Chapter 8 — The Imaginary Made Useful

This chapter introduces complex numbers and phasors as essential mathematical tools for analyzing AC circuits, where voltages and currents vary sinusoidally with time. It explains complex numbers, their rectangular and polar forms, and the algebraic operations of addition, subtraction, multiplication, and division. The chapter then introduces phasors, demonstrating how they represent sinusoidal signals in compact form and convert time-varying differential equations into algebraic equations. It also covers the visualization of phasors as rotating vectors and the calculation of time averages of time-harmonic signals.

Learning Objectives:

• Define complex numbers and explain their representation in rectangular and polar forms.
• Perform the basic algebraic operations (addition, subtraction, multiplication, division) with complex numbers.
• Explain the concept of a phasor and how it represents a sinusoidal signal.
• Convert between time-harmonic signals and their phasor representations.
• Use phasors to simplify AC circuit analysis, including differentiation and time averaging.

Up until now, the circuits examined in this book have been DC circuits, where voltages and currents remain constant over time. In the real world, however, many signals change with time, and a particularly important family of time-varying signals takes the form of sine and cosine waves. The electrical power delivered to a home is one familiar example: it is an AC (alternating current) signal that oscillates at 60 Hz in the United States.

How does one analyze circuits driven by oscillating signals? This is where complex numbers and phasors become indispensable. They are the mathematical tools that bring AC circuit analysis within reach.

Despite their name, complex numbers actually simplify work with AC signals. The complex plane is similar to the XY plots used when measuring IV curves in the laboratory, and that familiarity will help in visualizing the concepts that follow.

Sinusoidal signals and the Fourier Transform appeared in the Signals in Time and Signals in Frequency chapters, where signals were decomposed into sine and cosine components. The Circuits that Remember chapter showed how inductors and capacitors respond during transient events. Complex numbers and phasors tie those concepts together and provide a unified framework for understanding circuit behavior when signals vary continuously with time.

This chapter introduces complex numbers as a mathematical tool, then shows how they represent sinusoidal signals as phasors: rotating vectors that reduce AC circuit analysis to straightforward algebra. Instead of solving differential equations at every step, it becomes possible to apply algebraic operations and obtain results in the phasor domain, converting back to the time domain only at the end.

Section 1 — Complex Numbers

A complex number, typically written as z, combines two parts: a real part and an imaginary part. The imaginary part uses the special number $j$ (or $i$ in mathematics), defined as $j = \sqrt{-1}$. Although the term "imaginary" can seem off-putting, its applications in engineering are entirely real.

A complex number can be visualized in the complex plane. The horizontal axis represents real numbers and the vertical axis represents imaginary numbers. Think of a complex number as a vector[^fn1] starting at the origin and pointing to the coordinate $(a, b)$, where $a$ is the real part and $b$ is the imaginary part. That vector has two key characteristics:

[^fn1]: A vector is a quantity that has both magnitude and direction, represented by an arrow of a specific length pointing in a specific direction.

• Its length (magnitude) $r$, given by $r = \sqrt{a^{2} + b^{2}}$ (the Pythagorean theorem).
• Its angle to the positive real axis, $\theta$, given by $\theta = \tan^{-1}(b/a)$.

Figure: A complex number $z = a + jb = re^{j\theta}$ represented as a vector in the complex plane.

There are two common ways to represent a complex number, and proficiency with both is essential.

1. Rectangular form: The most direct representation,
   $$z = a + jb,$$
   where $a$ is the real part and $b$ is the imaginary part. For example, $3 + j4$ has a real part of 3 and an imaginary part of 4.

2. Polar form: A representation using magnitude and angle,
   $$z = re^{j\theta},$$
   where $r$ is the magnitude and $\theta$ is the angle. This form rests on Euler's formula:
   $$e^{j\theta} = \cos\theta + j\sin\theta.$$
   Here $e$ is the base of the natural logarithm (approximately 2.718). The polar form can also be written using the angle symbol: $r\angle\theta$. For example, $20e^{j\pi/4} = 20e^{j45^\circ} = 20\angle 45^\circ$.

The connection between complex exponentials and sinusoidal functions is precisely what makes complex numbers the right tool for representing AC signals.

Connecting the two forms: a complex number can be written as

$$z = a + jb = r\cos\theta + jr\sin\theta = re^{j\theta}.$$

The rectangular and polar forms are related by:

• Real part: $\mathcal{R}e\{z\} = a = r\cos\theta$
• Imaginary part: $\mathcal{I}m\{z\} = b = r\sin\theta$
• Magnitude: $r = \sqrt{a^2 + b^2}$
• Angle: $\theta = \tan^{-1}(b/a)$

Practical Tip: Learn to convert between rectangular and polar forms efficiently using a calculator. Most scientific calculators include a rectangular-to-polar conversion (often labeled R→P or x→r) and a polar-to-rectangular conversion (labeled P→R or r→x). Mastering these conversions will save time on homework and exams.

Example 1: Converting to Polar Form

Express the complex number $z = 3 + j4$ in polar form.

Solution:

Magnitude:
$$r = |z| = \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5.$$

Angle:
$$\theta = \tan^{-1}\!\left(\frac{b}{a}\right) = \tan^{-1}\!\left(\frac{4}{3}\right) \approx 0.93\,\text{radians} = 53.13^\circ.$$

Polar form using Euler's formula:
$$z = 5e^{j\,\tan^{-1}(4/3)} \approx 5e^{j\,0.93} = 5\angle 53.13^\circ.$$

Complex Conjugate

The complex conjugate of a number $z$, written $z^{*}$, is formed by changing the sign of the imaginary part:

$$z^{*} = (a + jb)^{*} = a - jb.$$

Geometrically, this is a reflection across the real axis in the complex plane. In polar form, the complex conjugate is:

$$z^{*} = \bigl(re^{j\theta}\bigr)^{*} = re^{-j\theta}.$$

Multiplying a complex number by its conjugate yields the square of the magnitude:

$$z \times z^{*} = (a + jb)(a - jb) = a^2 + b^2 = r^2,$$

so that $r = \sqrt{z\,z^{*}}$. This property will be particularly useful when computing power in AC circuits.

Example 2: Complex Conjugate

(a) What is the complex conjugate of $z = 3 + j4$?
$$z^* = 3 - j4.$$

(b) Find the magnitude of $z = 3 + j4$ using the complex conjugate.
$$z\,z^{*} = (3 + j4)(3 - j4) = 9 + 16 = 25, \qquad r = \sqrt{25} = 5.$$

(c) What is the complex conjugate of $z = 5e^{j\,\tan^{-1}(4/3)}$?
$$z^* = 5e^{-j\,\tan^{-1}(4/3)}.$$

Section 2 — Algebraic Operations with Complex Numbers

The two representations of a complex number each lend themselves to different operations. Addition and subtraction are most natural in rectangular form; multiplication and division are most natural in polar form.

Addition and Subtraction

Addition and subtraction follow directly in rectangular form: the real and imaginary parts are combined separately, just as vector components are combined in physics:

$$z_1 + z_2 = (a + jb) + (c + jd) = (a + c) + j(b + d),$$
$$z_1 - z_2 = (a + jb) - (c + jd) = (a - c) + j(b - d).$$

For example, $(3 + j4) + (2 - j5) = (3 + 2) + j(4 - 5) = 5 - j1$.

Multiplication

For multiplication, polar form is the more efficient choice. Multiplying two complex numbers in polar form requires only:

1. multiplying their magnitudes, and
2. adding their angles:

$$z_1 \cdot z_2 = r_1 e^{j\theta_1} \cdot r_2 e^{j\theta_2} = r_1 r_2\,e^{j(\theta_1 + \theta_2)}.$$

For example, $2e^{j30^\circ} \cdot 3e^{j45^\circ} = 6e^{j75^\circ}$.

Multiplication is also possible in rectangular form using the distributive property and the key identity $j^2 = -1$:

$$z_1 \cdot z_2 = (a + jb)(c + jd) = ac + jad + jbc + j^2 bd = (ac - bd) + j(ad + bc).$$

This rectangular approach requires more steps, which is why polar form is preferred for multiplication.

Division

Division in polar form is equally direct: divide the magnitudes and subtract the denominator's angle from the numerator's angle:

$$\frac{z_1}{z_2} = \frac{r_1 e^{j\theta_1}}{r_2 e^{j\theta_2}} = \frac{r_1}{r_2}\,e^{j(\theta_1 - \theta_2)}.$$

For example, $\dfrac{10e^{j60^\circ}}{2e^{j20^\circ}} = 5e^{j40^\circ}$.

In rectangular form, division requires eliminating the imaginary number from the denominator by multiplying numerator and denominator by the complex conjugate of the denominator:

$$\frac{z_1}{z_2} = \frac{a + jb}{c + jd} = \frac{(a + jb)(c - jd)}{(c + jd)(c - jd)} = \frac{(ac + bd) + j(bc - ad)}{c^2 + d^2}.$$

The denominator $c^2 + d^2$ is real, equal to the square of the magnitude of $z_2$.

Useful Shortcut: Dividing by $j$

A frequently encountered operation in circuit analysis is division by $j$. The result is:
$$\frac{1}{j} = -j.$$
This follows from multiplying numerator and denominator by $-j$:
$$\frac{1}{j} \cdot \frac{-j}{-j} = \frac{-j}{-j^2} = \frac{-j}{1} = -j,$$
or equivalently from polar form:
$$\frac{1}{j} = \frac{1}{e^{j\pi/2}} = e^{-j\pi/2} = -j.$$

Example 3: Multiplication and Division

Given $z_1 = 1 + j2$ and $z_2 = -2 - j$, calculate $z_1 \cdot z_2$ and $z_1 / z_2$.

Solution:

First, convert to polar form:
$$z_1 = 1 + j2 = \sqrt{5}\,e^{j\,1.107}, \qquad z_2 = -2 - j = \sqrt{5}\,e^{-j\,2.678}.$$

Multiplication in rectangular form:
$$z_1 \cdot z_2 = (1 + j2)(-2 - j) = -2 - j - j4 + 2 = -j5.$$

Multiplication in polar form:
$$z_1 \cdot z_2 = \bigl(\sqrt{5}\,e^{j\,1.107}\bigr)\bigl(\sqrt{5}\,e^{-j\,2.678}\bigr) = 5\,e^{-j\,1.571} = -j5.$$

Note: it is helpful to recall that $\pi \approx 3.142$, so $\pi/2 \approx 1.571$.

Division in rectangular form:
$$\frac{z_1}{z_2} = \frac{1 + j2}{-2 - j}\cdot\frac{-2 + j}{-2 + j} = \frac{-4 - j3}{5} = -0.8 - j0.6.$$

Division in polar form:
$$\frac{z_1}{z_2} = \frac{\sqrt{5}\,e^{j\,1.107}}{\sqrt{5}\,e^{-j\,2.677}} = e^{j\,3.784} = -0.8 - j0.6.$$

Section 3 — Phasors: A Compact Representation for AC Signals

The real power of complex numbers in electrical engineering comes from their role in representing sinusoidal signals. This section introduces phasors, the concept that bridges DC and AC circuit analysis and makes AC analysis almost as straightforward as the DC analysis covered in earlier chapters.

In the work on Fourier analysis and signal processing, sinusoidal signals appeared repeatedly as the fundamental building blocks of more complex waveforms. In AC circuit analysis, these sinusoidal signals take center stage, and they have a special name: time-harmonic signals.

What is a phasor? A phasor is a complex number that represents a sinusoidal voltage or current in compact form. Instead of working with the full time-varying function at every stage, all essential information is captured in a single complex number.

Why use phasors? They simplify AC circuit analysis in three important ways:

• Time-varying differential equations become algebraic equations.
• Multiple sinusoidal signals of the same frequency can be combined easily.
• Circuit behavior becomes easier to visualize and understand.

Begin with a time-harmonic signal (a sinusoid at constant frequency):

$$v(t) = V_o \cos(\omega t + \phi).$$

Here:

• $V_o$ is the amplitude (peak value) of the signal,
• $\omega$ is the angular frequency in radians per second ($\omega = 2\pi f$, where $f$ is frequency in hertz),
• $\phi$ is the phase offset, which shifts the wave earlier or later in time.

This signal is illustrated in the figure below.

Figure: A time-harmonic signal $v(t) = V_o\cos(\omega t + \phi)$. The signal oscillates sinusoidally; its average value over a full period is zero.

Now, using Euler's formula, the cosine can be written as:

$$v(t) = V_o\cos(\omega t + \phi) = \text{Re}\bigl\{V_o\,e^{j(\omega t + \phi)}\bigr\} = \text{Re}\bigl\{V_o\,e^{j\phi}\,e^{j\omega t}\bigr\} = \text{Re}\bigl\{\mathbf{V}\,e^{j\omega t}\bigr\}.$$

The quantity $\mathbf{V} = V_o e^{j\phi}$ is the phasor representing $v(t)$.[^fn2]

[^fn2]: When handwriting, a phasor is commonly written with a bar overhead, $\bar{V}$, rather than in bold.

The phasor $\mathbf{V}$ is a complex number that carries only the amplitude and phase information; it does not include the time-varying factor $e^{j\omega t}$. The angular frequency $\omega$ is assumed to be the same for all signals in the circuit and is therefore not stored in the phasor.

The relationship between the sinusoidal signal and its phasor is:

$$v(t) = V_o\cos(\omega t + \phi) \;\Longrightarrow\; \mathbf{V} = V_o\,e^{j\phi}.$$

To form a phasor from a sinusoidal signal:

• note the cosine convention: if the signal is given as a sine, convert it using $\sin(\theta) = \cos\!\left(\theta - \tfrac{\pi}{2}\right)$;
• take the amplitude $V_o$;
• take the phase offset $\phi$;
• combine them as $V_o e^{j\phi}$ or equivalently $V_o\angle\phi$.

To recover the time-varying signal from its phasor, multiply by $e^{j\omega t}$ and take the real part:

$$v(t) = \text{Re}\bigl\{\mathbf{V}\,e^{j\omega t}\bigr\} \;\Longleftarrow\; \mathbf{V} = V_o\,e^{j\phi}.$$

In practice, conversion back to the time domain is deferred until the very end of the analysis. All circuit operations are carried out in the phasor domain, where differential equations reduce to algebraic equations.

Visualizing Phasors

Think of a phasor as a rotating vector in the complex plane:

• The phasor $\mathbf{V} = V_o e^{j\phi}$ sits at angle $\phi$ from the positive real axis.
• As time progresses, this vector rotates counterclockwise at angular frequency $\omega$.
• The projection of the rotating vector onto the real axis gives the original cosine wave; the projection onto the imaginary axis gives the corresponding sine wave.

Figure: Phasor representation in the complex plane. As time increases, the phasor rotates counterclockwise at angular frequency $\omega$, as indicated by the red arrow. The actual time-domain signal is the projection of the rotating phasor onto the real axis, shown by the blue arrow.

This visualization explains why signals with different phase angles have different timing relationships: if one phasor leads another in angle, that signal reaches its peak earlier in time.

Working with Phasors in Practice

Example 4: Sine to Phasor Conversion

Convert the time-harmonic voltage $v(t) = 100\sin(120\pi t + \pi/3)$ to phasor form.

Step 1: Convert the sine function to a cosine using $\sin(\theta) = \cos(\theta - \pi/2)$:
$$v(t) = 100\sin\!\left(120\pi t + \tfrac{\pi}{3}\right) = 100\cos\!\left(120\pi t + \tfrac{\pi}{3} - \tfrac{\pi}{2}\right) = 100\cos\!\left(120\pi t - \tfrac{\pi}{6}\right).$$

Step 2: Identify the amplitude and phase:
- Amplitude: $100\,\text{V}$
- Phase: $-\pi/6\,\text{rad}$ ($= -30^\circ$)

Step 3: Write the phasor:
$$\mathbf{V} = 100e^{-j\pi/6} = 100\angle(-30^\circ)\;\text{V}.$$

The frequency information ($\omega = 120\pi\,\text{rad/s}$, i.e., 60 Hz) is not stored in the phasor; it is assumed to be known and shared by all signals in the circuit.

Example 5: Phasor Conversion for Current

Convert the time-harmonic current $i(t) = 10\sin(200\pi t - \pi/6)$ to phasor form.

Step 1: Convert sine to cosine:
$$i(t) = 10\sin\!\left(200\pi t - \tfrac{\pi}{6}\right) = 10\cos\!\left(200\pi t - \tfrac{\pi}{6} - \tfrac{\pi}{2}\right) = 10\cos\!\left(200\pi t - \tfrac{2\pi}{3}\right).$$

Step 2: Identify amplitude and phase:
- Amplitude: $10\,\text{A}$
- Phase: $-2\pi/3\,\text{rad}$ ($= -120^\circ$)

Step 3: Write the phasor:
$$\mathbf{I} = 10e^{-j2\pi/3} = 10\angle(-120^\circ)\;\text{A}.$$

The angular frequency $\omega = 200\pi\,\text{rad/s}$ corresponds to $f = 100\,\text{Hz}$.

Example 6: Interpreting the Phase Angle

A phasor voltage is $\mathbf{V} = 20\angle 60^\circ$. What does the angle represent?

The angle of a phasor ($\phi = 60^\circ$ here) is the phase offset of the associated sinusoidal signal. It indicates when the signal reaches its peak relative to a reference cosine wave:

• This phasor represents $v(t) = 20\cos(\omega t + 60^\circ)$.
• The positive angle means this signal leads a zero-phase reference cosine by $60^\circ$: it reaches its peak value earlier.

Conversely, $\mathbf{V} = 20\angle(-60^\circ)$ represents a signal that lags the reference by $60^\circ$, reaching its peak later.

This phase relationship is central to AC circuit analysis. In an RC circuit, for example, the current across a capacitor leads the voltage by $90^\circ$.

Example 7: Adding Phasors

Add $\mathbf{V}_1 = 10\angle 30^\circ$ and $\mathbf{V}_2 = 5\angle(-45^\circ)$. Express the result in rectangular form.

Addition is performed in rectangular form.

Step 1: Convert each phasor:
$$\mathbf{V}_1 = 10\angle 30^\circ = 10\cos 30^\circ + j\,10\sin 30^\circ = 5\sqrt{3} + j5,$$
$$\mathbf{V}_2 = 5\angle(-45^\circ) = 5\cos(-45^\circ) + j\,5\sin(-45^\circ) = \tfrac{5}{\sqrt{2}} - j\tfrac{5}{\sqrt{2}}.$$

Step 2: Add:
$$\mathbf{V}_1 + \mathbf{V}_2 = \!\left(5\sqrt{3} + \tfrac{5}{\sqrt{2}}\right) + j\!\left(5 - \tfrac{5}{\sqrt{2}}\right).$$

If these were two voltage sources in series, this sum would give the total phasor voltage. The result depends on both the amplitudes and the relative phases of the two signals.

Section 4 — Derivatives of Time-harmonic Signals

One of the most valuable properties of the phasor representation becomes apparent when taking derivatives of time-varying signals. This is especially important because capacitors and inductors are governed by differential relationships between voltage and current.

From calculus, the time derivative of a sinusoidal signal is:

$$\frac{d}{dt}\,V_o\cos(\omega t + \phi) = -\omega V_o\sin(\omega t + \phi).$$

Working directly with this result can be tedious. In the phasor domain, however, taking a derivative corresponds to a simple multiplication:

$$\frac{d}{dt}\,v(t) \;\Longrightarrow\; j\omega\,\mathbf{V}.$$

Whenever a time derivative of a sinusoidal signal is required, it suffices to multiply its phasor by $j\omega$.

Application to Circuit Components

This derivative property is central to the analysis of capacitors and inductors in the phasor domain:

For a capacitor: $i(t) = C\,\dfrac{dv(t)}{dt} \;\Longrightarrow\; \mathbf{I} = j\omega C\,\mathbf{V}$

For an inductor: $v(t) = L\,\dfrac{di(t)}{dt} \;\Longrightarrow\; \mathbf{V} = j\omega L\,\mathbf{I}$

These relationships lead directly to the concept of impedance, studied in the next chapter. The key result is that the phasor domain converts differential relationships into algebraic ones, making circuit analysis considerably more efficient.

Section 5 — Time Averages of Time-harmonic Signals

Phasors also simplify the calculation of time averages, which are essential for understanding power in AC circuits.

The average value of any periodic function $f(t)$ is found by integrating over one complete period $T$ and dividing by $T$:

$$\langle f(t) \rangle = \frac{1}{T}\int_{0}^{T} f(t)\,dt.$$

For a simple cosine wave $v(t) = V_o\cos(\omega t)$, this average is zero:

$$\langle v(t) \rangle = \frac{1}{T}\int_{0}^{T} V_o\cos(\omega t)\,dt = 0.$$

This result is expected: the positive and negative half-cycles cancel exactly, as shown in the harmonic cosine figure above.

Power in electrical circuits is proportional to the square of voltage or current: for a resistor, $p(t) = v^2(t)/R$. Squaring the cosine gives:

$$v(t)^2 = \bigl[V_o\cos(\omega t)\bigr]^2 = \frac{V_o^2}{2}\bigl[\cos(2\omega t) + 1\bigr].$$

The squared signal has a non-zero DC offset (the constant term in the brackets), as illustrated below. This is why AC power is still useful despite the voltage averaging to zero.

Figure: The squared time-harmonic signal $v(t)^2 = [V_o\cos(\omega t)]^2$. The non-zero average value is clearly visible.

The time average of the squared signal is:

$$\langle v(t)^2 \rangle = \frac{1}{T}\int_{0}^{T} v(t)^2\,dt = \frac{V_o^2}{2T}\int_{0}^{T}\bigl[\cos(2\omega t) + 1\bigr]\,dt = \frac{V_o^2}{2}.$$

Phasors allow this result to be obtained without integration. Given the phasor $\mathbf{V}$ of $v(t)$:

$$\langle v(t)^2 \rangle = \frac{V_o^2}{2} = \frac{\mathbf{V}\,\mathbf{V}^*}{2}.$$

Example 8: Average Power in a Resistor

A periodic voltage $v(t) = V_o\sin(\omega t + \phi)$ is applied to a resistor $R$. Find the average power dissipated.

Instantaneous power:
$$p(t) = \frac{v(t)^2}{R}.$$

Average power, using the phasor result $\langle v(t)^2 \rangle = V_o^2/2$:
$$\langle p \rangle = \frac{\langle v(t)^2 \rangle}{R} = \frac{V_o^2}{2R}.$$

For $v(t) = 10\sin(\omega t + \pi/3)$ applied to $R = 100\,\Omega$:
$$\langle p \rangle = \frac{10^2}{2 \times 100} = \frac{100}{200} = 0.5\,\text{W}.$$

This result also defines the RMS (root mean square) value of a sinusoid: since $\langle p \rangle = V_\text{rms}^2/R$ and $\langle p \rangle = V_o^2/(2R)$, it follows that $V_\text{rms} = V_o/\sqrt{2}$. The RMS value is the equivalent DC voltage that would deliver the same average power to the resistor.

Chapter Summary

This chapter has introduced two mathematical tools that are essential for AC circuit analysis: complex numbers and phasors. Complex numbers provide a compact algebraic framework for representing sinusoidal signals, while phasors exploit Euler's formula to convert time-varying differential equations into algebraic equations. Together, they extend the DC analysis techniques of earlier chapters to circuits driven by sinusoidal sources.

The key operations and their preferred forms are:

• Addition and subtraction: use rectangular form.
• Multiplication and division: use polar form.
• Conjugate: change the sign of the imaginary part (or negate the angle in polar form).

The phasor domain is especially powerful because differentiation with respect to time becomes multiplication by $j\omega$, eliminating differential equations from routine circuit analysis. Time averages of squared signals, which determine average power, follow directly from the phasor without integration.

The next chapter applies these tools to circuits containing resistors, capacitors, and inductors driven by sinusoidal sources, introducing the concept of impedance and the analysis of frequency-dependent circuit behavior.

Key Formulas — Chapter 8