Geometric problems with complex numbers

When you move, stretch or turn a picture on a computer, you act on its vertices. If the vertices are complex numbers, each move is one operation:

• translating by the vector $\displaystyle w$: $\displaystyle z\mapsto z+w$;
• stretching by $\displaystyle k>0$ from the origin: $\displaystyle z\mapsto kz$;
• reflecting in the real axis: $\displaystyle z\mapsto\bar{z}$;
• rotating about the origin by the angle $\displaystyle \theta$: $\displaystyle z\mapsto z\operatorname{cis}\theta$; about $\displaystyle z_0$: $\displaystyle z\mapsto z_0+(z-z_0)\operatorname{cis}\theta$.

For example, in a picture of a cat the ears are at $\displaystyle {A}={1+6i}$ and $\displaystyle {B}={2+6i}$, and the tail at $\displaystyle {C}={5+5i}$. Multiplying by $\displaystyle i$ turns the picture a quarter turn about the origin: $\displaystyle {iA}={-6+i}$, $\displaystyle {iB}={-6+2i}$, $\displaystyle {iC}={-5+5i}$.

As with vectors:

• the midpoint of the segment $\displaystyle z_1z_2$: $\displaystyle \frac{z_1+z_2}{2}$;
• the meeting point of the medians of the triangle $\displaystyle z_1z_2z_3$: $\displaystyle \frac{z_1+z_2+z_3}{3}$;
• $\displaystyle z_1z_2z_3z_4$ is a parallelogram exactly when $\displaystyle {z_1+z_3}={z_2+z_4}$: the diagonals have the same midpoint.

In a product the angles add, so a product can work out a sum of angles. The example below proves $\displaystyle {\tan^{-1}1+\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}}={\frac{\pi}{2}}$.

• The angles of $\displaystyle 1+i$, $\displaystyle 2+i$ and $\displaystyle 3+i$ are $\displaystyle \tan^{-1}1$, $\displaystyle \tan^{-1}\frac{1}{2}$ and $\displaystyle \tan^{-1}\frac{1}{3}$ – all in the first quadrant.
• The angle of the product is the sum of the angles.
• $\displaystyle {(1+i)(2+i)(3+i)}={(1+3i)(3+i)}={10i}$, and the angle of $\displaystyle 10i$ is $\displaystyle \frac{\pi}{2}$.
• The sum is between $\displaystyle 0$ and $\displaystyle \frac{3\pi}{2}$, so it is $\displaystyle \frac{\pi}{2}$ itself, not another $\displaystyle \frac{\pi}{2}+2k\pi$.

The cat’s points are $\displaystyle {A}={1+6i}$, $\displaystyle {B}={2+6i}$ and $\displaystyle {C}={5+5i}$.
Find the new point of the tail $\displaystyle C$ after the move, in algebraic form.

A reflection in the real axis, then a translation by the vector $\displaystyle -2+3i$.

A half turn about $\displaystyle A$.

A stretch by 2 from the origin, then a quarter turn about the origin clockwise.

$\displaystyle ABCD$ is a square, its vertices anticlockwise.
$\displaystyle {A}={1+2i}$ and $\displaystyle {B}={5+3i}$.
Find $\displaystyle C$ and $\displaystyle D$.

$\displaystyle C$

$\displaystyle D$

The vertices of a triangle are $\displaystyle 1+i$, $\displaystyle 5+2i$ and $\displaystyle 3+6i$.
Find the meeting point of the medians.

Prove that in a parallelogram the sum of the squares of the diagonals equals the sum of the squares of the sides: in the parallelogram with vertices $\displaystyle 0$, $\displaystyle z_1$, $\displaystyle z_1+z_2$, $\displaystyle z_2$, show that $\displaystyle {|z_1+z_2|^2+|z_1-z_2|^2}={2|z_1|^2+2|z_2|^2}$.
Hint: $\displaystyle {|w|^2}={w\cdot\bar{w}}$.

$\displaystyle z_1$, $\displaystyle z_2$ and $\displaystyle z_3$ are the vertices of an equilateral triangle, anticlockwise, and $\displaystyle {\omega}={\operatorname{cis}\frac{2\pi}{3}}$.
Explain why $\displaystyle {z_3-z_1}={(z_2-z_1)\operatorname{cis}\frac{\pi}{3}}$, and show that $\displaystyle {z_1+\omega z_2+\omega^2z_3}={0}$.