 1 + 2i \cdot 5i\)

\(= 3 + 15i + 2i + 10i^2 = 3 + 17i + 10(-1) = 3 + 17i - 10 = -7 + 17i\)

2.3 Complex Conjugate

The complex conjugate of \(z = a + bi\), denoted by \(\overline{z}\) or \(z^*\), is defined as:

\[ \overline{z} = a - bi \]

The complex conjugate is useful for division and has several important properties:

\(z \cdot \overline{z} = |z|^2 = a^2 + b^2\)
\(\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}\)
\(\overline{z_1 \cdot z_2} = \overline{z_1} \cdot \overline{z_2}\)
\(\overline{\overline{z}} = z\)

Example:

For \(z = 5 - 7i\), the complex conjugate is \(\overline{z} = 5 + 7i\)

Verifying that \(z \cdot \overline{z} = |z|^2\):

\((5 - 7i)(5 + 7i) = 25 + 35i - 35i - 49i^2 = 25 + 49 = 74 = 5^2 + 7^2 = |z|^2\)

2.4 Division

Division of complex numbers is performed by multiplying both numerator and denominator by the complex conjugate of the denominator:

\[ \frac{z_1}{z_2} = \frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} \]

Example:

\(\frac{3 + 2i}{1 + i} = \frac{(3 + 2i)(1 - i)}{(1 + i)(1 - i)} = \frac{3 - 3i + 2i - 2i^2}{1^2 - i^2} = \frac{3 - i + 2}{1 + 1} = \frac{5 - i}{2} = \frac{5}{2} - \frac{1}{2}i\)

For a more complex example:

\(\frac{1 + 2i}{3 + 4i} = \frac{(1 + 2i)(3 - 4i)}{(3 + 4i)(3 - 4i)} = \frac{3 - 4i + 6i - 8i^2}{9 + 16} = \frac{3 + 2i + 8}{25} = \frac{11 + 2i}{25} = \frac{11}{25} + \frac{2}{25}i\)

3. Modulus and Argument

3.1 Modulus

The modulus (or absolute value) of a complex number \(z = a + bi\) is the distance from the origin to the point \(z\) in the complex plane:

\[ |z| = \sqrt{a^2 + b^2} \]

Example:

The modulus of \(z = 3 + 4i\) is \(|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

The modulus of \(z = -1 - 2i\) is \(|z| = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}\)

3.2 Argument

The argument of a complex number \(z = a + bi\) is the angle in radians between the positive real axis and the line joining the origin to the point \(z\) in the complex plane.

For \(z \neq 0\), the argument is defined as:

\[ \arg(z) = \begin{cases} \arctan(b/a) & \text{if } a > 0 \\ \arctan(b/a) + \pi & \text{if } a < 0, b \geq 0 \\ \arctan(b/a) - \pi & \text{if } a < 0, b < 0 \\ \pi/2 & \text{if } a = 0, b > 0 \\ -\pi/2 & \text{if } a = 0, b < 0 \end{cases} \]

Example:

The argument of \(z = -3 + 3i\) is \(\arg(z) = \arctan(3/(-3)) + \pi = \arctan(-1) + \pi = -\pi/4 + \pi = 3\pi/4\)

The argument of \(z = i\) is \(\arg(z) = \pi/2\) (as it lies on the positive imaginary axis)

3.3 Properties of Modulus and Argument

Key properties for calculations:

\(|z_1 \cdot z_2| = |z_1| \cdot |z_2|\)
\(|z_1 / z_2| = |z_1| / |z_2|\)
\(\arg(z_1 \cdot z_2) = \arg(z_1) + \arg(z_2)\) (modulo \(2\pi\))
\(\arg(z_1 / z_2) = \arg(z_1) - \arg(z_2)\) (modulo \(2\pi\))
\(\arg(\overline{z}) = -\arg(z)\)

Example:

For \(z_1 = 1 + i\) and \(z_2 = -1 + i\):

\(|z_1| = \sqrt{1^2 + 1^2} = \sqrt{2}\) and \(|z_2| = \sqrt{1^2 + 1^2} = \sqrt{2}\)

\(|z_1 \cdot z_2| = |(1 + i)(-1 + i)| = |(-1 - i + i + i^2)| = |-1 - 1| = 2\)

\(|z_1| \cdot |z_2| = \sqrt{2} \cdot \sqrt{2} = 2\)

This verifies that \(|z_1 \cdot z_2| = |z_1| \cdot |z_2|\)

4. Polar Form and Euler's Formula

4.1 Polar Form

Using the modulus and argument, a complex number can be written in polar form:

\[ z = |z|(\cos(\arg(z)) + i\sin(\arg(z))) \]

We often denote the modulus as \(r = |z|\) and the argument as \(\theta = \arg(z)\), giving:

\[ z = r(\cos\theta + i\sin\theta) \]

Example:

The complex number \(z = 3 + 3i\) has modulus \(|z| = \sqrt{3^2 + 3^2} = 3\sqrt{2}\) and argument \(\arg(z) = \arctan(3/3) = \arctan(1) = \pi/4\).

In polar form: \(z = 3\sqrt{2}(\cos(\pi/4) + i\sin(\pi/4))\)

Converting from polar to rectangular form: \(z = 2e^{i\pi/3} = 2(\cos(\pi/3) + i\sin(\pi/3)) = 2 \cdot (1/2 + i\sqrt{3}/2) = 1 + i\sqrt{3}\)

4.2 Euler's Formula

Euler's formula establishes a profound connection between trigonometric functions and the complex exponential:

\[ e^{i\theta} = \cos\theta + i\sin\theta \]

Using Euler's formula, the polar form can be written more compactly as:

\[ z = re^{i\theta} \]

This exponential form is particularly useful for multiplication, division, and powers of complex numbers.

De Moivre's Formula:

For any complex number in polar form \(z = re^{i\theta}\) and any integer \(n\):

\[ z^n = r^ne^{in\theta} = r^n(\cos(n\theta) + i\sin(n\theta)) \]

Example using De Moivre's formula to find \((1+i)^{13}\):

First, express in polar form: \(1+i = \sqrt{2}e^{i\pi/4}\)

Then apply De Moivre: \((1+i)^{13} = (\sqrt{2})^{13}e^{i13\pi/4} = 2^{6.5}e^{i13\pi/4}\)

Simplify the argument: \(13\pi/4 = 3\pi + \pi/4\), so \(e^{i13\pi/4} = e^{i\pi/4}e^{3\pi i} = e^{i\pi/4}(-1)^3 = -e^{i\pi/4}\)

Therefore: \((1+i)^{13} = -2^{6.5}e^{i\pi/4} = -2^{6.5}(\cos(\pi/4) + i\sin(\pi/4)) = -2^{6.5} \cdot (1/\sqrt{2} + i/\sqrt{2}) = -64 - 64i\)

4.3 Multiplication and Division in Polar Form

In polar form, multiplication and division become simpler:

For \(z_1 = r_1e^{i\theta_1}\) and \(z_2 = r_2e^{i\theta_2}\):

\[ z_1 \cdot z_2 = r_1r_2e^{i(\theta_1 + \theta_2)} \] \[ \frac{z_1}{z_2} = \frac{r_1}{r_2}e^{i(\theta_1 - \theta_2)} \]

Example:

Let \(z_1 = 2e^{i\pi/6}\) and \(z_2 = 3e^{i\pi/4}\)

\(z_1 \cdot z_2 = 2 \cdot 3 \cdot e^{i(\pi/6 + \pi/4)} = 6e^{i5\pi/12}\)

\(\frac{z_1}{z_2} = \frac{2}{3}e^{i(\pi/6 - \pi/4)} = \frac{2}{3}e^{-i\pi/12}\)

5. Complex Numbers in Polynomials

5.1 Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra states that every non-constant polynomial with complex coefficients has at least one complex root. This means that any polynomial of degree \(n\) has exactly \(n\) roots when counted with multiplicity.

5.2 Finding Roots of Polynomials

For quadratic polynomials \(az^2 + bz + c\), the roots are given by:

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

When the discriminant \(b^2 - 4ac < 0\), the roots are complex conjugates.

Example:

Find the roots of \(p(z) = z^2 - 4z + 13\)

Using the quadratic formula: \(z = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i\)

The roots are \(z_1 = 2 + 3i\) and \(z_2 = 2 - 3i\) (complex conjugates)

5.3 Polynomial Division and Factorization

The polynomial division algorithm is a method to divide one polynomial by another, resulting in a quotient polynomial and a remainder.

Key facts about polynomials and their factors:

If \(r\) is a root of polynomial \(p(z)\), then \((z-r)\) is a factor of \(p(z)\)
A polynomial of degree \(n\) can be factored as \(p(z) = a_n(z-r_1)(z-r_2)...(z-r_n)\) where \(r_1, r_2, ..., r_n\) are the roots
For polynomials with real coefficients, complex roots always appear in conjugate pairs

Example: Polynomial Division

To divide \(p(z) = z^3 + z - 2\) by \(d(z) = z - 1\), we use the division algorithm:

Step 1: \(z^3 \div (z-1) = z^2 + z + 1\) with remainder 0

Step 2: \(z \div (z-1) = 1\) with remainder 1

Step 3: \(-2 \div (z-1) = 0\) with remainder -2

Combining these steps: \(p(z) = (z-1)(z^2 + z + 1) + (-1)\)

Example: Factoring a Polynomial with Complex Roots

For the polynomial \(p(z) = z^4 - 4z^2 + 16\), we can factor as follows:

Step 1: Substitute \(u = z^2\) to get \(p(u) = u^2 - 4u + 16\)

Step 2: Find the roots of this quadratic: \(u = \frac{4 \pm \sqrt{16 - 64}}{2} = \frac{4 \pm \sqrt{-48}}{2} = 2 \pm 2\sqrt{3}i\)

Step 3: Substitute back to find \(z^2 = 2 \pm 2\sqrt{3}i\), so \(z = \pm\sqrt{2 \pm 2\sqrt{3}i}\)

Step 4: Express as a product: \(p(z) = (z^2 - (2 + 2\sqrt{3}i))(z^2 - (2 - 2\sqrt{3}i))\)

6. Taylor Polynomials with Complex Variables

6.1 Approximating Functions with Taylor Series

Taylor polynomials can be extended to complex variables, allowing for approximation of complex functions:

\[ f(z) \approx f(z_0) + f'(z_0)(z-z_0) + \frac{f''(z_0)}{2!}(z-z_0)^2 + \cdots + \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n \]

This is particularly useful for computing values of transcendental functions with complex arguments.

Example:

The Taylor series for \(e^z\) around \(z_0 = 0\) is:

\[e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots\]

To approximate \(e^{1+i}\), we substitute \(z = 1+i\):

\[e^{1+i} \approx 1 + (1+i) + \frac{(1+i)^2}{2!} + \frac{(1+i)^3}{3!} + \frac{(1+i)^4}{4!}\]

Computing each term:

\((1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i\)

\((1+i)^3 = (1+i)(2i) = 2i + 2i^2 = 2i - 2 = -2 + 2i\)

\((1+i)^4 = (1+i)(-2+2i) = -2 - 2i + 2i + 2i^2 = -2 - 2 = -4\)

Substituting these values:

\[e^{1+i} \approx 1 + (1+i) + \frac{2i}{2} + \frac{-2+2i}{6} + \frac{-4}{24} = 1 + 1 + i + i - \frac{1}{3} + \frac{1}{3}i - \frac{1}{6} = \frac{3}{2} + \frac{7}{3}i\]

6.2 Applications in Complex Analysis

Taylor series in the complex plane have numerous applications, including:

Evaluating complex integrals via residue theory
Studying the analytic properties of complex functions
Approximating solutions to complex differential equations
Investigating convergence regions for power series

7. Differential Equations with Complex Solutions

7.1 First-Order Differential Equations

Complex numbers often emerge in the solutions to differential equations. For a first-order linear differential equation:

\[ \frac{dy}{dt} + p(t)y = q(t) \]

The general solution can include complex exponentials when the homogeneous equation has complex eigenvalues.

Example:

For the differential equation \(\frac{dy}{dt} + 3y = 2e^{2i t}\)

We try a particular solution of the form \(y_p(t) = Ae^{2i t}\) where \(A\) is a complex constant

Substituting into the equation:

\(2iAe^{2i t} + 3Ae^{2i t} = 2e^{2i t}\)

\(A(2i + 3) = 2\)

\(A = \frac{2}{3 + 2i} = \frac{2(3 - 2i)}{(3 + 2i)(3 - 2i)} = \frac{6 - 4i}{9 + 4} = \frac{6 - 4i}{13}\)

The particular solution is \(y_p(t) = \frac{6 - 4i}{13}e^{2i t}\)

The complete solution will also include the complementary solution \(y_c(t) = Ce^{-3t}\), giving \(y(t) = Ce^{-3t} + \frac{6 - 4i}{13}e^{2i t}\)

7.2 Second-Order Differential Equations

Second-order linear differential equations often have complex solutions, especially in systems involving oscillations:

\[ \frac{d^2y}{dt^2} + a\frac{dy}{dt} + by = f(t) \]

When the characteristic equation \(r^2 + ar + b = 0\) has complex roots \(r = \alpha \pm i\beta\), the solution contains terms like \(e^{\alpha t}\cos(\beta t)\) and \(e^{\alpha t}\sin(\beta t)\).

Example:

For the differential equation \(\frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 13y = 10\cos(t)\)

The characteristic equation is \(r^2 + 4r + 13 = 0\)

Using the quadratic formula: \(r = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3i\)

The complementary solution is \(y_c(t) = e^{-2t}(c_1\cos(3t) + c_2\sin(3t))\)

For the particular solution, we can use complex methods, considering \(10\cos(t) = 5e^{it} + 5e^{-it}\)

This leads to a particular solution of the form \(y_p(t) = A\cos(t) + B\sin(t)\)

After determining the constants, the complete solution is:

\(y(t) = e^{-2t}(c_1\cos(3t) + c_2\sin(3t)) + \cos(t) + \sin(t)\)

7.3 Complex Methods for Solving Differential Equations

Complex techniques can simplify the solution of differential equations with trigonometric forcing functions:

Replace \(\cos(\omega t)\) with \(\frac{e^{i\omega t} + e^{-i\omega t}}{2}\)
Replace \(\sin(\omega t)\) with \(\frac{e^{i\omega t} - e^{-i\omega t}}{2i}\)
Solve the resulting equation with complex exponentials
Extract the real or imaginary part as needed

Example: Using Complex Methods

For the differential equation \(y'' + 2y' + 5y = 10e^{-t}\cos(3t)\)

We can rewrite this as \(y'' + 2y' + 5y = 5e^{(-1+3i)t} + 5e^{(-1-3i)t}\)

Solving for the complex exponential inputs separately and then taking the real part of the result yields the solution more efficiently than direct methods.

8. Applications in Electrical Engineering

8.1 AC Circuit Analysis

Complex numbers are essential for analyzing alternating current (AC) circuits. The impedance \(Z\) of a circuit can be expressed as a complex number:

\[ Z = R + jX \]

where \(R\) is the resistance and \(X\) is the reactance. Note that in electrical engineering, \(j\) is commonly used instead of \(i\) to avoid confusion with current.

The impedance of basic components:

Resistor (R): \(Z_R = R\) (purely real)
Capacitor (C): \(Z_C = -j \frac{1}{\omega C}\) (purely imaginary, negative)
Inductor (L): \(Z_L = j\omega L\) (purely imaginary, positive)

Ohm's law for AC circuits is expressed as:

\[ V = ZI \]

where \(V\) and \(I\) are phasors representing voltage and current.

Example: RLC Series Circuit

In a series RLC circuit with \(R = 10\Omega\), \(L = 50\text{ mH}\), and \(C = 100\text{ μF}\) at frequency \(\omega = 100\text{ rad/s}\):

\(Z_R = 10\Omega\)

\(Z_L = j\omega L = j \cdot 100 \cdot 0.05 = j5\Omega\)

\(Z_C = -j \frac{1}{\omega C} = -j \frac{1}{100 \cdot 100 \times 10^{-6}} = -j100\Omega\)

Total impedance: \(Z = Z_R + Z_L + Z_C = 10 + j5 - j100 = 10 - j95\Omega\)

Magnitude: \(|Z| = \sqrt{10^2 + 95^2} \approx 95.5\Omega\)

Phase angle: \(\phi = \arctan(-95/10) \approx -84°\)

8.2 Phasors and Sinusoidal Signals

A sinusoidal signal \(v(t) = V_m\cos(\omega t + \phi)\) can be represented by the complex number (phasor):

\[ V = V_m e^{j\phi} \]

This simplifies calculations involving multiple sinusoidal signals of the same frequency.

Example: Adding Sinusoidal Signals

Consider two voltage sources: \(v_1(t) = 5\cos(60\pi t)\) and \(v_2(t) = 8\cos(60\pi t - \pi/3)\)

As phasors: \(V_1 = 5e^{j0} = 5\) and \(V_2 = 8e^{-j\pi/3} = 8(\cos(-\pi/3) + j\sin(-\pi/3)) = 8(1/2 - j\sqrt{3}/2) = 4 - j4\sqrt{3}\)

Sum: \(V = V_1 + V_2 = 5 + 4 - j4\sqrt{3} = 9 - j4\sqrt{3} \approx 9 - j6.93\)

Magnitude: \(|V| = \sqrt{9^2 + 6.93^2} \approx 11.33\)

Phase: \(\phi = \arctan(-6.93/9) \approx -37.6°\)

Therefore: \(v(t) = 11.33\cos(60\pi t - 37.6°)\)

8.3 Frequency Response and Bode Plots

Transfer functions in control systems and signal processing are often expressed as functions of the complex variable \(s = \sigma + j\omega\) (in the Laplace domain) or \(z = re^{j\theta}\) (in the Z-domain for discrete systems).

Example: Low-Pass Filter

A simple RC low-pass filter has the transfer function:

\[H(s) = \frac{1}{RCs + 1}\]

Substituting \(s = j\omega\) for frequency response analysis:

\[H(j\omega) = \frac{1}{RC(j\omega) + 1} = \frac{1}{1 + j\omega RC}\]

Magnitude response: \(|H(j\omega)| = \frac{1}{\sqrt{1 + (\omega RC)^2}}\)

Phase response: \(\angle H(j\omega) = -\arctan(\omega RC)\)

This complex function describes how the filter attenuates and phase-shifts sinusoidal inputs at different frequencies.

8.4 Power Calculations

The apparent power \(S\) in an AC circuit is a complex number:

\[ S = P + jQ \]

where \(P\) is the real power (measured in watts) and \(Q\) is the reactive power (measured in volt-amperes reactive).

For a load with impedance \(Z = R + jX\) and current \(I\):

\[S = VI^* = |I|^2 Z = |I|^2 (R + jX) = |I|^2 R + j|I|^2 X\]

where \(I^*\) is the complex conjugate of \(I\).

Example: Power Factor Correction

A load draws 100 kVA at a power factor of 0.8 lagging. This means:

\(S = 100e^{j\cos^{-1}(0.8)} = 100e^{j36.9°} = 80 + j60\) kVA

To improve the power factor to 0.95 lagging, a capacitor is added in parallel with the load.

The new desired apparent power is \(S' = P + jQ' = 80 + jQ'\) where \(Q'\) is calculated from:

\(\tan(\cos^{-1}(0.95)) = Q'/80\), giving \(Q' \approx 26.3\) kVAR

The capacitor must supply \(Q_C = Q - Q' = 60 - 26.3 = 33.7\) kVAR

9. Problem-Solving Strategies

9.1 General Approach to Complex Number Problems

When solving problems involving complex numbers, consider these strategies:

Choose the appropriate form: Rectangular form is best for addition/subtraction, while polar form is best for multiplication/division/powers
Convert between forms strategically: Don't convert if it makes the calculation more complicated
Use properties: Leverage properties of modulus, argument, and conjugates to simplify calculations
Visualize: Use the complex plane to gain geometric insight
Check your answer: Verify that your solution matches the original problem

9.2 Common Pitfalls and Misconceptions

Avoid these common mistakes when working with complex numbers:

Forgetting that \(\sqrt{a^2} = |a|\), not just \(a\) (important when working with complex values)
Incorrectly applying the laws of exponents with complex bases
Mixing up the rules for arguments when multiplying/dividing versus adding/subtracting
Forgetting that angles are only defined modulo \(2\pi\)
Applying real-number operations like "\(\sqrt{ab} = \sqrt{a}\sqrt{b}\)" without considering branch cuts in the complex plane

Example: Solving a Complex Equation

Solve the equation \(z^2 = i\)

Strategy: Use polar form to find the square root

Step 1: Express \(i\) in polar form \(i = e^{i\pi/2}\)

Step 2: Use the formula \(z^n = re^{i\theta} \implies z = r^{1/n}e^{i\theta/n + i2\pi k/n}\) for \(k = 0,1,...,n-1\)

Step 3: With \(n = 2\), we get \(z = e^{i\pi/4 + i\pi k}\) for \(k = 0,1\)

Step 4: This gives \(z_1 = e^{i\pi/4} = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\) and \(z_2 = e^{i5\pi/4} = -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\)

Verification: \(z_1^2 = (\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})^2 = \frac{1}{2} + i\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}} + i^2\frac{1}{2} = \frac{1}{2} + i - \frac{1}{2} = i\)

9.3 Computational Tools and Resources

Several computational tools can assist with complex number calculations:

Maple: Powerful symbolic mathematics software that handles complex numbers seamlessly
MATLAB/Python: Both provide built-in support for complex arithmetic and visualization
Scientific calculators: Many models allow direct entry and manipulation of complex numbers
Online resources: Websites like WolframAlpha can handle complex calculations and provide step-by-step solutions

10. Properties and Identities

10.1 Basic Properties

\(i^2 = -1\)
\(i^3 = -i\)
\(i^4 = 1\)
\(|z_1 \cdot z_2| = |z_1| \cdot |z_2|\)
\(|z_1 / z_2| = |z_1| / |z_2|\)
\(\arg(z_1 \cdot z_2) = \arg(z_1) + \arg(z_2)\) (modulo \(2\pi\))
\(\arg(z_1 / z_2) = \arg(z_1) - \arg(z_2)\) (modulo \(2\pi\))

10.2 Useful Identities

\(e^{i\pi} + 1 = 0\) (Euler's identity)
\(\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}\)
\(\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}\)
\(e^{i\theta} = \cos\theta + i\sin\theta\) (Euler's formula)
\(\sin(3\theta) = 3\cos^2\theta\sin\theta - \sin^3\theta\) (from De Moivre's formula)
\(\cos(3\theta) = \cos^3\theta - 3\cos\theta\sin^2\theta\) (from De Moivre's formula)

Conclusion

Complex numbers are an essential mathematical tool in engineering, particularly in electrical fields. They provide elegant solutions to problems involving oscillations, waves, and AC circuits. Understanding complex numbers and their operations is fundamental for advanced concepts in signal processing, control systems, and electromagnetic theory.

This comprehensive guide covers the key aspects of complex numbers, from basic operations to advanced applications in differential equations and electrical engineering. By mastering these concepts, you'll have a powerful mathematical tool at your disposal for solving a wide range of engineering problems.

Glossary of Terms

Complex number: A number of the form \(a + bi\) where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit
Imaginary unit: The number \(i\) where \(i^2 = -1\)
Real part: The coefficient \(a\) in the complex number \(a + bi\)
Imaginary part: The coefficient \(b\) in the complex number \(a + bi\)
Modulus: The absolute value of a complex number, equal to \(|z| = \sqrt{a^2 + b^2}\)
Argument: The angle \(\theta\) in the polar representation of a complex number
Complex conjugate: For \(z = a + bi\), the conjugate is \(\overline{z} = a - bi\)
Polar form: Representation of a complex number as \(z = re^{i\theta}\) or \(z = r(\cos\theta + i\sin\theta)\)
Euler's formula: The identity \(e^{i\theta} = \cos\theta + i\sin\theta\)
De Moivre's formula: The identity \((re^{i\theta})^n = r^ne^{in\theta} = r^n(\cos(n\theta) + i\sin(n\theta))\)
Phasor: A complex number representing a sinusoidal function in terms of magnitude and phase
Impedance: A complex number representing the opposition to current flow in an AC circuit