Functions of complex numbers

Perplexed-Girl doing math

\(\)This introduces the functions with complex arguments. The article Complex Numbers introduced a 2-dimensional number space called the complex-plane (\(\mathbb{C}\)-plane). The arithmetic functions, that we studied since first grade, gracefully extend from the one-dimensional number line onto this new \(\mathbb{C}\)-plane. Here we will introduce functions that operate on these complex numbers.

\(j\) We refer to the imaginary unit as “\(j\)”, to avoid confusion with electrical engineering, where the variable \(i\) is already used for current.

An overview of the functions is given for reference. We will proof the some of these functions in subsequent paragraphs.

Overview

Consider a complex number \(z\) expressed in either notation style

$$ z = x+jy=r\,(\cos\varphi+j\sin\varphi)=r\,\mathrm{e}^{j\varphi}\nonumber $$

As you wish

$$ \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align} z_1+z_2&=(x _1+x _2)+j\,(y _1+y _2) \\[6mu] z_1-z_2&=(x _1-x _2)+j\,(y _1-y _2) \\[6mu] z_1\,z_2&=r_1r_2\ \mathrm{e}^{j \cdot (\varphi_1+\varphi_2)} \\[6mu] \tfrac{1}{z} &= \tfrac{1}{r}\,\mathrm{e}^{-j\varphi}\\[6mu] \frac{z_2}{z_1} &= \frac{r_1}{r_2}\,\mathrm{e}^{j(\varphi _1-\varphi_2)}\\[6mu] z_1\parallelsum z_2 &= \frac{z_1\, z_2}{z_1+z_2}\\[6mu] \mathrm{e}^z &=\mathrm{e}^x\sin y + j\,\mathrm{e}^x\cos y\\[6mu] \ln z&= \ln r+j\,\varphi\\[6mu] {z_2}^{z_1} &= {r_1}^{x _2}\,\mathrm{e}^{-y_2\,\varphi_1}\,\mathrm{e}^{j \cdot (x _2\,\varphi_1+\,y_2\ln r_1)} \\[6mu] \sqrt[n]{z} &= \sqrt[n]{r}\,\mathrm{e}^{j\varphi/n} \end{align} $$

Circular based trigonometry

$$ \begin{align} \sin z &= \sin x\cosh y + j\,\cos x\sinh y \\[6mu] \cos z &= \cos x\cosh y + j\,\sin x\sinh y \\[6mu] \tan z &= \frac{\sin(2 x)}{\cosh(2 y) + \cos(2 x)} + j\,\frac{\sinh(2 y)}{\cosh(2 y) + \cos (2 x)} \\[6mu] \csc z &= {(\sin z)}^{-1} \\[6mu] \sec z &= {(\cos z)}^{-1} \\[6mu] \cot z &= {(\tan z)}^{-1} \\[6mu] \end{align} $$

Inverse circular based trigonometry

$$ \DeclareMathOperator{\asin}{asin} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\acos}{acos} \DeclareMathOperator{\atan}{atan} \DeclareMathOperator{\acsc}{acsc} \DeclareMathOperator{\asec}{asec} \DeclareMathOperator{\acot}{acot} \begin{align} \asin z &= \asin b +j\,\sgn(y)\ln\left(a + \sqrt{a^{\mathrm{e}}}-1\right), \quad a\geq1 \land b \text{ in } [\mathrm{rad}]\\[6mu] \acos z &= \acos b +j \sgn(y) \ln\left(a + \sqrt{a^{\mathrm{e}}}-1\right),\quad a\geq1 \land b \text{ in } [\mathrm{rad}]\\[6mu] \text{where}\quad a &= \tfrac{1}{2} \left( \sqrt{(x +1)^{2} + y ^{2} } + \sqrt{ (x -1)^{2} + y^{2}} \right),\nonumber \\[6mu] b &= \tfrac{1}{2} \left( \sqrt{(x +1)^{2} + y ^{2} } – \sqrt{ (x -1)^{2} + y^{2}} \right),\nonumber \\[6mu] \sgn(a) &= \begin{cases}-1 & a \lt 0\\[6mu]1 & a \geq 0\end{cases} \nonumber \\[6mu] \atan z &= \tfrac{1}{2}\left(\pi – \atan\left(\frac{1+ y}{x}\right) -\atan\left(\frac{1-y}{x}\right)\right) \\ &\quad +j\,\tfrac{1}{4}\,\ln\left( \frac{\left(\frac{1+y}{x}\right)^2 +1}{\left(\frac{1-y}{x}\right)^2 +1} \right) \\[6mu] \acsc z &= \asin(z^{-1}) \\[6mu] \asec z &= \acos(z^{-1}) \\[6mu] \acot z &= \atan(z^{-1}) \\[6mu] \end{align} $$

Hyperbolic based trigonometry

$$ \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\sech}{sech} \begin{align} \sinh z &= \cos y \sinh x + j\,\sin y\cosh x \\[6mu] \cosh z &= \cos y \cosh x + j\,\sin y\sinh x \\[6mu] \tanh z &= \frac{\sinh(2y)}{\cosh(2x)} +j\,\frac{\sin(2 y)}{\cosh(2 x) + \cos(2y)}\\[6mu] \csch z &= {(\sinh z)}^{-1} \\[6mu] \sech z &= {(\cosh z)}^{-1} \\[6mu] \coth z &= {(\tanh z)}^{-1} \\[6mu] \end{align} $$

Inverse hyperbolic based trigonometry

$$ \DeclareMathOperator{\asin}{asin} \DeclareMathOperator{\acos}{acos} \DeclareMathOperator{\atan}{atan} \DeclareMathOperator{\acsc}{acsc} \DeclareMathOperator{\asec}{asec} \DeclareMathOperator{\acot}{acot} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\asinh}{asinh} \DeclareMathOperator{\acosh}{acosh} \DeclareMathOperator{\atanh}{atanh} \DeclareMathOperator{\acsch}{acsch} \DeclareMathOperator{\asech}{asech} \DeclareMathOperator{\acoth}{acoth} \begin{align} \asinh z &= -j \asin(jz) \\[6mu] \acosh z &= j \acos z \\[6mu] \atanh z &= -j \atan(jz) \\[6mu] \acsch z &= j \acsc(jz) \\[6mu] \asech z &= -j \asec z \\[6mu] \acoth z &= j \acot(jz) \\[6mu] \end{align} $$

This table formed the basis of software like Complex Arithmetic for HP-41cv/cx.

Proofs

Without further ado, we introduce the proofs for some common complex functions

Addition

Consider adding the numbers \(z_1\) and \(z_2\) in cartesian form

$$ z_1+z_2 = (x_1+i\,y_1)+(x_2+i\,y_2) $$

Thus

$$ \shaded{z_1+z_2=(x_1+x_2)+i(y_1+y_2)} $$

This can be visualized similar to adding real numbers by putting the vectors head to tail

Visualization of complex addition

Subtraction

Consider subtracting the numbers \(z_1\) and \(z_2\) in cartesian form

$$ z_1-z_2 = (x_1+i\,y_1) – (x_2+i\,y_2) $$

So that

$$ \shaded{ z_1-z_2=(x_1-x_2)+i(y_1-y_2) } $$

This can be visualized similar to the subtraction of real numbers by rotating the subtrahend by \(\pi\) and the putting them head to tail

Visualization of complex addition

Multiplication

Consider the product \(z_1\,z_2\) in polar form, using the trig identities

$$ \begin{align} \cos\alpha\cos\beta-\sin\alpha\sin\beta&=\sin(\alpha+\beta) \nonumber \\ \sin\alpha\cos\beta+\cos\alpha\sin\beta&=\sin(\alpha+\beta) \nonumber \end{align}\nonumber $$

$$ \require{enclose} \begin{align} z_1\,z_2&=r_1\enclose{phasorangle}{\small\varphi_1}\ r_2\enclose{phasorangle}{\small\varphi_2}\\ &=r_1 (\cos\varphi_1+i\sin\varphi_1)\ r_2 (\cos\varphi_2+i\sin\varphi_2)\nonumber\\ &=r_1r_2\,((\cos\varphi_1\cos\varphi_2-\sin\varphi_1\sin\varphi_2)+i\,(\cos\varphi_1\sin\varphi_2+\sin\varphi_1\cos\varphi_2)) \end{align} $$

From what follows that

$$ \shaded{ z_1\,z_2=r_1r_2\,(\cos(\varphi_1+\varphi_2)+i\,\sin(\varphi_1+\varphi_2) } $$

where

$$ \begin{align} |z_1\,z_2|&=|z_1|\,|z_2|\\[6mu] \angle(z_1\,z_2)&=\angle z_1+\angle z_2 \end{align} $$

This can be visualized as adding the angles and multiplying the lengths of the vectors

Visualization of complex addition

Division

Consider the quotient \(\frac{z_1}{z_2}\) in polar form

$$ \require{enclose} \begin{align} \frac{z_1}{z_2}&=\frac{r_1\enclose{phasorangle}{\small\varphi_1}}{r_2\enclose{phasorangle}{\small\varphi_2}} =\frac{r_1 \left({\cos \varphi_1 + i \sin \varphi_1}\right)} {r_2 \left({\cos \varphi_2 + i \sin \varphi_2}\right)}\nonumber\\[6mu] &=\frac{r_1}{r_2}\,\frac{\cos\varphi_1+i\sin\varphi_1}{\cos\varphi_2+i\sin\varphi_2}\ \frac{\cos \varphi_2 – i \sin \varphi_2}{\cos \varphi_2 – i \sin \varphi_2},&\text{product rule}\nonumber\\[6mu] &=\frac{r_1}{r_2}\,\frac{\cos \left({\varphi_1 – \varphi_2}\right) + i \sin \left({\varphi_1 – \varphi_2}\right)}{\cos \left({\varphi_2 – \varphi_2}\right) + i \sin \left({\varphi_2 – \varphi_2}\right)}\nonumber\\[6mu] &=\frac{r_1}{r_2}\,\frac{\cos \left({\varphi_1 – \varphi_2}\right) + i \sin \left({\varphi_1 – \varphi_2}\right)} {\cos 0 + i \sin 0}\\[6mu] \end{align} $$

So that

$$ \shaded{ \frac{z_1}{z_2}=\frac{r_1}{r_2}{\Large(}{\cos \left({\varphi_1 – \varphi_2}\right) + i \sin \left({\varphi_1 – \varphi_2}\right)}{\Large)} } $$

where

$$ \begin{align} \left|\frac{z_1}{z_2}\right|&=\frac{|z_1|}{|z_2|}\\[6mu] \angle\frac{z_1}{z_2}&=\angle z_1-\angle z_2 \end{align} $$

This can be visualized as subtracting the angles and dividing the lengths of the vectors

Visualization of complex addition

\(n\)th power

Consider the power \(z^n\) in polar form, where \(n\in\mathbb{Z}^+\}\)

$$ z^n=(r(\cos\varphi+i\sin\varphi))^n\nonumber\\ $$

Using Euler’s formula

$$ \cos\phi+i\sin\phi=\mathrm{e}^{i\phi}\nonumber $$

$$ \require{enclose} \begin{align} z^n&=\left(r\enclose{phasorangle}{\small\varphi}\right)^n\\ &=\left(r\,\mathrm{e}^{i\varphi}\right)^n\nonumber\\ &=r^n\,\mathrm{e}^{in\varphi},&\text{Euler’s formula} \end{align} $$

So that

$$ \shaded{ z^n=r^n\,(\cos(n\varphi)+i\sin(n\varphi)) } $$

where

$$ \begin{align} |z^n| &= {|z|}^n\\[6mu] \angle(z^n) &= n\angle z \end{align} $$

This can be visualized multiplying the angles with \(n\) and taking the \(n\)th power of the length of the vector

Visualization of complex power with real exponent

\(n\)th root

Consider the \(n\)th root \(\sqrt[n]{z}\) in polar form, where \(n\in\mathbb{Z}^+\}\)

$$ \require{enclose} \begin{align} \sqrt[n]{z}&=\sqrt[n]{r\enclose{phasorangle}{\small\varphi}} =\left(r\,\mathrm{e}^{i\varphi}\right)^{\frac{1}{n}},&\text{Euler’s formula}\nonumber\\ &=r^{\frac{1}{n}}\,\mathrm{e}^{i(\varphi+2k\pi)/n},\quad k\in\mathbb{Z},&\text{Euler’s formula} \end{align} $$

Therefore

$$ \shaded{ \sqrt[n]{z}=\sqrt[n]{r}\,\left(\cos\frac{\varphi+2k\pi}{n}+i\sin\frac{\varphi+2k\pi}{n}\right),\quad k\in\mathbb{Z} } \label{eq:root} $$

where

$$ \begin{align} |\sqrt[n]{z}|&=\sqrt[n]{|z|},\\[6mu] \angle\,\sqrt[n]{z}&=\frac{\angle z+2k\pi}{n},\quad k\in\mathbb{Z} \end{align} $$

This can be visualized dividing the angles by \(n\) and taking the \(n\)th root of the length of the vector. The other vectors will be separated by \(\frac{2\pi}{n}\) radians.

Visualization of complex root with real exponent

Wait a minute

Depending on how we measure the angle \(\varphi\), we get different answers? Correct, because adding \(2k\pi\) to \(\varphi\) still maps to the same complex number, but may give a different function value.

In comparison, the functions that we saw described do not produce different results when adding extra rotations to the angle. Other multivalued functions are \(\log{z}\), \(\mathrm{arcsin}z\) and \(\mathrm{arccos}z\).

In general:

the \(n\)th root has \(n\) values,
because when we add \(2k\pi\) to the angle \(\varphi\), for \(k\in\mathbb{Z}\), we may get different results.

The big question becomes: how do we define the angle \(\varphi\)?

Different ways of measuring \(\varphi\)

Multi-valued

Even real valued functions can have multiple values. Remember \(\sqrt{1}=\{-1,1\}\)? Using equation \(\eqref{eq:root}\), we find the function values that we are familiar with.

$$ \begin{align} \sqrt{1}&=\sqrt{\cos\varphi+i\sin\varphi},&\text{polar notation}\nonumber\\ &=\cos\frac{\varphi+2k\pi}{2}+i\sin\frac{\varphi+2k\pi}{2},&\text{equation }\eqref{eq:root}\nonumber\\ &=\cos\frac{\varphi+2k\pi}{2},&k\in\mathbb{Z}\nonumber\\ &=\left\{1,-1\right\} \end{align} $$

all roots have magnitude \(1\), but their angles \(\varphi\) are \(\pi\) apart.

Similarly, the cube root \(\sqrt[3]{1}\) has three roots, two of which are complex. All roots have magnitude \(1\), but their angles \(\varphi\) are \(\frac{2\pi}{3}\) apart.

$$ \require{enclose} \begin{align} y_1&=1\,\enclose{phasorangle}{0}=\cos0+i\sin0=1\nonumber\\[8mu] y_2&=1\,\enclose{phasorangle}{\small\tfrac{2\pi}{3}}=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}=-\tfrac{1}{2}+\tfrac{1}{2}\sqrt{3}i\nonumber\\ y_3&=1\,\enclose{phasorangle}{\small\tfrac{4\pi}{3}}=\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}=-\tfrac{1}{2}-\tfrac{1}{2}\sqrt{3}i\nonumber \end{align}\nonumber $$

Making it single-valued

For real valued arguments, we conventionally choose \(\varphi\) in the range \([0,2\pi)\) where the function is single-valued and where we find a positive function value. This default single-value is called the principal value.

Besides that the function \(\sqrt[n]{z}\) is not differentiable at \(0\), it has no discontinuities. To make the function single-valued, we can limit the range of \(\varphi\) similar to what we usually do for real valued arguments. The technical term for this is branch cut. We then only express \(\varphi\) so that it doesn’t cross the branch cut. Some common branch cuts are shown in the table below. In the table \(\mathbb{R}^-\) stands for the negative real axis.

Example branch cuts and their effect on the function value
Branch cut Range for \(\varphi\) Effect Consistent with
just under \(\mathbb{R}^-\) \((-\pi,\pi]\) \(\Re(z)\geq0\) Sqrt of real numbers
just under \(\mathbb{R}^+\) \([0,2\pi)\) \(\Im(z)\geq0\) Phase shift in waves

No matter where you define the branch cut, when \(z\) approaches a point on the branch cut from opposite sides, either the real or imaginary part of the function value abruptly changes signs. In practice, the best place for the branch cut depends on the application. For instance, it there is already a discontinuity at the point \(-1\), you may as well put the branch cut just under \(\mathbb{R}^-\).

Real and Imaginary part of \(\sqrt{z}\) as function of \(\varphi\)

We will use the \(\mathbb{C}\)-plane extensively as we explore the physic fields of electronics and domain transforms.

Complex numbers

Perplexed-Girl doing math

\(\)Instead of projecting the future merits of complex numbers, we will introduce them in an intuitive way. We draw a parallel to negative numbers that have been universally accepted around the same time.

We start this writing with a review of concepts that should be evident. Nevertheless, I encourage you to read through them, as we build on these concepts while introducing complex numbers.

Review

Arithmetic gives us tools to manipulate numbers. It allows us to transform one number into another using transformations such as negation, addition, subtraction, multiplication and division.

Positive numbers

In first grade, we learned the concept of the number line and how numbers can be represented by vectors starting at \(0\). We visualized addition by putting these vectors head to tail, where the net length and direction is the answer.

Number line animation for \(5+3=8\)

Soon thereafter, we learned how to subtract numbers by rotating the subtrahend (the value that you subtract) vector and then putting the head to tail.

Number line animation for \(5-3=2\)

We will expand on this as we discuss negative numbers and imaginary numbers. Before we introduce such numbers, let’s also refresh on the concept of equations with squares and square roots.

Square and square root

When we solve the equation \(2x^2=8\), we look for a transformation (\(\times x\)) that, when applied twice, turns the number \(2\) into \(8\).

$$ 2 \times x \times x = 8 $$

As shown in the animation below, the two solutions \(x=2\) and \(x=-2\), both satisfy the equation \(2x^2=8\).

Number line animation for \(2 \times 2 \times 2=8\)

Negative numbers

Negative numbers have lingered around since 200 BC, but with mathematics based on geometrical ideas such as length and count, there was little place for negative numbers. After all, how can a pillar be less than nothing in height? How could you own something less than nothing?

Even a hundred years after the invention of algebra in 1637, the answer to \(4=4x+20\) would be thought of as absurd as illustrated by the quotes:

Negative numbers darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple. Francis Maseres, British mathematician (1757)
To really obtain an isolated negative quantity, it would be necessary to cut off an effective quantity from zero, to remove something of nothing: impossible operation. Lazare Carnot, French mathematician (1803)

Acceptance

Some mathematicians in the 17th century discovered that negative numbers did have their use in solving cubic and quadratic equations, provided they didn’t worry about the meaning of these negative numbers. While the intermediate steps of their calculations may involve negative numbers, the solutions were typically real positive numbers.

Only in the 19th century were negative numbers truly accepted when mathematicians started to approach mathematics in terms of logical definitions.

Physical meaning has given way to algebraic use.

The English mathematician, John Wallis (1616 – 1703) is credited with giving meaning to negative numbers by inventing the number line. Even today, the number line helps students as they actively construct mathematical meaning, number sense and an understanding of number relationships. We learned to use negative numbers without a thinking about the thousands of years it took to develop the principle.

You were probably introduced to the concepts of absolute value and direction through the geometric representation on this number line. With negative numbers so embedded in our mathematics, we accept the solution to \(3-5\) without a second thought.

Number line animation for \(3-5=-2\)

In general, we learned that the negative symbol represents the “opposite” of a number. To change the sign of a number, we rotate its vector 180 degrees (\(\pi\)) around the point \(0\). We will extend on this important concept as we introduce imaginary numbers.

Imaginary numbers

First off, imaginary numbers are called such because for a long time they were poorly understood and regarded by some as fictitious. Even Euler, who used them extensively, once wrote:

Because all conceivable numbers are either greater than zero or less than 0 or equal to 0, then it is clear that the square roots of negative numbers cannot be included among the possible numbers [real numbers]. Consequently we must say that these are impossible numbers. And this circumstance leads us to the concept of such number, which by their nature are impossible, and ordinarily are called imaginary or fancied numbers, because they exist only in imagination. Leonhard Euler, Swiss mathematician and physicist, Introduction to Algebra pg.594

History

Around the same time that mathematicians were struggling with the concept of negative numbers, they also came across square roots of negative numbers.

Girolamo Cardano, an Italian mathematician who published the solution to cubic and quartic equations, studied a problem that in modern algebra would be expressed as \(x^2-10x+40=0\,\land\,y=10-x\)

Find two numbers whose sum is equal to 10 and whose product is equal to 40. Translation of Ars Magna chapter 37, pg 219 (1545) by Girolamo Cardano

Cardano states that it is impossible to find the two numbers. Nonetheless, he says, we will proceed. He goes on to provide two objects that satisfy the given condition. Cardano found a sensible answer by working through his algorithm, but he called these numbers “ficticious” because not only did they disappear during the calculation, but they did not seem to have any real meaning.

This subtlety results from arithmetic of which this final point is as I have said as subtle as it is useless. Translation of Ars Magna chapter 37, pg 219 (1545) by Girolamo Cardano

Acceptance

In 1849, Carl Friedrich Gauß, produced a rigorous proof for complex numbers what gave a big boost to the acceptance of these numbers in the mathematical community.

With this historic perspective, let’s see how these imaginary numbers fit into what we know about number theory.

We learned that the negative symbol represents the “opposite” of a number. On the number line we can represent this by rotating the vector that represents the number 180° (\(\pi\)) around the origin \(0\).

Let’s dive straight in and consider the equation

$$ \begin{align} x^2&=-1\\ \Rightarrow\quad 1\times x\times x&=-1\label{eq:1tomin1} \end{align} $$

What multiplication with \(x\), when applied twice, turn \(1\) into \(-1\)? Multiplying twice by a positive number gives a positive result. Same for a negative number.

Time to take a step back: we said that a negation represents a rotation of \(\pi\) around the origin. What if we rotate the vector \(1\) by \(\frac{\pi}{2}\) twice, and worry about its meaning later.

Number line animation multiplying \(1\) twice so that we get \(-1\)

Indeed, twice rotating the vector \(1\) around the origin by \(\frac{\pi}{2}\) gives us \(-1\). All that we have left to do, is to find a name for the vector where \(1\) is rotated by \(\frac{\pi}{2}\). To credit Euler’s “imaginary or fancied numbers”, we call it \(i\). The first multiplication turns \(1\) into \(i\), and the second multiplication turns \(i\) into \(-1\).

More over, rotating in the opposite direction works as well. The first multiplication turns \(1\) into \(-i\), and the second multiplication turns \(-i\) into \(-1\). So there are two square roots of \(-1\): \(i\) and \(-i\). We have two solutions to \(x^2=-1\)!

Alternate number line animation multiplying \(1\) twice so that we get \(-1\)

Interpretation of “\(i\)”

What is the meaning of this mysterious value \(i\)? The first rotation turned \(1\) into \(i\), so the rotation is a visualization of multiplying with \(i\). Rotating \(i\) once more turns \(i\) into \(-1\).

Number line animation for \(1 \times i \times i=-1\)

Substituting \(x=i\) in \(\eqref{eq:1tomin1}\) implies that \(i\times i=-1\) or written as

$$ \shaded{ i^2=-1 } $$

By introducing an axis perpendicular to the number line, we have extended or number space to a two dimensional plane called \(\mathbb{C}\). This \(\mathbb{C}\)-plane includes the real numbers from the real number line, along with imaginary numbers on the \(i\)-axis and every combination thereof. We call this new number set “Complex Numbers“.

Notation

By introducing \(i\), we have added a dimension to the number line. With that come three new notation forms that each have their own use case. We will express the complex number \(z\) as shown below in these forms.

Point in complex plane as cartesian, polar and exponential form
Relation between cartesian, polar and exponential forms

Cartesian form

Each complex number has a real part \(x\) and an imaginary part \(y\), where \(x\) and \(y\) are real numbers. Using \(i\) as the imaginary unit, we can denote a complex number \(z\) as

$$ \shaded{z=x+iy} $$

The point \(z\) can be specified by its rectangular coordinates \((x,y)\), where \(x\) and \(y\) are the signed distances to the imaginary \(y\) and real \(x\)-axis. This \(xy\)-plane is commonly called the complex plane \(\mathbb{C}\).

This cartesian form is a logical extension of the number line and will prove useful when adding or subtracting complex numbers.

Polar form

The same point \(z\) can be specified by its polar coordinates \((r,\varphi)\), where \(r\) is the distance to the origin and \(\varphi\) is the angle of the vector, in radians, with the positive \(x\)-axis. With \(r\in\mathbb{R}^+\) and \(\varphi\in\mathbb{R}\), we can describe point \(z\) as

$$ \require{enclose} \shaded{ r\enclose{phasorangle}{\small\varphi}=r\,(\cos\varphi+i\sin\varphi) } $$

here \(r\) corresponds to modulus \(|z|\), and \(\varphi\) is called the argument. The value \(z=0\) was excluded because the angle \(\varphi\) is not defined that that point.

The polar form simplifies the arithmetic when used in multiplication or powers of complex numbers.

From the illustration, it is clear how to convert from cartesian to polar form

$$ \shaded{ \begin{align} x&=r\cos\varphi\nonumber\\ y&=r\sin\varphi\nonumber \end{align}} $$ and back $$ \shaded{\begin{align} r&=\sqrt{x^2+y^2}\nonumber\\[6mu] \varphi&=\mathrm{atan2}(y,x)\nonumber\\[10mu] \end{align} } $$

Here \(\mathrm{atan2(y,x)}\) prevents negative signs in \(\arctan\frac{y}{x}\) from canceling each other out. Otherwise, we would not be able to distinguish \(\varphi\) in the 1st from that in the 3rd quadrant, or \(\varphi\) in the 2nd from that in the 4th.

$$ \begin{align} \mathrm{atan2}(y,x) &= \begin{cases} \arctan\left(\frac{y}{x}\right) & x\gt0\nonumber\\ \arctan\left(\frac{y}{x}\right)+\pi & x\lt0 \land y\geq0\nonumber\\ \arctan\left(\frac{y}{x}\right)-\pi & x\lt0 \land y\lt0\nonumber\\ \frac{\pi}{2} & x= 0 \land y\gt0\nonumber\\ -\frac{\pi}{2} & x= 0 \land y\lt0\nonumber\\ \text{undefined} & x= 0 \land y = 0\nonumber \end{cases} \end{align} $$

Consider complex number \(z\) with angle \(\varphi_0\). If you make any integer number of rotation rotate around the origin, you will be back the your initial starting point. Since a full rotation corresponds to an angle of \(2\pi\), the same point \(z\) can be described as

$$ \require{enclose} r\enclose{phasorangle}{\small\varphi+2k\pi}=r\,(\cos(\varphi+2k\pi)+i\sin(\varphi+2k\pi)),\quad k\in\mathbb{Z} $$

We will the effects of this further when discussing multi-valued functions such as square root.

Exponential form

Euler’s formula was introduced in a separate write-up as

$$ \mathrm{e}^{i\varphi} = \cos\varphi+j\sin\varphi \nonumber $$

Using Euler’s formula we can rewrite the polar form of a complex number into its exponential form

$$ \shaded{ z=r\,\mathrm{e}^{i\varphi} } $$

Similar to the polar form, the angle can be expressed in infinite different ways

$$ \begin{align} z &= r\,\mathrm{e}^{i(\varphi+2k\pi)}, & k\in\mathbb{Z} \end{align} $$

This exponential form is often preferred over the polar form, because it reduces the need for trigonometry.

What next?

My follow-up article Complex Functions introduces functions that operate on complex numbers. Such functions include addition, subtraction, multiplication to the most obscure trigonometry functions.

We will use the \(\mathbb{C}\)-plane extensively as we explore electronics and domain transforms. From here on

we will refer to the imaginary unit as “\(j\)”, to avoid confusion with electronics where the variable \(i\) is already used for electrical current.

The same Leonard Euler that once said these numbers “only exist in our imagination” also used imaginary numbers to unite trigonometry and analysis in his most beautiful formula.

Copyright © 1996-2022 Coert Vonk, All Rights Reserved