# Complex Numbers in Geometry

In plane geometry, complex numbers can be used to represent points, and thus other geometric objects as well such as lines, circles, and polygons. They are somewhat similar to Cartesian coordinates in the sense that they are used to algebraically prove geometric results, but they are especially useful in proving results involving circles and/or regular polygons (unlike Cartesian coordinates, which are useful for proving results involving lines).

#### Contents

## Basics

A point in the plane can be represented by a complex number

\[z = x + yi,\]

which corresponds to the Cartesian point \((x,y)\). Therefore, the \(x\)-axis is renamed the **real axis** and the \(y\)-axis is renamed the **imaginary axis**, or **imaginary line**. This can also be converted into a **polar coordinate** \((r,\theta)\), which represents the complex number

\[(r,\theta) = re^{i\theta},\]

which, intuitively speaking, means rotating the point \((r,0)\) an angle of \(\theta\) about the origin. By Euler's formula, this is equivalent to

\[(r,\theta) = re^{i\theta}=r\cos\theta + ri\sin\theta,\]

which means that the polar coordinate \((r,\theta)\) corresponds to the Cartesian coordinate \((r\cos\theta,r\sin\theta).\)

Incidentally, this immediately illustrates why complex numbers are so useful for circles and regular polygons: these involve heavy use of rotations, which are easily expressed using complex numbers.

In particular, a rotation of \(\theta\) about the origin sends \(z \rightarrow ze^{i\theta}\) for all \(\theta.\)

In comparison, rotating Cartesian coordinates involves heavy calculation and (generally) an ugly result.

Additionally, each point \(z=a+bi\) has an associated **conjugate** \(\overline{z}=a-bi\). It satisfies the properties

- \(\overline{w+z}=\overline{w}+\overline{z}\)
- \(\overline{wz}=\overline{w} \cdot \overline{z}\)
- \(|z|^2=z \cdot \overline{z}.\)

Geometrically, the conjugate can be thought of as the reflection over the real axis. This implies two useful facts: if \(z\) is real, \(z = \overline{z}\), and if \(z\) is pure imaginary, \(z = -\overline{z}\).

## Lines in Complex Geometry

*In this and the following sections, a capital letter denotes a point and the analogous lowercase letter denotes the complex number associated with it.*

Though lines are less nice in complex geometry than they are in coordinate geometry, they still have a nice characterization:

The points \(A,B,C\) are collinear if and only if \(\frac{a-b}{b-c}\) is real, or equivalently, if and only if

\[\frac{a-b}{\ \overline{a}-\overline{b}\ }=\frac{a-c}{\ \overline{a}-\overline{c}\ }.\]

There are two similar results involving lines. This is the one for parallel lines:

Lines \(AB\) and \(CD\) are parallel if and only if \(\frac{a-b}{c-d}\) is real, or equivalently, if and only if

\[\frac{a-b}{\ \overline{a}-\overline{b}\ } = \frac{c-d}{\ \overline{c}-\overline{d}\ }.\]

The following is the result for perpendicular lines:

Lines \(AB\) and \(CD\) are perpendicular if and only if \(\frac{a-b}{c-d}\) is pure imaginary, or equivalently, if and only if

\[\frac{a-b}{\ \overline{a}-\overline{b}\ } = -\frac{c-d}{\ \overline{c}-\overline{d}\ }.\]

Additionally, there is a nice expression of reflection and projection in complex numbers:

Let \(W\) be the reflection of \(Z\) over \(AB\). Then

\[w = \frac{(a-b)\overline{z}+\overline{a}b-a\overline{b}}{\overline{a}-\overline{b}}\]

and the projection of \(Z\) onto \(AB\) is \(\frac{w+z}{2}\).

However, it is easy to express the intersection of two lines in Cartesian coordinates. In complex coordinates, this is not quite the case:

Lines \(AB\) and \(CD\) intersect at the point

\[\frac{\big(\overline{a}b-a\overline{b}\big)(c-d)-(a-b)\big(\overline{c}d-c\overline{d}\big)}{\big(\overline{a}-\overline{b}\big)(c-d)-(a-b)\big(\overline{c}-\overline{d}\big)},\]

which is impractical to use in all but a few specific situations (e.g. when one of the points is at 0).

## The Unit Circle

The unit circle is of special interest in the complex plane, as points \(z\) on the complex plane satisfy the key property that

\[z = \frac{1}{\overline{z}},\]

which is a consequence of the fact that \(|z|=1\). This means that

in general, complex geometry is most useful when there is a primary circle in the problem that can be set to the unit circle.

For instance, some of the formulas from the previous section become significantly simpler. Reflection and projection, for instance, simplify nicely:

If \(A,B\) lie on the unit circle, the reflection of \(z\) across \(AB\) is \(a+b-ab\overline{z}\). The projection of \(z\) onto \(AB\) is thus \(\frac{1}{2}(z+a+b-ab\overline{z})\).

Also, the intersection formula becomes practical to use:

If \(A,B,C,D\) lie on the unit circle, lines \(AB\) and \(CD\) intersect at

\[\frac{ab(c+d)-cd(a+b)}{ab-cd}.\]

## Triangle Centers

Triangles in complex geometry are extremely nice when they can be placed on the unit circle; this is generally possible, by setting the triangle's circumcircle to the unit circle. This immediately implies the following obvious result:

Suppose \(A,B,C\) lie on the unit circle. Then the circumcenter of \(ABC\) is 0.

This is because the circumcenter of \(ABC\) coincides with the center of the unit circle.

The centroid is also easy to compute:

Suppose \(A,B,C\) lie on the unit circle. Then the centroid of \(ABC\) is \(\frac{a+b+c}{3}\).

This also illustrates the similarities between complex numbers and vectors.

More interestingly, we have the following theorem:

Suppose \(A,B,C\) lie on the unit circle. Then the orthocenter of \(ABC\) is \(a+b+c.\)

Proof:Let \(h = a + b +c\). Recall from the "lines" section that \(AH\) is perpendicular to \(BC\) if and only if \(\frac{h-a}{b-c}\) is pure imaginary. This is equal to \(\frac{b+c}{b-c}\) since \(h=a+b+c\). Then

\[\overline{\left(\frac{b+c}{b-c}\right)} = \frac{\overline{b}+\overline{c}}{\ \overline{b}-\overline{c}\ }.\]

Since \(B,C\) are on the unit circle, \(\overline{b}=\frac{1}{b}\) and \(\overline{c}=\frac{1}{c}\). So

\[\overline{\left(\frac{b+c}{b-c}\right)} = \frac{\overline{b}+\overline{c}}{\ \overline{b}-\overline{c}\ } = \frac{\frac{1}{b}+\frac{1}{c}}{\frac{1}{b}-\frac{1}{c}}=\frac{b+c}{c-b},\]

which implies \(\overline{\left(\frac{b+c}{b-c}\right)}=-\left(\frac{b+c}{b-c}\right)\). From the intro section, this implies that \(\left(\frac{b+c}{b-c}\right)\) is pure imaginary, so \(AH\) is perpendicular to \(BC\).

By similar logic, \(BH\) is perpendicular to \(AC\) and \(CH\) to \(AB\), so \(H\) is the orthocenter, as desired. \(_\square\)

It is also possible to find the incenter, though it is considerably more involved:

Suppose \(A,B,C\) lie on the unit circle, and let \(I\) be the incenter. Let \(D,E,F\) be the feet of the angle bisectors from \(A,B,C,\) respectively. Then there exist complex numbers \(x,y,z\) such that \(a=x^2, b=y^2, c=z^2, d=-yz, e=-xz, f=-xy\). Then

\[I = -(xy+yz+zx).\]

## Other Properties

There are two other properties worth noting before attempting some problems. The first is the tangent line through the unit circle:

Let \(W\) lie on the unit circle. Then \(Z\) lies on the tangent through \(W\) if and only if

\[z+w^2\overline{z}=2w.\]

This is especially useful in the case of two tangents:

Let \(X,Y\) be points on the unit circle. Their tangents meet at the point \(\frac{2xy}{x+y},\) the harmonic mean of \(x\) and \(y\).

Proof:Let \(Z\) be the intersection point. Then \(z+x^2\overline{z}=2x\) and \(z+y^2\overline{z}=2y\), so

\[\big(x^2-y^2\big)\overline{z}=2(x-y) \implies (x+y)\overline{z}=2 \implies \overline{z}=\frac{2}{x+y}.\]

Thus, \(z=\overline{\left(\frac{2}{x+y}\right)}=\frac{2}{\overline{x}+\overline{y}}\). Since \(x,y\) lie on the unit circle, \(\overline{x}=\frac{1}{x}\) and \(\overline{y}=\frac{1}{y}\), so \(z=\frac{2xy}{x+y},\) as desired. \(_\square\)

The second result is a condition on cyclic quadrilaterals:

Points \(A,B,C,D\) lie on a circle if and only if

\[\large\frac{\frac{c-a}{c-b}}{\hspace{3mm} \frac{d-a}{d-b}\hspace{3mm} }\]

is real.

## Illustrative Problems

This section contains Olympiad problems as examples, using the results of the previous sections.

A point \(A\) is taken inside a circle. For every chord of the circle passing through \(A,\) consider the intersection point of the two tangents at the endpoints of the chord. Find the locus of these intersection points.

WLOG assume that \(A\) is on the real axis. Let \(P,Q\) be the endpoints of a chord passing through \(A\). From the previous section, the tangents through \(p\) and \(q\) intersect at \(z=\frac{2}{\overline{p}+\overline{q}}\). It is also true since \(P,A,Q\) are collinear, that

\[ \begin{align*} \frac{p-a}{\overline{p}-\overline{a}}&=\frac{a-q}{a-\overline{q}} \\ \\ \frac{p-a}{\frac{1}{p}-a}&=\frac{a-q}{a-\frac{1}{q}} \\ \\ pa-\frac{p}{q}+\frac{a}{q}&=\frac{a}{p}-\frac{q}{p}+aq \\ \\ p^2aq-p^2+ap&=aq-q^2+apq^2 \\ \\ ap-aq+p^2aq-apq^2&=p^2-q^2 \\ \\ a+apq&=p+q \\ \\ a&=\frac{p+q}{pq+1}. \end{align*} \]

Now note that

\[\text{Re}(z)=\frac{z+\overline{z}}{2}=\frac{1}{p+q}+\frac{1}{\overline{p}+\overline{q}}=\frac{pq+1}{p+q}=\frac{1}{a},\]

so \(z\) must lie on the vertical line through \(\frac{1}{a}\).

For any point on this line, connecting the two tangents from the point to the unit circle at \(P\) and \(Q\) allows the above steps to be reversed, so every point on this line works; hence, the desired locus is this line. More formally, the locus is a line perpendicular to \(OA\) that is a distance \(\frac{1}{OA}\) from \(O\). \(_\square\)

The following application of what we have learnt illustrates the fact that complex numbers are more than a tool to solve or "bash" geometry problems that have otherwise "beautiful" synthetic solutions, they often lead to the most beautiful and unexpected of solutions.

Consider a polygonal line \(P_0P_1...P_n\) such that \(\angle P_0P_1P_2=\angle P_1P_2P_3=...=\angle P_{n-2}P_{n-1}P_{n}\), all measured clockwise. If \(P_0P_1>P_1P_2>...>P_{n-1}P_{n}\), \(P_0\) and \(P_n\) cannot coincide.

Proof:Let us consider complex coordinates with origin at \(P_0\) and let the line \(P_0P_1\) be the x-axis. Let \(\alpha\) be the angle between any two consecutive segments and let \(a_1>a_2>...>a_n\) be the lengths of the segments. If we set \(z=e^{i(\pi-\alpha)}\), then the coordinate of \(P_{n}\) is \(a_1+a_2z+...+a_{n}z^{n-1}\). We must prove that this number is not equal to zero.

Using the Abel Summation lemma, we obtain,

\(a_1+a_2z+...+a_{n-1}z^n=(a_1-a_2) + (a_2-a_3)(1+z) + (a_3-a_4)(1+z+z^2) + ... + a_{n}(1+z+...+z^{n-1})\).

If \(\alpha\) is zero, then this quantity is a strictly positive real number, and we are done.

If not, multiply by \((1-z)\) to get \((a_1-a_2)(1-z) + (a_2-a_3)(1-z^2) + (a_3-a_4)(1-z^3) + ... + a_{n}(1-z^n)\). This expression cannot be zero. Indeed, since \(\mid z\mid=1\), by the triangle inequality, we have,

\(\mid (a_1-a_2)z + (a_2-a_3)z^2 + (a_3-a_4)z^3 + ... + a_{n}z^n \mid < (a_1-a_2) + (a_2-a_3) + (a_3-a_4) + ... + a_{n}\).

The conclusion follows.

## See Also

**Cite as:**Complex Numbers in Geometry.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/complex-numbers-in-geometry/