6.1 Complex numbers, polar form, and geometry

Why complex numbers live in a plane

The equation $x^2+1=0$ has no real solution. Introducing a symbol $i$ with $i^2=-1$ is not only an algebraic trick; it also suggests a geometric extension of the real line. Multiplication by $-1$ rotates a real number by $180^\circ$. Since $i^2=-1$, multiplication by $i$ should behave like a $90^\circ$ anticlockwise rotation. Thus $i\cdot 1$ is naturally placed at the point $(0,1)$ rather than somewhere on the real line.

This motivates the formal construction: a complex number is a point in the plane, but with addition and multiplication chosen so that the rule $i^2=-1$ is built into the arithmetic.

Definition

Complex numbers as ordered pairs

The set of complex numbers is

$$\mathbb C=\{(x,y):x,y\in\mathbb R\}.$$

For $(x_1,y_1),(x_2,y_2)\in\mathbb C$, define

$$(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2),$$

and

$$(x_1,y_1)(x_2,y_2)=(x_1x_2-y_1y_2,\;x_1y_2+x_2y_1).$$

The real number $x$ is identified with $(x,0)$, and $i$ means $(0,1)$.

These definitions are chosen so that the informal expression $x+iy$ is not just notation but actually agrees with the ordered-pair construction. Indeed, for real $x$ and $y$,

$$(x,0)+(0,1)(y,0)=(x,0)+(0,y)=(x,y).$$

So from now on we write

$$z=x+iy$$

instead of $(x,y)$. In particular, the pair rule gives $(0,1)^2=(-1,0)$, so $i^2=-1$ holds in the constructed system. Expanding in this notation gives

$$\begin{aligned} (x_1+iy_1)(x_2+iy_2) &=x_1x_2+i x_1y_2+i x_2y_1+i^2y_1y_2\\ &=(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1). \end{aligned}$$

Thus the familiar expansion reproduces both coordinates of the defined product; the minus sign comes from $i^2$ in the product $(iy_1)(iy_2)$.

Definition

Real and imaginary parts

If $z=x+iy$ with $x,y\in\mathbb R$, then

$$\operatorname{Re}(z)=x,\qquad \operatorname{Im}(z)=y.$$

The number is real when $\operatorname{Im}(z)=0$, and it is purely imaginary when $\operatorname{Re}(z)=0$.

Equality of complex numbers is equality of ordered pairs. Thus

$$a+ib=c+id \quad\Longleftrightarrow\quad a=c\text{ and }b=d.$$

This simple point is often the reason complex equations can be solved by comparing real and imaginary parts.

Checkpoint

If $z=(3,-5)$ in the ordered-pair construction, what are $z$, $\operatorname{Re}(z)$, and $\operatorname{Im}(z)$ in $x+iy$ notation?

Use the identification $(x,y)=x+iy$.

Solution · Answer

$z=3-5i$, $\operatorname{Re}(z)=3$, and $\operatorname{Im}(z)=-5$.

Arithmetic and field structure

The ordered-pair rules make $\mathbb C$ behave like a number system. Addition and multiplication are commutative and associative, multiplication distributes over addition, $0=0+i0$ is the additive identity, and $1=1+i0$ is the multiplicative identity.

Theorem

Complex numbers form a field

For every $z=x+iy\in\mathbb C$, the additive inverse is

$$-z=(-x)+i(-y).$$

For every non-zero $z=x+iy$, the multiplicative inverse is

$$z^{-1}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}.$$

Together with the usual commutative, associative, identity, and distributive laws, these inverse formulas say that $\mathbb C$ is a field.

The inverse formula is worth deriving once. If $z=x+iy\ne0$, then $x^2+y^2\gt 0$: this is equivalent to $z\ne0$, since a sum of two real squares is zero exactly when both coordinates are zero. Hence

$$(x+iy)\left(\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}\right) =\frac{x^2+y^2}{x^2+y^2} +i\frac{-xy+xy}{x^2+y^2} =1.$$

Because non-zero complex numbers have multiplicative inverses, there are no zero divisors in $\mathbb C$: if $zw=0$, then $z=0$ or $w=0$.

Subtraction and division

For $z=a+ib$ and $w=c+id$,

$$z-w=(a-c)+i(b-d).$$

If $w\ne0$, then division means multiplication by the inverse of $w$:

$$\frac{z}{w} =\frac{a+ib}{c+id} =(a+ib)\left(\frac{c}{c^2+d^2}-i\frac{d}{c^2+d^2}\right).$$

After multiplying out,

$$\frac{a+ib}{c+id} =\frac{ac+bd}{c^2+d^2} +i\frac{bc-ad}{c^2+d^2}.$$

Worked example

A division calculation

Express

$$\frac{-1+4i}{6-3i}$$

in the form $a+bi$.

Using the inverse of $6-3i$,

$$\frac{-1+4i}{6-3i} =(-1+4i)\frac{6+3i}{6^2+(-3)^2}.$$

Therefore

$$\frac{-1+4i}{6-3i} =\frac{(-1)(6)+4(-3)}{45} +i\frac{4(6)-(-1)(-3)}{45} =-\frac25+\frac{7}{15}i.$$

Checkpoint

Find the multiplicative inverse of $2-i$.

Use the inverse formula for $x+iy$.

Solution · Answer

Here $x=2$ and $y=-1$, so

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

The Argand plane

The same ordered pair $(x,y)$ can be read geometrically as a point or vector in the coordinate plane. For $z=x+iy$, the horizontal coordinate is the real part and the vertical coordinate is the imaginary part. This plane is called the complex plane or Argand plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis.

Addition is vector addition:

$$(x_1+iy_1)+(x_2+iy_2) =(x_1+x_2)+i(y_1+y_2).$$

Multiplication is less visible in rectangular coordinates, but it becomes clear after we introduce modulus and argument.

Conjugate and modulus

Definition

Complex conjugate and modulus

For $z=x+iy$, define the conjugate and modulus by

$$\overline z=x-iy,\qquad |z|=\sqrt{x^2+y^2}.$$

The conjugate is reflection across the real axis. The modulus is the distance from the origin to the point representing $z$.

The basic identities are

$$\overline{\overline z}=z,\qquad z=\overline z\Longleftrightarrow z\in\mathbb R,$$

and

$$\operatorname{Re}(z)=\frac12(z+\overline z), \qquad \operatorname{Im}(z)=\frac{1}{2i}(z-\overline z).$$

Conjugation respects arithmetic:

$$\overline{z_1\pm z_2}=\overline{z_1}\pm\overline{z_2}, \qquad \overline{z_1z_2}=\overline{z_1}\,\overline{z_2}, \qquad \overline{\left(\frac{z_1}{z_2}\right)} =\frac{\overline{z_1}}{\overline{z_2}}$$

when $z_2\ne0$.

The modulus is tied to conjugation by

$$|z|^2=z\overline z.$$

Indeed, $(x+iy)(x-iy)=x^2+y^2$, so $z\overline z$ is always a non-negative real number. For $z\ne0$, this rewrites the inverse as $z^{-1}=\overline z/|z|^2$: reflect across the real axis, then scale by $1/|z|^2$. Likewise, multiplying numerator and denominator by the conjugate of a nonzero denominator makes that denominator a positive real number. Also,

$$|z|=|-z|=|\overline z|, \qquad |z_1z_2|=|z_1||z_2|, \qquad \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$$

for $z_2\ne0$.

Worked example

Parallelogram law

For $z_1,z_2\in\mathbb C$,

$$|z_1+z_2|^2+|z_1-z_2|^2=2|z_1|^2+2|z_2|^2.$$

Use $|u|^2=u\overline u$:

$$\begin{aligned} |z_1+z_2|^2+|z_1-z_2|^2 &=(z_1+z_2)(\overline{z_1}+\overline{z_2}) +(z_1-z_2)(\overline{z_1}-\overline{z_2})\\ &=2z_1\overline{z_1}+2z_2\overline{z_2}\\ &=2|z_1|^2+2|z_2|^2. \end{aligned}$$

Argument and polar form

Let $z=x+iy\ne0$. An argument of $z$ is any angle from the positive real axis to the point $(x,y)$, measured anticlockwise. If $\theta$ is one argument, then all arguments are

$$\theta+2k\pi,\qquad k\in\mathbb Z.$$

The principal argument is the unique argument in the interval

$$-\pi\lt\operatorname{Arg}(z)\le\pi.$$

If $r=|z|$, then

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

where $\theta$ is any argument of $z$. This is the polar form of $z$. The zero complex number has no argument: it has no direction from the origin. All arguments describe one direction modulo $2\pi$; the principal argument selects a single representative, including $\pi$ but excluding $-\pi$.

Worked example

Finding polar form

Let

$$z=-2+2\sqrt3\,i.$$

Then

$$|z|=\sqrt{(-2)^2+(2\sqrt3)^2}=4.$$

Since

$$\cos\theta=-\frac12,\qquad \sin\theta=\frac{\sqrt3}{2},$$

the point lies in quadrant II and $\theta=2\pi/3$. Thus

$$\operatorname{Arg}(z)=\frac{2\pi}{3}, \qquad \arg(z)=\frac{2\pi}{3}+2k\pi, \quad k\in\mathbb Z,$$

and

$$z=4\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right).$$

Polar form also explains conjugates and inverses:

$$\overline z=r(\cos(-\theta)+i\sin(-\theta)), \qquad z^{-1}=\frac1r(\cos(-\theta)+i\sin(-\theta)).$$

In particular, if $|z|=1$, then $z^{-1}=\overline z$.

Checkpoint

What is the principal argument of $-1-i$?

Locate the point in the complex plane.

Solution · Answer

The point is in quadrant III with reference angle $\pi/4$, so an argument is $5\pi/4$. The principal argument must lie in $(-\pi,\pi]$, hence

$$\operatorname{Arg}(-1-i)=-\frac{3\pi}{4}.$$

Multiplication as rotation and scaling

Suppose $z_1,z_2\ne0$, so both moduli are positive and both arguments exist. Write

$$z_1=r_1(\cos\theta_1+i\sin\theta_1), \qquad z_2=r_2(\cos\theta_2+i\sin\theta_2).$$

Using the angle-sum formulas,

$$z_1z_2 =r_1r_2\bigl(\cos(\theta_1+\theta_2) +i\sin(\theta_1+\theta_2)\bigr).$$

Thus multiplication multiplies moduli and adds arguments:

$$|z_1z_2|=|z_1||z_2|, \qquad \arg(z_1z_2)=\arg(z_1)+\arg(z_2)$$

up to multiples of $2\pi$.

Concept lensGeometric

A complex number as an object and as a multiplier

Algebraically, $w=u+iv$ is an object with two real coordinates, and the product $wz$ is determined by the multiplication rule. Fixing $w$ and allowing $z$ to vary gives another viewpoint: $z\mapsto wz$ is a transformation of the whole plane.

For $w\ne0$, write $w=\rho(\cos\theta+i\sin\theta)$ with $\rho>0$. The same product now means rotation about the origin through $\theta$, followed by scaling by $\rho$. For example, $i(x+iy)=-y+ix$ sends $(x,y)$ to $(-y,x)$: the algebraic coordinate rule is exactly a quarter-turn anticlockwise. For nonzero input, arguments add modulo $2\pi$; the input $z=0$ stays at the origin and needs no argument. If the multiplier is $w=0$, every point instead maps to $0$. That map collapses the plane and has no inverse; it is not a rotation followed by a positive scaling.

Explore multiplication as scale and rotation

Fix a nonzero multiplier and vary its modulus or argument. Predict the product before changing a control: its distance from the origin is multiplied by the modulus, and its direction turns through the argument. The figure lets you compare that prediction with the Cartesian coordinates.

Angle of w = 90°

|w| = 1

wz ≈ (-1, 1); |wz| = 1√2

The solid vector is $z=1+i$; the dashed vector is $wz$. Changing the angle of $w$ rotates $z$ by that angle. Changing $|w|$ scales its length. At $w=i$, the exact product is $-1+i$: a quarter-turn anticlockwise. Angles add modulo $2\pi$.

For the same nonzero operands, division reverses the second rotation and divides the lengths:

$$\frac{z_1}{z_2} =\frac{r_1}{r_2} \bigl(\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)\bigr).$$

Common mistake

Principal arguments do not simply add

It is correct that arguments add modulo $2\pi$. It is not always correct that $\operatorname{Arg}(z_1z_2)=\operatorname{Arg}(z_1)+\operatorname{Arg}(z_2)$. After adding angles, the result may need to be moved back into $(-\pi,\pi]$. For example, two arguments of $3\pi/4$ add to $3\pi/2$, whose principal representative is $-\pi/2$.

Checkpoint

If $z=2(\cos(\pi/6)+i\sin(\pi/6))$ and $w=3(\cos(-\pi/4)+i\sin(-\pi/4))$, what is the polar form of $zw$?

Multiply moduli and add arguments.

Solution · Answer

The modulus is $2\cdot3=6$, and the argument is $\pi/6-\pi/4=-\pi/12$. Therefore

$$zw=6\left(\cos\left(-\frac{\pi}{12}\right) +i\sin\left(-\frac{\pi}{12}\right)\right).$$

De Moivre's theorem

Integral powers are defined by repeated multiplication. If $z\ne0$, then $z^0=1$ and $z^{-n}=(z^{-1})^n$ for positive integers $n$.

Theorem

De Moivre's theorem

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

$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta).$$

For positive $n$, the result follows by induction from the multiplication rule in polar form. The case $n=0$ is immediate. For negative $n$, write $n=-m$, use the positive case for $m$, and use the fact that a unit complex number has inverse equal to its conjugate. Its modulus is one, so this inverse always exists. This statement concerns integer powers; it does not assign a unique value to a fractional complex power.

Worked example

Triple-angle formulas

By De Moivre's theorem,

$$\cos(3\theta)+i\sin(3\theta) =(\cos\theta+i\sin\theta)^3.$$

Expanding the right-hand side gives

$$(\cos^3\theta-3\cos\theta\sin^2\theta) +i(3\cos^2\theta\sin\theta-\sin^3\theta).$$

Comparing real and imaginary parts,

$$\cos(3\theta)=4\cos^3\theta-3\cos\theta,$$

and

$$\sin(3\theta)=3\sin\theta-4\sin^3\theta.$$

Checkpoint

Use De Moivre's theorem to simplify $(\cos\theta+i\sin\theta)^4$.

Raise the unit polar form to the fourth power.

Solution · Answer

$$(\cos\theta+i\sin\theta)^4=\cos(4\theta)+i\sin(4\theta).$$

Euler's formula and exponential form

The power series for sine, cosine, and the exponential suggest what $e^{i\theta}$ should mean:

$$e^{i\theta} =1+i\theta-\frac{\theta^2}{2!}-i\frac{\theta^3}{3!} +\frac{\theta^4}{4!}+\cdots =\cos\theta+i\sin\theta.$$

This formal comparison motivates the notation; it is not a proof of complex series convergence or of the validity of rearranging their terms. For real $\theta$, we use it here to write

$$e^{i\theta}=\cos\theta+i\sin\theta.$$

A non-zero complex number can then be written in exponential form as

$$z=re^{i\theta},$$

where $r=|z|$ and $\theta$ is an argument of $z$.

This notation compresses multiplication and De Moivre's theorem:

$$r_1e^{i\theta_1}\,r_2e^{i\theta_2} =r_1r_2e^{i(\theta_1+\theta_2)}, \qquad (e^{i\theta})^n=e^{in\theta}.$$

It also gives

$$\cos\theta=\frac12(e^{i\theta}+e^{-i\theta}), \qquad \sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta}).$$

Putting $\theta=\pi$ gives Euler's identity:

$$e^{i\pi}+1=0.$$

Triangle inequality

Theorem

Triangle inequality

For all $z_1,z_2\in\mathbb C$,

$$|z_1+z_2|\le |z_1|+|z_2|.$$

Equality holds exactly when either input is zero, or both are nonzero and point in the same direction. Equivalently, $z_1=t z_2$ or $z_2=t z_1$ for some real $t ≥ 0$.

The proof is a good example of how conjugates avoid coordinate expansion:

$$\begin{aligned} |z_1+z_2|^2 &=(z_1+z_2)(\overline{z_1}+\overline{z_2})\\ &=|z_1|^2+2\operatorname{Re}(z_1\overline{z_2})+|z_2|^2\\ &\le |z_1|^2+2|z_1\overline{z_2}|+|z_2|^2\\ &=(|z_1|+|z_2|)^2. \end{aligned}$$

Taking square roots gives the inequality because both sides are non-negative.

The only inequality above is $\operatorname{Re}(u)\le|u|$, with $u=z_1\overline{z_2}$. Equality holds precisely when $u$ is a non-negative real number: its imaginary part must vanish and its real part must be non-negative. If $z_2\ne0$, then

$$z_1=\frac{u}{|z_2|^2}z_2.$$

The scalar is non-negative, giving the asserted direction condition whenever both inputs are nonzero. If either input is zero, equality holds directly. Conversely, a non-negative multiple in either order makes the two lengths add, so these conditions are sufficient as well as necessary.

Common mistake

Modulus is not linear

Usually $|z_1+z_2|\ne |z_1|+|z_2|$. The triangle inequality gives an upper bound. If both vectors are nonzero, equality requires the same direction; equality also holds whenever either vector is zero. For example, $1$ and $i$ do not satisfy equality.

Polynomial roots over $\mathbb C$

The construction of complex numbers began because $x^2+1=0$ has no real solution. The deeper closure statement is that, once we allow complex coefficients and complex roots, every non-constant polynomial has a root.

Theorem

Fundamental theorem of algebra

Let $p(z)\in\mathbb C[z]$ have degree $n ≥ 1$ and leading coefficient $a_n\ne0$. Then the equation

$$p(z)=0$$

has at least one solution in $\mathbb C$.

The theorem is not proved in this unit, but its consequences are central. If $\alpha$ is a root, then $p(z)=(z-\alpha)q(z)$ for a polynomial $q$ whose degree is one smaller. Repeating this factorization gives

$$p(z)=a_n(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_n),$$

where the $n$ roots are counted with multiplicity: a repeated linear factor contributes its number of occurrences, not just one distinct root.

When the polynomial has real coefficients, non-real roots come in conjugate pairs. Indeed, if

$$p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0,\qquad a_k\in\mathbb R,$$

and $p(\alpha)=0$, then taking conjugates of the equation gives $p(\overline\alpha)=0$.

This explains why real polynomials factor into real linear and real quadratic pieces:

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

Worked example

Conjugate roots in a real quadratic

For

$$x^2+2x+5=0,$$

the discriminant is $4-20=-16$. Hence

$$x=\frac{-2\pm\sqrt{-16}}2=-1\pm2i.$$

The two roots are conjugates. Their product factor is

$$(x-(-1+2i))(x-(-1-2i))=x^2+2x+5.$$

Checkpoint

Let $\omega=e^{2\pi i/5}$. Why is $\omega^3$ also a fifth root of unity?

Use the law of indices.

Solution · Answer

Because

$$(\omega^3)^5=\omega^{15}=(\omega^5)^3=1^3=1.$$

Thus $\omega^3$ is a root of $z^5-1=0$.

Roots of unity and solving $z^n=z_0$

Let $n$ be a positive integer. Polar form makes roots of equations such as $z^n=1$ visible. Define

$$\omega=e^{2\pi i/n} =\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}.$$

Then $\omega^n=1$, and

$$1,\omega,\omega^2,\ldots,\omega^{n-1}$$

are the $n$ distinct $n$-th roots of unity. Therefore

$$z^n-1=(z-1)(z-\omega)(z-\omega^2)\cdots(z-\omega^{n-1}).$$

Over the real numbers, conjugate roots are paired into quadratic factors. For example, if $n=5$, then

$$z^5-1 =(z-1) \left(z^2-2\cos\frac{2\pi}{5}\,z+1\right) \left(z^2-2\cos\frac{4\pi}{5}\,z+1\right).$$

The same idea solves $z^n=z_0$. If $z_0=r_0e^{i\theta_0}$ with $r_0\gt 0$, then write a possible root as $z=re^{i\theta}$ with $r\gt 0$. Moduli and arguments then give

$$r^n=r_0,\qquad n\theta=\theta_0+2k\pi,\quad k\in\mathbb Z.$$

Thus the solutions are

$$\alpha_k=r_0^{1/n} e^{i(\theta_0+2k\pi)/n}, \qquad k=0,1,\ldots,n-1.$$

Each root has the same modulus $r_0^{1/n}$; the arguments are equally spaced by $2\pi/n$. Two listed indices give the same root only if their difference is a multiple of $n$, so the indices $0,1,\ldots,n-1$ give distinct roots. Every integer index reduces to one of these modulo $n$, proving that the list is exhaustive. If $z_0=0$, the only root is $z=0$, and the nonzero polar-root formula is not used.

Worked example

Cube roots of one

For $n=3$,

$$\omega=e^{2\pi i/3}=-\frac12+\frac{\sqrt3}{2}i.$$

The three cube roots of one are

$$1,\quad \omega,\quad \omega^2 =-\frac12-\frac{\sqrt3}{2}i.$$

They lie on the unit circle at angles $0$, $2\pi/3$, and $4\pi/3$, forming an equilateral triangle.

Checkpoint

For pairwise distinct points $z_1,z_2,z_3$, how can the ratio $(z_1-z_3)/(z_2-z_3)$ detect collinearity?

Think about its argument.

Solution · Answer

The ratio compares two directed vectors from $z_3$. If the ratio is real, its argument is $0$ or $\pi$, so the two vectors point in the same or opposite direction. Therefore the three points lie on one line.

Geometry with complex ratios

Differences of complex numbers represent directed vectors. For distinct points $z_1,z_2,z_3$, write

$$\frac{z_1-z_3}{z_2-z_3}=re^{i\theta}.$$

Then $r=|z_1-z_3|/|z_2-z_3|$, and $\theta$ is the oriented angle from the vector $z_2-z_3$ to the vector $z_1-z_3$. Thus one complex ratio records both a side-length ratio and an angle.

This gives a compact collinearity test:

Theorem

Collinearity test

Distinct points $z_1,z_2,z_3\in\mathbb C$ are collinear if and only if

$$\frac{z_1-z_3}{z_2-z_3}\in\mathbb R.$$

Equivalently,

$$\begin{vmatrix} 1&1&1\\ z_1&z_2&z_3\\ \overline{z_1}&\overline{z_2}&\overline{z_3} \end{vmatrix}=0.$$

The real-ratio condition says the two directed vectors have argument difference $0$ or $\pi$; that is exactly the condition that they lie on the same line.

For two nondegenerate triangles, complex ratios also encode oriented similarity:

$$\triangle z_1z_2z_3\sim \triangle w_1w_2w_3 \quad\Longleftrightarrow\quad \frac{z_1-z_3}{z_2-z_3} = \frac{w_1-w_3}{w_2-w_3},$$

when the orientation is the same. The equality means the corresponding side-ratio and included angle both agree.

Worked example

Equilateral triangle criterion

Let $\omega=e^{2\pi i/3}$. For a counter-clockwise triangle with vertices $z_1,z_2,z_3$, the triangle is equilateral exactly when it is similar to $1,\omega,\omega^2$. One compact form of the condition is

$$z_1+z_2\omega+z_3\omega^2=0.$$

This is useful because it turns a geometric statement about equal side lengths and $60^\circ$ angles into one algebraic equation.

For four distinct complex points, define the cross ratio

$$(z_1,z_2;z_3,z_4) =\frac{(z_1-z_3)(z_2-z_4)} {(z_2-z_3)(z_1-z_4)}$$

for four distinct points. At this level, the important lesson is not to memorize every identity at once, but to notice the recurring strategy: translation is represented by differences, scaling and rotation by multiplication, and shape information by ratios.

Locus equations and transformations

Many plane loci can be written cleanly in complex notation. The equation

$$|z-z_0|=r$$

is a circle when $r\gt 0$, centered at $z_0$ with radius $r$. Squaring gives

$$(z-z_0)(\overline z-\overline{z_0})=r^2,$$

which is often easier to expand.

For distinct endpoints $z_1\ne z_2$, similarly,

$$|z-z_1|=|z-z_2|$$

is the perpendicular bisector of the segment joining $z_1$ and $z_2$, because it is the set of points equally distant from the two endpoints. A line through $z_1$ and $z_2$ can be described by the real-ratio condition

$$\frac{z-z_2}{z_1-z_2}\in\mathbb R.$$

For three fixed non-collinear points $z_1,z_2,z_3$ and a point $z\notin\{z_1,z_2,z_3\}$, membership in their circle is equivalent to

$$(z_1,z_2;z_3,z)\in\mathbb R.$$

Add the three fixed vertices separately to describe the entire circle. In particular, the displayed rational expression has a pole at $z=z_1$; the four-distinct-point definition does not evaluate the test at any fixed vertex.

Worked example

An Apollonius circle

Show that

$$|z-1|=3|z+1|$$

represents a circle. As a distance ratio this is $|z-1|/|z+1|=3$, with $z\ne-1$; the original equality already excludes that point. Squaring and expanding,

$$(z-1)(\overline z-1)=9(z+1)(\overline z+1).$$

After collecting terms,

$$z\overline z+\frac54z+\frac54\overline z+1=0.$$

Completing the square gives

$$\left|z+\frac54\right|=\frac34.$$

So the locus is the circle centered at $-5/4$ with radius $3/4$. At $z=-1$, the distance to this center is $1/4$, not $3/4$, so the completed circle introduces no excluded denominator point.

The following transformation discussion is optional. Complex transformations act on points of the complex plane, with a restricted domain when division is involved. The basic transformations are:

• scaling: $f(z)=rz$, with real $r\ne0$;
• translation: $f(z)=z+a$;
• rotation: $f(z)=e^{i\theta}z$;
• inversion: $f(z)=1/z$, for $z\ne0$;
• reflection in the real axis: $f(z)=\overline z$.

For example, $f(z)=z^2$ sends $z=re^{i\theta}$ to

$$f(z)=r^2e^{2i\theta},$$

so it squares distances from the origin and doubles arguments.

Translations, rotations, and conjugation preserve distances. Translations, rotations, and scalings preserve the three-point ratio

$$\frac{z_1-z_3}{z_2-z_3}.$$

This explains why they preserve similarity. Cross ratios are even more stable: they are preserved by scalings, translations, rotations, and inversions, hence by Mobius transformations wherever the finite expressions are defined

$$f(z)=\frac{az+b}{cz+d},\qquad ad-bc\ne0.$$

The finite formula requires $cz+d\ne0$. The whole-line and whole-circle preservation statement uses the extended plane $\mathbb C\cup\{\infty\}$, where every line includes infinity. If $c\ne0$, set $f(-d/c)=\infty$ and $f(\infty)=a/c$; if $c=0$, set $f(\infty)=\infty$. With these conventions, Mobius transformations send lines and circles to lines and circles. Restricting to finite inputs and finite outputs can omit a point of the resulting locus. One algebraic form for the finite line or circle is

$$A z\overline z+Bz+\overline B\,\overline z+C=0,$$

where $A,C\in\mathbb R$ and $|B|^2-AC\gt 0$. Under this condition, $A\ne0$ gives a genuine circle, while $A=0$ gives a line.

Quick checks

Checkpoint

What geometric object is described by $|z-1|=|z+3|$?

Read it as an equal-distance condition.

Solution · Answer

It is the perpendicular bisector of the segment joining $1$ and $-3$. Since those points lie on the real axis, the midpoint is $-1$, and the perpendicular bisector is the vertical line $\operatorname{Re}(z)=-1$.

Exercises

1. Prove that the embedding $\phi:\mathbb R\to\mathbb C$, $\phi(x)=x+i0$, preserves addition and multiplication.
2. Express $(3-2i)/(1+i)$ in the form $a+bi$.
3. Prove that $z\overline z=|z|^2$ and use it to derive the inverse formula for non-zero $z$.
4. Find the modulus, all arguments, principal argument, and polar form of $-\sqrt3+i$.
5. Prove $|1-\overline{z_1}z_2|^2-|z_1-z_2|^2=(1-|z_1|^2)(1-|z_2|^2)$.
6. Use De Moivre's theorem to express $\cos(5\theta)$ and $\sin(5\theta)$ in terms of powers of $\cos\theta$ and $\sin\theta$.
7. Prove by induction that $|\sum_{k=1}^n z_k|\le \sum_{k=1}^n |z_k|$.
8. For $\theta\ne2k\pi$, use the geometric sum for $e^{i\theta}$ to derive the formula $\sum_{k=1}^n \cos(k\theta)=\sin(n\theta/2)\cos((n+1)\theta/2)/\sin(\theta/2)$.
9. Factor $z^6-1$ over $\mathbb C$ and then group the non-real conjugate roots to factor it over $\mathbb R$.
10. Solve $z^4=16$ in polar form.
11. Show that if $p(z)\in\mathbb R[z]$ and $p(\alpha)=0$, then $p(\overline\alpha)=0$.
12. For pairwise distinct points $z_1,z_2,z_3$, prove that they are collinear exactly when $(z_1-z_3)/(z_2-z_3)$ is real.
13. Show directly that $|z-1|=3|z+1|$ is the circle $|z+5/4|=3/4$.
14. Let $f(z)=az+b$ with $a\ne0$. Prove that $(f(z_1)-f(z_3))/(f(z_2)-f(z_3))=(z_1-z_3)/(z_2-z_3)$ for distinct $z_1,z_2,z_3$.

Guided solutions

Solution · Model solution 1

For real $a,b$, $\phi(a+b)=(a+b)+i0=(a+i0)+(b+i0)=\phi(a)+\phi(b)$. Also $\phi(ab)=ab+i0=(a+i0)(b+i0)=\phi(a)\phi(b)$. Thus real arithmetic is preserved inside $\mathbb C$.

Solution · Model solution 2

Multiply by the conjugate of the denominator:

$$\frac{3-2i}{1+i} =\frac{(3-2i)(1-i)}{(1+i)(1-i)} =\frac{1-5i}{2} =\frac12-\frac52 i.$$

Solution · Model solution 3

If $z=x+iy$, then $z\overline z=(x+iy)(x-iy)=x^2+y^2=|z|^2$. If $z\ne0$, then $|z|^2\gt 0$, so $z^{-1}=\overline z/|z|^2=(x-iy)/(x^2+y^2)$.

Solution · Model solution 4

The modulus is $2$. Since $\cos\theta=-\sqrt3/2$ and $\sin\theta=1/2$, the point is in quadrant II with principal argument $5\pi/6$. Hence all arguments are $5\pi/6+2k\pi$, and $-\sqrt3+i=2(\cos(5\pi/6)+i\sin(5\pi/6))$.

Solution · Model solution 5

Expand both squared moduli using conjugates:

$$|1-\overline{z_1}z_2|^2 =(1-\overline{z_1}z_2)(1-z_1\overline{z_2}) =1-\overline{z_1}z_2-z_1\overline{z_2}+|z_1|^2|z_2|^2,$$

while

$$|z_1-z_2|^2 =|z_1|^2-z_1\overline{z_2}-\overline{z_1}z_2+|z_2|^2.$$

Subtracting cancels the mixed terms and gives $(1-|z_1|^2)(1-|z_2|^2)$.

Solution · Model solution 6

Expand $(\cos\theta+i\sin\theta)^5=\cos(5\theta)+i\sin(5\theta)$ by the binomial theorem. The real terms are those with even powers of $i$, and the imaginary terms are those with odd powers of $i$:

$$\cos(5\theta)=\cos^5\theta-10\cos^3\theta\sin^2\theta +5\cos\theta\sin^4\theta,$$$$\sin(5\theta)=5\cos^4\theta\sin\theta -10\cos^2\theta\sin^3\theta+\sin^5\theta.$$

Solution · Model solution 7

The case $n=1$ is immediate. If the result holds for $n$, then

$$\left|\sum_{k=1}^{n+1}z_k\right| = \left|\left(\sum_{k=1}^{n}z_k\right)+z_{n+1}\right| \le \left|\sum_{k=1}^{n}z_k\right|+|z_{n+1}| \le \sum_{k=1}^{n+1}|z_k|.$$

Solution · Model solution 8

Since $e^{i\theta}\ne1$,

$$\sum_{k=1}^n e^{ik\theta} =e^{i\theta}\frac{e^{in\theta}-1}{e^{i\theta}-1}.$$

Rewrite numerator and denominator by factoring half-angles:

$$e^{in\theta}-1 =e^{in\theta/2}(e^{in\theta/2}-e^{-in\theta/2}),$$

and similarly for $e^{i\theta}-1$. This gives

$$\sum_{k=1}^n e^{ik\theta} = \frac{\sin(n\theta/2)}{\sin(\theta/2)} e^{i(n+1)\theta/2}.$$

Taking real parts gives the stated cosine-sum formula.

Solution · Model solution 9

Over $\mathbb C$, let $\omega=e^{2\pi i/6}$. Then

$$z^6-1=\prod_{k=0}^{5}(z-\omega^k).$$

The roots are $1$, $-1$, $e^{\pi i/3}$, $e^{-\pi i/3}$, $e^{2\pi i/3}$, and $e^{-2\pi i/3}$. Pairing conjugates gives

$$z^6-1=(z-1)(z+1) \left(z^2-2\cos\frac{\pi}{3}z+1\right) \left(z^2-2\cos\frac{2\pi}{3}z+1\right),$$

so over $\mathbb R$,

$$z^6-1=(z-1)(z+1)(z^2-z+1)(z^2+z+1).$$

Solution · Model solution 10

Write $16=16e^{2\pi i m}$. The fourth roots have modulus $2$ and arguments $(2k\pi)/4=k\pi/2$, so the solutions are

$$2,\quad 2i,\quad -2,\quad -2i.$$

Solution · Model solution 11

Write $p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_0$ with all $a_j$ real. If $p(\alpha)=0$, then conjugating both sides gives

$$a_n\overline\alpha^n+a_{n-1}\overline\alpha^{n-1} +\cdots+a_0=0,$$

because conjugation respects sums and products and $\overline{a_j}=a_j$. This is exactly $p(\overline\alpha)=0$.

Solution · Model solution 12

If the points are collinear, the vectors $z_1-z_3$ and $z_2-z_3$ have the same or opposite direction, so their argument difference is $0$ or $\pi$; the quotient is real. Conversely, if the quotient is real, its argument is $0$ or $\pi$, so the two vectors are parallel and the three points lie on a line.

Solution · Model solution 13

Squaring gives

$$(z-1)(\overline z-1)=9(z+1)(\overline z+1).$$

Expanding and collecting terms gives

$$z\overline z+\frac54z+\frac54\overline z+1=0.$$

This is

$$\left(z+\frac54\right)\left(\overline z+\frac54\right)=\frac9{16},$$

hence $|z+5/4|=3/4$.

Solution · Model solution 14

Since $f(z_j)=az_j+b$,

$$\frac{f(z_1)-f(z_3)}{f(z_2)-f(z_3)} =\frac{a(z_1-z_3)}{a(z_2-z_3)} =\frac{z_1-z_3}{z_2-z_3}.$$

The translation part cancels, and the non-zero scaling/rotation factor $a$ cancels.