Applications of Complex Numbers and Vectors

In one sentence

Points of the plane $\leftrightarrow$ complex numbers, and therefore:

the distance between two points is $∣ Z_{1} Z_{2} ∣ = ∣ z_{1} - z_{2} ∣$ , and $∣ z_{1} - z_{2} ∣^{2} = ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} - 2 Re (z_{1} \overset{z}{ˉ}_{2})$ ;
turning $A B$ about $A$ counterclockwise by $θ$ gives $a + (b - a) (cos θ + i sin θ)$ : rotation is multiplication;
$A, B, C$ are collinear $⟺ \frac{c - a}{b - a}$ is real; $A B ⊥ A C ⟺ \frac{c - a}{b - a}$ is purely imaginary;
four points are concyclic (or collinear) $⟺ \frac{z _{3} - z _{1}}{z _{4} - z _{1}} / \frac{z _{3} - z _{2}}{z _{4} - z _{2}}$ is real.

The formulas are not the hard part. The hard part is translating the geometric statement you want to prove into an equation. The four “stops” below are four kinds of translation.

Squares on two sides: BG and CE are equal and perpendicular

Step 1 / 4

The figure
On the sides $A B$ and $A C$ of $△ A B C$ , construct the squares $A B D E$ and $A C F G$ outward. We want to prove $B G = C E$ and $B G ⊥ C E$ .

$BG = CE, BG ⊥ CE$
Translate the squares into multiplication
Compute the two vectors
The ratio is exactly i

Stop 1 · Rotation: equilateral triangles

The third vertex of an equilateral triangle $A B T$ is $B$ turned about $A$ by $\pm 6 0^{\circ}$ . The result in the figure at the top of the page is Napoleon’s theorem: construct equilateral triangles outward on the three sides, and their three centers form an equilateral triangle. To prove it with complex numbers, compute the three centers $x, y, z$ and check that $z - x = (y - x) (cos 6 0^{\circ} \pm i sin 6 0^{\circ})$ ; the simplification uses $ω^{2} + ω + 1 = 0$ again and again.

Predict first

An equilateral triangle has vertices $A = 0$ and $B = 2$ , and its third vertex is in the upper half-plane. Which complex number is it?

1 + i

1 + 3 i

3 + i

2 i

Choose an answer first; then you can see the result.

Stop 2 · Concyclic: a ratio is real

The argument of $\frac{z _{3} - z _{1}}{z _{4} - z _{1}}$ is the directed angle $∠ Z_{4} Z_{1} Z_{3}$ , and the argument of $\frac{z _{3} - z _{2}}{z _{4} - z _{2}}$ is $∠ Z_{4} Z_{2} Z_{3}$ . The two angles being equal (same side) or differing by $18 0^{\circ}$ (opposite sides) is exactly “inscribed angles on the same arc are equal / opposite angles are supplementary”. Then the quotient has argument $0^{\circ}$ or $18 0^{\circ}$ , so it is real.

All four points sit on lattice points. Drag

Z_{4}

onto the dashed circle (for example to

(- 1.5, 2)

(- 2.5, 0)

) and watch the imaginary part of

λ

Stop 3 · Area: the cross product hides in the imaginary part

The area of $△ Z_{1} Z_{2} Z_{3}$ is $\frac{1}{2} Im (\overline{(z_{2} - z_{1})} (z_{3} - z_{1}))$ . If $z_{2} - z_{1} = p + q i$ and $z_{3} - z_{1} = r + s i$ , the imaginary part inside is $p s - q r$ , the “cross product” of the vectors; it is positive when the three vertices go counterclockwise.

One problem, several solutions

Find the area of the triangle with vertices $1 + i$ , $4 + 2 i$ , $2 + 5 i$ .

Complex numbers (cross product)

$z_{2} - z_{1} = 3 + i$ , $z_{3} - z_{1} = 1 + 4 i$ . $\overline{(3 + i)} (1 + 4 i) = (3 - i) (1 + 4 i) = 7 + 11 i$ , so the area is $\frac{1}{2} \times 11 = \frac{11}{2}$ .

CostOne step, and it works for a triangle in any position.

Cut and subtract

The enclosing rectangle $[1, 4] \times [1, 5]$ has area $12$ . Subtract three right triangles: $\frac{1}{2} \cdot 3 \cdot 1 + \frac{1}{2} \cdot 2 \cdot 3 + \frac{1}{2} \cdot 1 \cdot 4 = 1.5 + 3 + 2 = 6.5$ . The area is $12 - 6.5 = 5.5$ .

The same triangle shifted by

(- 3, - 3)

; the area does not change.

CostNo new formula, but you need a picture and three right triangles to subtract.

Stop 4 · Collinearity and dividing points

If $P$ divides $A B$ in the ratio $λ$ ( $A P = λ P B$ , $λ \neq = - 1$ ), then $p = \frac{a + λb}{1 + λ}$ . Three points $A, B, C$ are collinear $⟺$ there are real numbers $λ_{1}, λ_{2}, λ_{3}$ , not all zero, with $λ_{1} + λ_{2} + λ_{3} = 0$ and $λ_{1} a + λ_{2} b + λ_{3} c = 0$ . This is the same fact as “ $x + y = 1$ in $A D = x A B + y A C$ ” from the exercises of the station Adding and Subtracting Vectors.

Practice

Exercise 1Basic

The vertices of the square $A B C D$ go counterclockwise, with $A = 1 + i$ and $B = 4 + 2 i$ . Which complex number is $D$ ?

Exercise 2Basic

Find the area of the triangle with vertices $0$ , $3 + i$ , $1 + 4 i$ .

Exercise 3Advanced

The four points $0$ , $4$ , $4 + 2 i$ , $y i$ ( $y \neq = 0$ ) are concyclic. Find $y$ .

Exercise 4Challenge

On $A = 0$ and $B = 6$ , construct the equilateral triangle $A B C$ downward (in the lower half-plane), and let $O$ be its center. Find $∣ o ∣^{2}$ .

In one sentence

Squares on two sides: BG and CE are equal and perpendicular

Stop 1 · Rotation: equilateral triangles

Stop 2 · Concyclic: a ratio is real

Stop 3 · Area: the cross product hides in the imaginary part

Complex numbers (cross product)

Cut and subtract

Stop 4 · Collinearity and dividing points

Practice

Mistake Museum · Exhibits from this stop

Rotating the wrong way

Calling the points concyclic because a ratio is real