In this section, we return to our study of complex numbers which were first introduced in Section ComplexZeros. Recall that a complex number is a number of the form \(z = a + bi\) where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit defined by \(i = \sqrt {-1}\).

The number \(a\) is called the real part of \(z\), denoted \(\text {Re}(z)\), while the real number \(b\) is called the imaginary part of \(z\), denoted \(\text {Im}(z)\). From Intermediate Algebra, we know that if \(z = a + bi = c + di\) where \(a\), \(b\), \(c\) and \(d\) are real numbers, then \(a = c\) and \(b = d\), which means \(\text {Re}(z)\) and \(\text {Im}(z)\) are well-defined. ^[1] (‘Well-defined’ means that no matter how we express \(z\), the number \(\text {Re}(z)\) is always the same, and the number \(\text {Im}(z)\) is always the same. In other words, \(\text {Re}\) and \(\text {Im}\) are functions of complex numbers.)

To start off this section, we associate each complex number \(z = a+bi\) with the point \((a,b)\) on the Cartesian (rectangular) coordinate plane. In this case, the \(x\)-axis is relabeled as the real axis, which corresponds to the real number line as usual, and the \(y\)-axis is relabeled as the imaginary axis, which is demarcated in increments of the imaginary unit \(i\). The plane determined by these two axes is called the complex plane.

Since the ordered pair \((a,b)\) gives the rectangular coordinates associated with \(z = a+bi\), the expression \(z=a+bi\) is called the rectangular form of the complex number \(z\).

We could just as easily associate \(z\) with a pair of polar coordinates \((r,\theta )\). Although it is not as straightforward as the definitions of \(\text {Re}(z)\) and \(\text {Im}(z)\), we give \(r\) and \(\theta \) special names in relation to \(z\) below.

Some remarks about Definition modulusargumentdefn are in order. We know from Section PolarCoordinates that every point in the plane has infinitely many polar coordinate representations \((r,\theta )\) which means it’s worth our time to make sure the quantities ‘modulus’, ‘argument’ and ‘principal argument’ are well-defined.

Concerning the modulus, if \(z = 0\) then the point associated with \(z\) is the origin. In this case, the only \(r\)-value which can be used here is \(r=0\). Hence for \(z= 0\), \(|z| = 0\) is well-defined.

If \(z \neq 0\), then the point associated with \(z\) is not the origin, and there are two possibilities for \(r\): one positive and one negative. However, we stipulated \(r \geq 0\) in our definition so this pins down the value of \(|z|\) to one and only one number. Thus the modulus is well-defined in this case, too. ^[2] (In case you’re wondering, the use of the absolute value notation \(|z|\) for modulus will be explained shortly.)

Even with the requirement \(r \geq 0\), there are infinitely many angles \(\theta \) which can be used in a polar representation of a point \((r,\theta )\). If \(z \neq 0\) then the point in question is not the origin, so all of these angles \(\theta \) are coterminal. Since coterminal angles are exactly \(2\pi \) radians apart, we are guaranteed that only one of them lies in the interval \((-\pi , \pi ]\), and this angle is what we call the principal argument of \(z\), \(\text {Arg}(z)\).

The set \(\text {arg}(z)\) of all arguments of \(z\) can be described as \(\text {arg}(z) = \left \{ \text {Arg}(z) + 2\pi k \, | \, \text {$k$ is an integer} \right \}\). Note that since \(\text {arg}(z)\) is a set, we will write ‘\(\theta \in \text {arg}(z)\)’ to mean ‘\(\theta \) is in ^[3] (Recall the symbol being used here, ‘\(\in \),’ is the mathematical symbol which denotes membership in a set. See Section AppSetTheory.) the set of arguments of \(z\)’.

If \(z=0\) then the point in question is the origin, which we know can be represented in polar coordinates as \((0,\theta )\) for any angle \(\theta \). In this case, we have \(\text {arg}(0) = (-\infty , \infty )\) and since there is no one value of \(\theta \) which lies \((-\pi , \pi ]\), we leave \(\text {Arg}(0)\) undefined. It is time for an example.

Now that we’ve had practice computing the modulus of a complex number, we state some properties below.

To prove the first three properties in Theorem modprops, suppose \(z = a + bi\) where \(a\) and \(b\) are real numbers. To determine \(|z|\), we find a polar representation \((r,\theta )\) with \(r \geq 0\) for the point \((a,b)\).

From Section PolarCoordinates, we know \(r^2 = a^2 + b^2\) so that \(r = \pm \sqrt {a^2+b^2}\). Since we require \(r \geq 0\), then it must be that \(r = \sqrt {a^2 +b^2}\), which means \(|z| = \sqrt {a^2+b^2}\). Using the distance formula, we find the distance from \((0,0)\) to \((a,b)\) is also \(\sqrt {a^2+b^2}\), establishing the first property. ^[4] (Since the absolute value \(|x|\) of a real number \(x\) can be viewed as the distance from \(x\) to \(0\) on the number line, this first property justifies the notation \(|z|\) for modulus. We leave it to the reader to show that if \(z\) is real, then the definition of modulus coincides with absolute value so the notation \(|z|\) is unambiguous.)

For the second property, note that since \(|z|\) is a distance, \(|z| \geq 0\). Furthermore, \(|z| = 0\) if and only if the distance from \(z\) to \(0\) is \(0\), and the latter happens if and only if \(z = 0\), which is what we were asked to show. ^[5] (This may be considered by some to be a bit of a cheat, so we work through the underlying Algebra to see this is true. We know \(|z| = 0\) if and only if \(\sqrt {a^2+b^2} = 0\) if and only if \(a^2+b^2 = 0\), which is true if and only if \(a = b = 0\). The latter happens if and only if \(z = a + bi =0\). There.)

For the third property, we note that since \(a = \text {Re}(z)\) and \(b = \text {Im}(z)\), \(z = \sqrt {a^2+b^2} = \sqrt {\text {Re}(z)^2 + \text {Im}(z)^2}\).

To prove the product rule, suppose \(z = a + bi\) and \(w = c + di\) for real numbers \(a\), \(b\), \(c\) and \(d\). Then \(zw = (a+bi)(c+di)\). After the usual arithmetic ^[6] (See Example complexnumberarithmetic in Section AppCmpNums for a review of complex number arithmetic.) we get \(zw = (ac-bd) + (ad+bc)i\). Therefore,

Hence \(|zw| = |z| |w|\) as required.

Now that the Product Rule has been established, we use it and the Principle of Mathematical Induction ^[7] (See Section Induction for a review of this technique.) to prove the power rule. Let \(P(n)\) be the statement \(\left |z^{n}\right | = |z|^n\). Then \(P(1)\) is true since \(\left |z^{1}\right | = |z| = |z|^1\).

Next, assume \(P(k)\) is true. That is, assume \(\left |z^{k}\right | = |z|^k\) for some \(k \geq 1\). Our job is to show that \(P(k+1)\) is true, namely \(\left |z^{k+1}\right | = |z|^{k+1}\). As is customary with induction proofs, we first try to reduce the problem in such a way as to use the Induction Hypothesis.

Hence, \(P(k+1)\) is true, which means \(\left |z^{n}\right | = |z|^{n}\) is true for all natural numbers \(n\).

Like the Power Rule, the Quotient Rule can also be established with the help of the Product Rule. We assume \(w \neq 0\) (so \(|w| \neq 0\)) and we get

Hence, the proof really boils down to showing \(\left | \frac {1}{w} \right | = \frac {1}{|w|}\). This is left as an exercise.

Next, we characterize the argument of a complex number in terms of its real and imaginary parts.

To prove Theorem argprops, suppose \(z = a + bi\) for real numbers \(a\) and \(b\). By definition, \(a = \text {Re}(z)\) and \(b = \text {Im}(z)\), so the point associated with \(z\) is \((a,b) = \left (\text {Re}(z), \text {Im}(z)\right )\). From Section PolarCoordinates, we know that if \((r,\theta )\) is a polar representation for \(\left (\text {Re}(z), \text {Im}(z)\right )\), then \(\tan (\theta ) = \frac {\text {Im}(z)}{\text {Re}(z)}\), provided \(\text {Re}(z) \neq 0\).

If \(\text {Re}(z) = 0\) and \(\text {Im}(z) > 0\), then \(z\) lies on the positive imaginary axis. Since we take \(r > 0\), we have that \(\theta \) is coterminal with \(\frac {\pi }{2}\), and the result follows. If \(\text {Re}(z) = 0\) and \(\text {Im}(z) < 0\), then \(z\) lies on the negative imaginary axis, and a similar argument shows \(\theta \) is coterminal with \(-\frac {\pi }{2}\).

The last property in the theorem was already discussed in the remarks following Definition modulusargumentdefn.

Our next goal is to completely marry the Geometry and the Algebra of the complex numbers. To that end, consider the figure below.

We know from Theorem polarrectangularconversion that \(a = r\cos (\theta )\) and \(b = r\sin (\theta )\). Making these substitutions for \(a\) and \(b\) gives \(z = a + bi = r\cos (\theta ) + r \sin (\theta ) i = r \left [\cos (\theta ) + i \sin (\theta )\right ]\).

The expression ‘\(\cos (\theta ) + i\sin (\theta )\)’ is abbreviated \(\text {cis}(\theta )\) so we can write \(z = r\text {cis}(\theta ) = |z| \text {cis}(\theta )\).

Since there are infinitely many choices for \(\theta \in \text {arg}(z)\), there infinitely many polar forms for \(z\), so we used the indefinite article ‘a’ in Definition polarformcomplex. It is time for an example.

The following theorem summarizes the advantages of working with complex numbers in polar form.

The proof of Theorem prodquotpolarcomplex requires a healthy mix of definition, arithmetic and identities. We first start with the product rule.

We now focus on the quantity in brackets on the right hand side of the equation.

Putting this together with our earlier work, we get \(zw = |z| |w| \text {cis}(\alpha + \beta )\), as required.

Next take aim at the Power Rule, better known as DeMoivre’s Theorem. ^[8] (Compare this proof with the proof of the Power Rule in Theorem modprops.) We proceed by induction on \(n\). Let \(P(n)\) be the sentence \(z^{n} = |z|^{n} \text {cis}(n \theta )\). Then \(P(1)\) is true, since \(z^{1} = z = |z| \text {cis}(\theta ) = |z|^{1} \text {cis}(1\cdot \theta )\).

We now assume \(P(k)\) is true, that is, we assume \(z^{k} = |z|^{k} \text {cis}(k \theta )\) for some \(k \geq 1\). Our goal is to show that \(P(k+1)\) is true, or that \(z^{k+1} = |z|^{k+1} \text {cis}((k+1)\theta )\). We have

Hence, assuming \(P(k)\) is true, we have that \(P(k+1)\) is true, so by the Principle of Mathematical Induction, \(z^{n} = |z|^{n} \text {cis}(n \theta )\) for all natural numbers \(n\).

The last property in Theorem prodquotpolarcomplex to prove is the quotient rule. Assuming \(|w| \neq 0\) we have

Next, we multiply both the numerator and denominator of the right hand side by \((\cos (\beta ) - i \sin (\beta ))\) which is the complex conjugate of \((\cos (\beta ) + i \sin (\beta ))\) to get

If we let the numerator be \(N = \left [\cos (\alpha ) + i \sin (\alpha )\right ] \left [\cos (\beta ) - i \sin (\beta )\right ]\) and simplify we get

If we call the denominator \(D\) then we get

Putting it all together, we get

and we are done. The next example makes good use of Theorem prodquotpolarcomplex.

Some remarks are in order. First, the reader may not be sold on using the polar form of complex numbers to multiply complex numbers – especially if they aren’t given in polar form to begin with.

Indeed, a lot of work was needed to convert the numbers \(z\) and \(w\) in Example polararithmeticex into polar form, compute their product, and convert back to rectangular form – certainly more work than is required to multiply out \(zw = (2\sqrt {3} + 2i)(-1 + i\sqrt {3})\) the old-fashioned way.

However, Theorem prodquotpolarcomplex pays huge dividends when computing powers of complex numbers. Consider how we computed \(w^{5}\) above and compare that to using the Binomial Theorem, Theorem BinomialTheorem, to accomplish the same feat by expanding \((-1 + i\sqrt {3})^{5}\).

Moreover, division is tricky in the best of times, and we saved ourselves a lot of time and effort using Theorem prodquotpolarcomplex to find and simplify \(\frac {z}{w}\) using their polar forms as opposed to starting with \(\frac {2\sqrt {3} + 2i}{-1 + i\sqrt {3}}\), rationalizing the denominator, and so forth.

There is geometric reason for studying these polar forms and we would be derelict in our duties if we did not mention the Geometry hidden in Theorem prodquotpolarcomplex.

Take the product rule, for instance. If \(z = |z| \text {cis}(\alpha )\) and \(w = |w| \text {cis}(\beta )\), the formula \(zw = |z||w| \text {cis}(\alpha + \beta )\) can be viewed geometrically as a two step process.

The multiplication of \(|z|\) by \(|w|\) can be interpreted as magnifying ^[9] (Assuming \(|w| > 1\).) the distance \(|z|\) from \(z\) to \(0\), by the factor \(|w|\). Adding the argument of \(w\) to the argument of \(z\) can be interpreted geometrically as a rotation of \(\beta \) radians counter-clockwise. ^[10] (Assuming \(\beta > 0\).)

Focusing on \(z\) and \(w\) from Example polararithmeticex, we can arrive at the product \(zw\) by plotting \(z\), doubling its distance from \(0\) (since \(|w| = 2\)), and rotating \(\frac {2\pi }{3}\) radians counter-clockwise. The sequence of diagrams below attempts to describe this process geometrically.

We may also visualize division similarly. Here, the formula \(\frac {z}{w} = \frac {|z|}{|w|} \text {cis}(\alpha - \beta )\) may be interpreted as shrinking ^[11] (Again, assuming \(|w| > 1\).) the distance from \(0\) to \(z\) by the factor \(|w|\), followed up by a clockwise ^[12] (Again, assuming \(\beta > 0\).) rotation of \(\beta \) radians.

In the case of \(z\) and \(w\) from Example polararithmeticex, we arrive at \(\frac {z}{w}\) by first halving the distance from \(0\) to \(z\), then rotating clockwise \(\frac {2\pi }{3}\) radians as shown below.

Our last goal of the section is to reverse DeMoivre’s Theorem to extract roots of complex numbers.

Unlike Definition principalnthrootdefn in Section AppRealNumberArithmetic, we do not specify one particular prinicpal \(n^{\text {th}}\) root, hence the use of the indefinite article ‘an’ as in ‘an \(n^{\text {th}}\) root of \(z\)’. Using this definition, both \(4\) and \(-4\) are square roots of \(16\), while \(\sqrt {16}\) means the principal square root of \(16\) as in \(\sqrt {16}= 4\).

Suppose we wish to find all complex third (cube) roots of \(8\). Algebraically, we are trying to solve \(w^{3} = 8\). We know that there is only one real solution to this equation, namely \(w = \sqrt [3]{8} = 2\), but if we take the time to rewrite this equation as \(w^3 - 8 = 0\) and factor, we get \((w-2)\left (w^2 + 2w + 4\right ) = 0\).

Solving \(w^2 + 2w + 4=0\) gives two more cube roots \(w = -1 \pm i \sqrt {3}\), for a total of three cube roots of \(8\). Per Theorem complexfactorization, since the degree of \(p(w) = w^3 -8\) is three, there are three complex zeros, counting multiplicity. Since we have found three distinct zeros, we know we have found all of the zeros, so there are exactly three distinct cube roots of \(8\).

Let us now solve this same problem using the machinery developed in this section. To do so, we express \(z = 8\) in polar form. Since \(z=8\) lies \(8\) units away on the positive real axis, we get \(z = 8 \text {cis}(0)\). If we let \(w = |w| \text {cis}(\alpha )\) be a polar form of \(w\), the equation \(w^3 = 8\) becomes

The complex number on the left hand side of the equation corresponds to the point with polar coordinates \(\left (|w|^3, 3\alpha \right )\), while the complex number on the right hand side corresponds to the point with polar coordinates \((8,0)\). Since \(|w| \geq 0\), so is \(|w|^3\), which means \(\left (|w|^3, 3\alpha \right )\) and \((8,0)\) are two polar representations corresponding to the same complex number, both with positive \(r\) values.

From Section PolarCoordinates, we know \(|w|^3 = 8\) and \(3\alpha = 0 + 2\pi k\) for integers \(k\). Since \(|w|\) is a real number, we solve \(|w|^3 = 8\) by extracting the principal cube root to get \(|w| = \sqrt [3]{8} = 2\).

As for \(\alpha \), we get \(\alpha = \frac {2\pi k}{3}\) for integers \(k\). This produces three distinct points with polar coordinates corresponding to \(k = 0\), \(1\) and \(2\): specifically \((2,0)\), \(\left (2, \frac {2\pi }{3}\right )\) and \(\left (2, \frac {4\pi }{3}\right )\).

The point \((2,0)\) corresponds to the complex number \(w_0 = 2 \text {cis}(0)\), the point \(\left (2, \frac {2\pi }{3}\right )\) corresponds to the complex number \(w_1 = 2 \text {cis}\left (\frac {2\pi }{3}\right )\), and the point \(\left (2, \frac {4\pi }{3}\right )\) corresponds to the complex number \(w_2 = 2 \text {cis}\left (\frac {4\pi }{3}\right )\). Converting to rectangular form, we find \(w_0 = 2\), \(w_1 = -1 + i\sqrt {3}\) and \(w_2 = -1-i\sqrt {3}\).

While this process seems a tad more involved than our previous factoring approach, this procedure can be generalized to find, for example, all of the fifth roots of \(32\). (Try using Chapter PolynomialFunctions techniques on that!)

If we start with a generic complex number in polar form \(z = |z| \text {cis}(\theta )\) and solve \(w^{n} = z\) in the same manner as above, we arrive at the following theorem.

The proof of Theorem nthrootscomplexthm breaks into to two parts: first, showing that each \(w_k\) is an \(n^{\text {th}}\) root, and second, showing that the set \(\left \{ w_k \, | \, k = 0, 1, \ldots , (n-1)\right \}\) consists of \(n\) different complex numbers.

To show \(w_k\) is an \(n^{\text {th}}\) root of \(z\), we use DeMoivre’s Theorem to show \(\left (w_k\right )^n = z\).

Since \(k\) is a whole number, \(\cos (\theta + 2\pi k) = \cos (\theta )\) and \(\sin (\theta + 2\pi k) = \sin (\theta )\). Hence, it follows that \(\text {cis}(\theta + 2\pi k) = \text {cis}(\theta )\), so \(\left (w_k\right )^n = r \text {cis}(\theta ) = z\), as required.

To show that the formula in Theorem nthrootscomplexthm generates \(n\) distinct numbers, we assume \(n \geq 2\) (or else there is nothing to prove) and note that the modulus of each of the \(w_k\) is the same, namely \(\sqrt [n]{r}\).

Therefore, the only way any two of these polar forms correspond to the same number is if their arguments are coterminal – that is, if the arguments differ by an integer multiple of \(2\pi \).

Suppose \(k\) and \(j\) are whole numbers between \(0\) and \((n-1)\), inclusive, with \(k \neq j\). Since \(k\) and \(j\) are different, let’s assume for the sake of argument that \(k > j\). Then \(\left ( \frac {\theta }{n} + \frac {2\pi }{n} k \right ) - \left ( \frac {\theta }{n} + \frac {2\pi }{n} j \right ) = 2\pi \left (\frac {k-j}{n}\right )\).

For \(2\pi \left (\frac {k-j}{n}\right )\) to be an integer multiple of \(2\pi \), \((k-j)\) must be a multiple of \(n\). But because of the restrictions on \(k\) and \(j\), \(0 < k - j \leq n-1\). (Think this through.) Hence, \((k-j)\) is a positive number less than \(n\), so it cannot be a multiple of \(n\).

As a result, \(w_k\) and \(w_j\) are different complex numbers, and we are done. By Theorem complexfactorization, we know there at most \(n\) distinct solutions to \(w^{n} = z\), and we have just found all \(n\) of them.

We illustrate Theorem nthrootscomplexthm in the next example.

Having done some computations with Theorem nthrootscomplexthm, it’s time to take a step back to look at things geometrically.

Essentially, Theorem nthrootscomplexthm says that to find the \(n^{\text {th}}\) roots of a complex number, we first take the \(n^{\text {th}}\) root of the modulus and divide the argument by \(n\). This gives the first root \(w_0\).

Each successive root is found by adding \(\frac {2\pi }{n}\) to the argument, which amounts to rotating \(w_0\) by \(\frac {2\pi }{n}\) radians. The result of these actions produces \(n\) roots, spaced equally around the complex plane.

As an example of this, we plot our answers to number fourthrootsneg16 in Example nthrootscomplexex below.

We have only glimpsed at the beauty of the complex numbers in this section. The complex plane is without a doubt one of the most important mathematical constructs ever devised. Coupled with Calculus, it is the venue for incredibly important Science and Engineering applications. ^[13] (For more on this, see the beautifully written epilogue to Section ComplexZeros found on page ?? .)