Complex Numbers

A complex number extends real arithmetic and represents a point in a plane. These two interpretations reinforce each other: addition combines displacements, while multiplication combines rotations and changes of scale. The aim of this chapter is to move between these descriptions without losing a sign, a root, or a condition.

We use the real-number operations from Number Systems, Algorithms, and Recursion, elementary trigonometry, and the triangle inequality. MIT’s introductory complex-variable notes provide a companion treatment of complex algebra, geometry, and exponentials [1][1] J. Orloff, “Topic 1: Complex Algebra and the Complex Plane,” 2018. MIT OpenCourseWare, 18.04 Complex Variables with Applications, Spring 2018. https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/resources/mit18_04s18_topic1/.

Construct complex arithmetic

No real number squares to $-1$. To extend the number system consistently, take ordered pairs of real numbers and define their operations.

DefinitionComplex numbers

The set $\mathbb C$ is $\mathbb R^2$ with addition and multiplication given by

$$\begin{aligned} (a,b)+(c,d)&=(a+c,b+d),\\ (a,b)(c,d)&=(ac-bd,ad+bc). \end{aligned}$$

Identify a real number $a$ with $(a,0)$ and set $i=(0,1)$. Then $i^2=(-1,0)=-1$, and $(a,b)=a+bi$. Equality means equality of both coordinates.

For $z=a+bi$, the real part is $\operatorname{Re}z=a$ and the imaginary part is $\operatorname{Im}z=b$. Both parts are real numbers; the imaginary part is $b$, not $bi$. Real numbers have $b=0$, and nonzero purely imaginary numbers have $a=0$, $b\ne0$.

The pair rules yield the familiar formulas

$$\begin{aligned} &(a+bi)+(c+di)\\ &\quad=(a+c)+(b+d)i,\\ &(a+bi)(c+di)\\ &\quad=(ac-bd)+(ad+bc)i. \end{aligned}$$

Thus ordinary expansion works, followed by replacing $i^2$ with $-1$. For example, $(2+i)(3-2i)=8-i$. Powers of $i$ repeat every four steps:

$$i^0=1,\quad i^1=i,\quad i^2=-1,\quad i^3=-i.$$

This cycle also handles negative integer powers: $i^{-1}=-i$, since $i(-i)=1$.

Conjugation, modulus, and division

DefinitionConjugate and modulus

For $z=a+bi$, define

$$\overline z=a-bi,\qquad |z|=\sqrt{a^2+b^2}.$$

The square root here is the nonnegative real square root. In particular, $|z|\ge0$, and $|z|=0$ exactly when $z=0$.

Multiplication gives the key identity $z\overline z=|z|^2$. If $z\ne0$, this positive real number supplies an inverse:

$$\frac1z=\frac{\overline z}{|z|^2} =\frac{a-bi}{a^2+b^2}.$$

To divide by a complex number, multiply numerator and denominator by the conjugate of the denominator. For example,

$$\begin{aligned} \frac{3+4i}{1+2i} &=\frac{(3+4i)(1-2i)}{5}\\ &=\frac{11}{5}-\frac25i. \end{aligned}$$

Conjugation respects addition and multiplication:

$$\begin{aligned} \overline{z+w}&=\overline z+\overline w,\\ \overline{zw}&=\overline z\,\overline w,\\ \overline{\overline z}&=z. \end{aligned}$$

Also, $z+\overline z=2\operatorname{Re}z$ and $z-\overline z=2i\operatorname{Im}z$. These identities follow by writing out the coordinates; they often avoid a much longer expansion.

Why these operations form a field

Proof

Addition inherits commutativity and associativity from real coordinate addition. Its identity is $(0,0)$, and the additive inverse of $(a,b)$ is $(-a,-b)$. The multiplication formula is symmetric in the two pairs, so multiplication is commutative, with identity $(1,0)$.

For $z=a+bi$, $w=c+di$, and $u=e+fi$, expanding either $(zw)u$ or $z(wu)$ gives real part

$$ace-bde-adf-bcf$$

and imaginary part

$$acf-bdf+ade+bce.$$

This proves associativity. Distributivity follows by distributing the real coordinates in the same way. The inverse formula above gives a multiplicative inverse for every nonzero element. These are the field axioms.

In particular, $zw=0$ implies $z=0$ or $w=0$: if $z\ne0$, multiply by its inverse. This justifies solving a factored equation one factor at a time.

Being a field does not make $\mathbb C$ an ordered field. In an ordered field every nonzero square is positive; applying that to $i$ would make $-1$ positive. Complex inequalities therefore compare real quantities such as moduli or real parts, rather than assigning every complex number a compatible position in an order.

RemarkAn equation has roots; a square-root symbol needs a convention

The equation $z^2=-5$ has two roots, $\pm i\sqrt5$. If $\sqrt{-5}$ denotes the principal complex square root, its value is $i\sqrt5$. Do not extend the real product rule for square roots without conditions:

$$\sqrt{-1}\sqrt{-1}=-1\ne1=\sqrt{(-1)(-1)}.$$

We will obtain all roots systematically from polar form.

Read arithmetic as geometry

Represent $z=x+iy$ by the point $(x,y)$. Its modulus is its distance from the origin; $|z-w|$ is the distance between two points. Addition is vector addition, subtraction is displacement, and conjugation reflects a point across the real axis.

Complex condition	Geometric meaning
$	z-a
$	z-a
$\operatorname{Re}z>c$, $c\in\mathbb R$	Open half-plane to the right of $x=c$
$	z-a

For radius zero, the circle equation and closed-disk inequality both reduce to the single point $a$. A negative radius gives no solution.

The product $z\overline w$ connects complex algebra to Euclidean geometry. If $z=x+iy$ and $w=u+iv$, then

$$\operatorname{Re}(z\overline w)=xu+yv.$$

This is the ordinary dot product of the two corresponding real vectors. Consequently,

$$|z+w|^2=|z|^2+|w|^2+2\operatorname{Re}(z\overline w).$$

TheoremModulus of a product and quotient

For all $z,w\in\mathbb C$,

$$|zw|=|z||w|.$$

If $w\ne0$, then $|z/w|=|z|/|w|$.

Proof

Using conjugation and commutativity,

$$\begin{aligned} |zw|^2&=zw\overline z\,\overline w\\ &=(z\overline z)(w\overline w)\\ &=|z|^2|w|^2. \end{aligned}$$

Both proposed moduli are nonnegative, so equality of squares gives the result. Apply it to $z=(z/w)w$ to obtain the quotient rule.

TheoremTriangle and reverse triangle inequalities

For all complex $z,w$,

$$|z+w|\le|z|+|w|,$$

and

$$\bigl||z|-|w|\bigr|\le|z-w|.$$

In either inequality, equality holds exactly when one number is zero or the two nonzero numbers point in the same direction.

Proof

For any complex $v$, $\operatorname{Re}v\le|v|$, with equality exactly when $v$ is a nonnegative real number. Therefore

$$\begin{aligned} |z+w|^2 &=|z|^2+|w|^2+2\operatorname{Re}(z\overline w)\\ &\le(|z|+|w|)^2. \end{aligned}$$

Equality means $z\overline w$ is nonnegative real. If $w\ne0$, the identity $z/w=z\overline w/|w|^2$ shows that this means $z=tw$ for some real $t\ge0$.

The reverse inequality follows by applying the triangle inequality to $z=(z-w)+w$ and then exchanging $z,w$. For its equality condition, observe that

$$\begin{aligned} &|z-w|^2-(|z|-|w|)^2\\ &\quad=2\bigl(|z||w|-\operatorname{Re}(z\overline w)\bigr). \end{aligned}$$

The difference vanishes under exactly the same condition.

Polar form: magnitude and direction

For $z\ne0$, let $r=|z|>0$. An angle $\theta$ describing its direction satisfies

$$z=r(\cos\theta+i\sin\theta).$$

We abbreviate $\cos\theta+i\sin\theta$ as $\operatorname{cis}\theta$. The angle is not unique: adding an integer multiple of $2\pi$ gives the same point.

ConventionArguments and the principal argument

We use $\arg z$ for the set of all arguments and $\operatorname{Arg}z$ for the unique argument in $(-\pi,\pi]$. Thus

$$\arg z=\{\operatorname{Arg}z+2\pi k:k\in\mathbb Z\}.$$

Neither $\arg0$ nor $\operatorname{Arg}0$ is defined. Although $0\operatorname{cis}\theta=0$ for every angle, the origin has no direction.

For $z=x+iy\ne0$, find the angle using both $\cos\theta=x/r$ and $\sin\theta=y/r$. The ratio $y/x$ alone loses quadrant information and is undefined when $x=0$.

| $z$ | $|z|$ | $\operatorname{Arg}z$ | | --- | --- | --- | | $-1/2$ | $1/2$ | $\pi$ | | $-3+3i$ | $3\sqrt2$ | $3\pi/4$ | | $-\pi i$ | $\pi$ | $-\pi/2$ | | $-2\sqrt3-2i$ | $4$ | $-5\pi/6$ |

Multiplication rotates and scales

The trigonometric addition formulas give

$$\operatorname{cis}\alpha\operatorname{cis}\beta =\operatorname{cis}(\alpha+\beta).$$

Hence, for $z=r\operatorname{cis}\alpha$ and $w=s\operatorname{cis}\beta$ with $r,s>0$,

$$zw=rs\operatorname{cis}(\alpha+\beta),$$

and

$$\frac zw=\frac rs\operatorname{cis}(\alpha-\beta).$$

Multiplication by a fixed nonzero $w$ scales every distance by $s$ and rotates every vector by $\beta$. In particular, multiplication by $i$ rotates counterclockwise through $\pi/2$.

The principal argument of a product must be brought back into $(-\pi,\pi]$:

$$\operatorname{Arg}(zw) \equiv\operatorname{Arg}z+\operatorname{Arg}w\pmod{2\pi}.$$

This is a congruence, not an unrestricted equality of real numbers. For $z=w=\operatorname{cis}(3\pi/4)$, the sum of principal arguments is $3\pi/2$, whereas $\operatorname{Arg}(zw)=-\pi/2$.

For example, division in polar form gives

$$\begin{aligned} \frac{1+i}{\sqrt3-i} &=\frac{\sqrt2\operatorname{cis}(\pi/4)}{2\operatorname{cis}(-\pi/6)}\\ &=\frac{\sqrt2}{2}\operatorname{cis}\frac{5\pi}{12}. \end{aligned}$$

Exponentials and integer powers

We can extend the real exponential by defining, for real $x,y$,

$$e^{x+iy}=e^x(\cos y+i\sin y).$$

At $x=0$ this is Euler’s formula $e^{iy}=\cos y+i\sin y$. It makes polar form $z=re^{i\theta}$. The factor $i$ in the exponent is essential: $re^\theta$ is a positive real number when $r>0$ and $\theta$ is real.

The real exponential law and the angle-addition formulas show that, for all complex $u,v$,

$$e^{u+v}=e^ue^v.$$

This verifies that the definition preserves multiplication of exponentials. The power-series approach gives the same extension once convergence of those series is available; it is not needed for the algebra here.

In particular,

$$|e^{x+iy}|=e^x,\qquad \overline{e^z}=e^{\overline z}.$$

The exponential is never zero and is periodic in the imaginary direction:

$$e^u=e^v\quad\Longleftrightarrow\quad u-v\in2\pi i\mathbb Z.$$

Indeed, equality of moduli first forces equal real parts. Equality of both sine and cosine then forces the imaginary parts to differ by an integer multiple of $2\pi$.

TheoremDe Moivre's formula

For every real $\theta$ and integer $n$,

$$(\operatorname{cis}\theta)^n=\operatorname{cis}(n\theta).$$

More generally, for $r>0$,

$$(re^{i\theta})^n=r^ne^{in\theta}.$$

Proof

For positive $n$, repeatedly apply the product rule; this can also be written as induction. For $n=0$, both sides are $1$. For negative $n$, use the reciprocal and the fact that $e^{i\theta}e^{-i\theta}=1$.

The restriction $r>0$ makes negative powers valid. If the base is zero, positive integer powers are still zero; no negative power is defined, and we do not assign a value to $0^0$ here. Noninteger exponents need a separate convention, discussed in the final exploration.

Find every complex root

TheoremThe nth roots of a nonzero complex number

Let $n\ge1$ be an integer and $a=re^{i\theta}\ne0$, where $r>0$. The equation $z^n=a$ has exactly $n$ distinct roots:

$$z_k=r^{1/n}\exp\left(i\frac{\theta+2\pi k}{n}\right),$$

where $k=0,1,\ldots,n-1$ and $r^{1/n}$ is the positive real root.

Proof

Write any solution as $z=\rho e^{i\phi}$, where $\rho>0$. Comparing moduli gives $\rho^n=r$, so $\rho=r^{1/n}$. Comparing directions gives

$$n\phi=\theta+2\pi k$$

for some integer $k$. Conversely, each angle of this form gives a solution. Two indices give the same point exactly when their difference is divisible by $n$. The listed indices therefore give all solutions without repetition.

The roots lie on a circle of radius $r^{1/n}$, separated by an angle $2\pi/n$. Changing the chosen argument of $a$ only reorders the roots. If $a=0$, the only root is zero, with multiplicity $n$ in the polynomial $z^n$.

For instance, $\sqrt2+i\sqrt2=2e^{i\pi/4}$ has cube roots

$$2^{1/3}e^{i(\pi/12+2\pi k/3)},\qquad k=0,1,2.$$

Roots of unity

Set $\omega=e^{2\pi i/n}$. The $n$th roots of unity are $1,\omega,\ldots,\omega^{n-1}$. They are closed under multiplication because exponents can be reduced modulo $n$.

For $n\ge2$, $\omega\ne1$, and the finite geometric-sum formula gives

$$\sum_{k=0}^{n-1}\omega^k =\frac{1-\omega^n}{1-\omega}=0.$$

For $n=1$, the sum is $1$. The zero sum for $n\ge2$ expresses the symmetry of equally spaced points around the origin.

A root of unity is primitive of order $n$ if its smallest positive power equal to $1$ is its $n$th power. The root $\omega^k$ has order $n/\gcd(n,k)$, so it is primitive exactly when $\gcd(n,k)=1$: its $j$th power is $1$ precisely when $n$ divides $kj$.

Quadratics and polynomial roots

For $Az^2+Bz+C=0$ with complex coefficients and $A\ne0$, completing the square gives

$$(2Az+B)^2=B^2-4AC.$$

Choose a square root $s$ of the discriminant. The solutions are

$$z=\frac{-B\pm s}{2A},\qquad s^2=B^2-4AC.$$

The two choices of $s$ give the same solution set. A zero discriminant gives one root of multiplicity two. For example,

$$2z^2-z+1=0 \quad\Longrightarrow\quad z=\frac{1\pm i\sqrt7}{4}.$$

TheoremConjugate roots require real coefficients

If a polynomial $P$ has real coefficients, then

$$P(\overline z)=\overline{P(z)}.$$

Consequently, every nonreal root occurs with its conjugate, with the same multiplicity.

Proof

Conjugation preserves sums and products and fixes every real coefficient. Applying it term by term proves the identity and the root implication. If $P(t)=(t-z)^mQ(t)$ with $Q(z)\ne0$, conjugating the coefficients of this factorization gives a factor $(t-\overline z)^m$ whose remaining factor is nonzero at $\overline z$. Thus the multiplicities agree.

For example, if $-2+i$ is a root of

$$P(z)=2z^3+9z^2+14z+5,$$

then $-2-i$ is another. Their factors multiply to

$$(z+2-i)(z+2+i)=z^2+4z+5.$$

Polynomial division, or direct multiplication, now verifies

$$P(z)=(z^2+4z+5)(2z+1).$$

The remaining root is $-1/2$. Checking the leading coefficient is essential: the product of three monic linear factors alone would omit the factor $2$.

TheoremFundamental theorem of algebra

Every complex polynomial of degree $n\ge1$ factors into $n$ linear factors over $\mathbb C$, counting multiplicity. Its leading coefficient remains as the prefactor.

ProofA minimum-modulus argument

Let $p(z)=a_nz^n+\cdots+a_0$ with $a_n\ne0$. The triangle inequality gives

$$|p(z)|\ge |a_n||z|^n-\sum_{j<n}|a_j||z|^j\longrightarrow\infty \quad\text{as }|z|\to\infty.$$

Thus the continuous function $|p|$ attains a global minimum at some $z_0$: outside a sufficiently large disk it exceeds $|p(0)|+1$, while inside the closed disk it attains a minimum. This compactness assertion follows by taking convergent subsequences of the bounded real and imaginary coordinates, using real completeness.

Suppose $c=p(z_0)\ne0$. Expand around $z_0$ and let $m\ge1$ be the first nonzero positive-degree coefficient:

$$\frac{p(z_0+w)}c=1+b_mw^m+\sum_{j>m}b_jw^j, \qquad b_m\ne0.$$

For $0<t<1$, the already proved complex-root formula supplies $w$ with $w^m=-t/b_m$. Then $|w|=(t/|b_m|)^{1/m}$. For $|w|\le1$, the remaining finite sum has modulus at most $C|w|^{m+1}$, where $C=\sum_{j>m}|b_j|$. Choose $t$ small enough that this is less than $t/2$. It follows that

$$\left|\frac{p(z_0+w)}c\right| \le1-t+t/2<1,$$

contradicting minimality. Hence $p$ has a root. Polynomial division extracts its linear factor; induction on degree extracts all $n$ factors and preserves the leading coefficient.

Over $\mathbb R$, conjugate factors combine into real quadratics. For example,

$$z^4+1=(z^2-\sqrt2z+1)(z^2+\sqrt2z+1).$$

Both quadratics have negative discriminants, so this is a factorization into irreducible real quadratics.

Exercises

Try each problem before opening its solution. Cartesian form is usually best for linear equations, polar form for products and roots, and conjugation for modulus identities and real-coefficient polynomials.

ExerciseParts and powers of i

Show that $\operatorname{Re}(iz)=-\operatorname{Im}z$. Simplify

$$3i^{11}+6i^3+\frac8{i^{20}}+i^{-1}.$$

Solution

If $z=a+bi$, then $iz=-b+ai$. The four terms are $-3i$, $-6i$, $8$, and $-i$, so the expression is $8-10i$.

ExerciseSolve without losing a zero root

Solve each equation over $\mathbb C$:

$$\begin{aligned} iz&=4-zi,\\ \frac z{1-z}&=1-5i,\\ 8z^2+(2-i)z&=0,\\ z^2+16&=0. \end{aligned}$$

Solution

The first gives $2iz=4$, hence $z=-2i$.

The second requires $z\ne1$. Rearranging gives $(2-5i)z=1-5i$, so

$$z=\frac{1-5i}{2-5i}=\frac{27-5i}{29}.$$

This value is not $1$, so it is admissible. The third factors as $z(8z+2-i)=0$, giving $z=0$ or $z=(-2+i)/8$. Dividing by $z$ at the start would lose a solution. The last equation gives $z=\pm4i$.

ExerciseA complex linear system

Solve and check both equations:

$$\begin{aligned} (1-i)z_1+3z_2&=2-3i,\\ iz_1+(1+2i)z_2&=1. \end{aligned}$$

Solution

The determinant is $(1-i)(1+2i)-3i=3-2i\ne0$. Elimination gives

$$\begin{aligned} z_1&=\frac{5+i}{3-2i}=1+i,\\ z_2&=\frac{-2-3i}{3-2i}=-i. \end{aligned}$$

Substitution gives $(1-i)(1+i)-3i=2-3i$ and $i(1+i)+(1+2i)(-i)=1$.

ExerciseThree real multiplications

Compute $(a+bi)(c+di)$ using three real multiplications, together with additions and subtractions. Explain why this does not by itself prove a runtime improvement on every computer.

Solution

Compute

$$\begin{aligned} p&=ac,\qquad q=bd,\\ t&=(a+b)(c+d). \end{aligned}$$

The real part is $p-q$, and the imaginary part is $t-p-q$. This saves one multiplication but uses more additions. Runtime and floating-point error also depend on the implementation and hardware, so an operation count alone is not a benchmark.

ExerciseModuli without expansion

Find the moduli of

$$\frac{1+2i}{-2-i},\qquad (1+i)(2-3i)(4i-3),$$

and

$$\frac{i(2+i)^3}{(1-i)^2},\qquad \frac{(\pi+i)^{100}}{(\pi-i)^{100}}.$$

Solution

Apply the product and quotient rules. The answers, in order, are

$$1,\qquad 5\sqrt{26},\qquad \frac{5\sqrt5}{2},\qquad1.$$

Every denominator is nonzero. For the last expression, numerator and denominator have equal moduli because their bases are conjugates.

ExerciseTranslate loci

Describe the sets given by $|z-1|=|z+1|$, $|z-(1+i)|\le2$, and $|z-1|=\operatorname{Re}z+1$.

Solution

Write $z=x+iy$. The first equation reduces to $(x-1)^2+y^2=(x+1)^2+y^2$, hence $x=0$, the imaginary axis. The second is the closed disk of radius $2$ centred at $(1,1)$.

For the third, the right side must be nonnegative. Squaring gives

$$(x-1)^2+y^2=(x+1)^2,$$

so $y^2=4x$. Conversely, this equation implies $x\ge0$, hence $x+1>0$, so taking the nonnegative square root recovers the original equation. The locus is the entire parabola $y^2=4x$.

ExerciseDot products and the parallelogram law

Prove the parallelogram identity

$$|z+w|^2+|z-w|^2=2|z|^2+2|w|^2.$$

For nonzero $z,w$, express perpendicularity using $z\overline w$.

Solution

Expand the two squared moduli. Their cross terms are $2\operatorname{Re}(z\overline w)$ and its negative, so they cancel. Since $\operatorname{Re}(z\overline w)$ is the real vector dot product, the vectors are perpendicular exactly when it is zero. A complex product $zw$ itself is not the dot product.

ExerciseWeighted points in the unit disk

Let $\lambda_1,\ldots,\lambda_n\ge0$, with $\Lambda=\sum_{k=1}^n\lambda_k>0$, and let $|z_k|\le1$. Prove

$$\left|\frac{\sum_{k=1}^n\lambda_kz_k}{\Lambda}\right|\le1.$$

When does equality hold?

Solution

Repeated use of the triangle inequality gives

$$\left|\sum_k\lambda_kz_k\right| \le\sum_k\lambda_k|z_k|\le\Lambda.$$

Divide by the positive number $\Lambda$. Equality holds exactly when all $z_k$ with $\lambda_k>0$ are the same point on the unit circle. To see necessity, write the weighted average as $u$ with $|u|=1$ and take the real part after multiplying the sum by $\overline u$. Each $\operatorname{Re}(z_k\overline u)\le1$; a positive weighted average of these numbers is $1$ only if each active term equals $1$. Together with $|z_k|\le1$, this forces $z_k=u$.

ExercisePrincipal arguments after multiplication

Find the modulus and principal argument of each number:

$$(1-i)(-\sqrt3+i),\qquad(\sqrt3-i)^2,$$

and

$$\frac{-1+\sqrt3i}{2+2i}.$$

Solution

Add or subtract the angles of the factors, then reduce into $(-\pi,\pi]$. The respective pairs are

$$\begin{gathered} \left(2\sqrt2,\frac{7\pi}{12}\right),\qquad \left(4,-\frac\pi3\right),\\ \left(\frac1{\sqrt2},\frac{5\pi}{12}\right). \end{gathered}$$

For example, the first angle is $-\pi/4+5\pi/6=7\pi/12$. All arguments of each number are its listed principal argument plus $2\pi k$.

ExerciseExponentials in Cartesian form

Convert the following to $a+bi$:

$$e^{-i\pi/4},\qquad \frac{e^{1+3\pi i}}{e^{-1+\pi i/2}},\qquad 2e^{3+i\pi/6}.$$

Also simplify $(e^{3i}-e^{-3i})/(2i)$ and find $|e^z|$ for $z=4e^{i\pi/3}$.

Solution

The first three values are

$$\frac{1-i}{\sqrt2},\qquad ie^2,\qquad \sqrt3e^3+ie^3.$$

Euler’s formula reduces the difference quotient to the real number $\sin3$. Finally, $z=2+2\sqrt3i$, so $|e^z|=e^2$; its modulus depends on the real part of $z$, not on $|z|$.

ExercisePowers and a geometric sum

Write $a_n=(1+i)^n$ in polar form for integers $n\ge0$. Compute its first $N$ terms’ sum, indexed from $0$ to $N-1$, for $N\ge0$.

Solution

De Moivre’s formula gives

$$a_n=2^{n/2}e^{in\pi/4}.$$

For $N\ge1$, the geometric-sum identity gives

$$\sum_{n=0}^{N-1}(1+i)^n=\frac{(1+i)^N-1}{i}.$$

For $N=0$, the empty sum is zero, and the displayed right side is also zero. The points rotate by $\pi/4$ and increase their distance from the origin by a factor $\sqrt2$ at each step.

ExerciseEnumerate roots without duplicates

Find all fourth roots of $-16$, all fifth roots of $1$, and all square roots of $i-1$.

Solution

The three solution sets are, respectively,

$$\left\{2e^{i(\pi/4+k\pi/2)}:k=0,1,2,3\right\},$$$$\left\{e^{2\pi ik/5}:k=0,1,2,3,4\right\},$$

and

$$\left\{2^{1/4}e^{i(3\pi/8+k\pi)}:k=0,1\right\}.$$

The first set can also be written $\{\sqrt2+i\sqrt2,-\sqrt2+i\sqrt2,-\sqrt2-i\sqrt2,\sqrt2-i\sqrt2\}$.

ExerciseA roots-of-unity filter

Let $n\ge1$, $\omega=e^{2\pi i/n}$, and $m\in\mathbb Z$. Evaluate

$$\sum_{k=0}^{n-1}\omega^{mk}.$$

Solution

If $n$ divides $m$, every term is $1$, so the sum is $n$. Otherwise $\omega^m\ne1$, and the geometric-sum formula gives zero because $(\omega^m)^n=1$. This includes negative $m$ and the case $n=1$. Thus equally spaced complex phases distinguish divisibility by $n$, an identity that later appears in the discrete Fourier transform.

ExerciseProducts of roots of unity

Suppose $u^m=1$ and $v^n=1$, where $m,n\ge1$. Prove $(uv)^{mn}=1$. Does this require $m,n$ to be coprime? If $u,v$ are primitive of orders $m,n$ and $\gcd(m,n)=1$, show that $uv$ is primitive of order $mn$.

Solution

Commutativity gives $(uv)^{mn}=(u^m)^n(v^n)^m=1$, without a coprimality assumption.

For the stronger claim, suppose $(uv)^k=1$ for $k>0$. Raising to the $n$th power yields $u^{kn}=1$. Thus $m$ divides $kn$, and coprimality implies $m$ divides $k$. Similarly, $n$ divides $k$, hence $mn$ divides $k$. Since the $mn$th power is already $1$, the order is exactly $mn$.

ExerciseFactor over the complex numbers

Factor each polynomial completely:

$$z^2+z+3,\qquad z^3+3z^2+7z+5,$$

and

$$z^3-iz^2-4z+4i.$$

Solution

The quadratic has roots $r_\pm=(-1\pm i\sqrt{11})/2$. Its factorization is

$$z^2+z+3=(z-r_+)(z-r_-).$$

For the real cubic, $-1$ is a root, and division gives

$$\begin{aligned} &z^3+3z^2+7z+5\\ &\quad=(z+1)(z^2+2z+5). \end{aligned}$$

Its complete factorization is $(z+1)(z+1-2i)(z+1+2i)$.

For the complex-coefficient cubic, group the terms:

$$\begin{aligned} &z^3-iz^2-4z+4i\\ &\quad=(z-i)(z^2-4)\\ &\quad=(z-i)(z-2)(z+2). \end{aligned}$$

Its root $i$ does not imply a root $-i$, since its coefficients are not all real.

ExerciseRecover a real cubic

A real-coefficient cubic $P$ has roots $2+i$ and $1$, and $P(0)=10$. Find $P$.

Solution

The conjugate root is $2-i$. Therefore

$$P(z)=A(z^2-4z+5)(z-1)$$

with nonzero real $A$. At zero, $-5A=10$, so $A=-2$. Expanding gives

$$P(z)=-2z^3+10z^2-18z+10.$$

ExerciseCheck the hypothesis before pairing roots

Suppose $1+i$ is a root of

$$P(z)=z^3+az^2+bz+10-6i,$$

where $a,b\in\mathbb R$. Find $a,b$. Is $1-i$ also a root?

Solution

Using $(1+i)^2=2i$ and $(1+i)^3=-2+2i$ gives

$$P(1+i)=(8+b)+i(2a+b-4).$$

Equating the real and imaginary parts to zero gives $b=-8$ and $a=6$. The constant coefficient is complex, so the conjugate-root theorem does not apply. Direct substitution gives $P(1-i)=-12i\ne0$.

ExerciseRecover every exponent

For $a=re^{i\theta}\ne0$, solve $e^w=a$. Explain what happens if $a=0$.

Solution

Write $w=x+iy$. Comparing moduli gives $e^x=r$, hence $x=\ln r$. Comparing directions gives $y=\theta+2\pi k$. Therefore all solutions are

$$w=\ln r+i(\theta+2\pi k),\qquad k\in\mathbb Z.$$

There is no solution when $a=0$, since a complex exponential is never zero.

Exploration: what does a fractional power mean?

Integer powers are unambiguous because they use repeated multiplication and inverses. A fractional power can mean a chosen principal value or a set of algebraic values. State the convention before using exponent laws.

Principal values, all values, and a failed exponent law

For $z\ne0$, define the pointwise principal logarithm by

$$\operatorname{Log}z=\ln|z|+i\operatorname{Arg}z.$$

A principal value of a real power is

$$z^\alpha_{\mathrm{principal}} =\exp\bigl(\alpha\operatorname{Log}z\bigr).$$

This selects one value using our argument convention. It does not make every real exponent law valid. For instance,

$$\bigl((-1)^2\bigr)^{1/2}_{\mathrm{principal}}=1,$$

whereas $(-1)^{2(1/2)}_{\mathrm{principal}}=-1$. Taking an intermediate principal argument discarded a full turn. This pointwise logarithm is also discontinuous across the negative real axis; choosing a continuous logarithm on a suitable domain belongs to complex analysis.

For a positive rational exponent $m/n$ in lowest terms, another convention is the set

$$\{u^m:u^n=z\}.$$

Writing $z=re^{i\theta}$ yields the $n$ distinct values

$$r^{m/n}\exp\left(i\frac{m(\theta+2\pi k)}n\right).$$

Here $k=0,1,\ldots,n-1$.

These are exactly the roots of $v^n=z^m$. Indeed, those roots have angles $(m\theta+2\pi j)/n$, and multiplication by $m$ permutes the residue classes modulo $n$ because $\gcd(m,n)=1$.

The lowest-terms condition matters. For $m=n=2$ and $z=1$, squaring the square roots of $1$ gives only $\{1\}$, while $v^2=1^2$ has roots $\{1,-1\}$. Reduce the exponent before using this description.

For example, the two algebraic values of $(1-i)^{3/2}$ are

$$\pm2^{3/4}e^{-3\pi i/8}.$$

The principal value is the one with the plus sign, obtained from $\operatorname{Arg}(1-i)=-\pi/4$. Both values square to $(1-i)^3=-2-2i$.

References

1. [1] J. Orloff, “Topic 1: Complex Algebra and the Complex Plane,” 2018. MIT OpenCourseWare, 18.04 Complex Variables with Applications, Spring 2018. https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/resources/mit18_04s18_topic1/ ↩