---
tag: ZJA9Z7263W
kind: proof
name: Proof of the inscribed angle theorem by complex numbers
msc: [51M04]
proves: 0GA1V8NN8C
uses: [060W5CBZP6, 03YN70NK0Z, 09FTAET5Z4, 0FTH2E9TAS, 0KW2X0TFBX, 0BRXFNQ5JV, 04R3CY41VJ, 0WM7DJCM6G, 0S52TNNTH9, 00KDZRW4C0]
sources:
  - url: https://static1.squarespace.com/static/5fe101b108d85d5e817a934a/t/60acfa50392cb622fbbb2151/1621949009648/Complex_Numbers_in_Geometry_Polhem.pdf
    locator: Theorem 3.2
---

## Proof

We identify the plane with the [[complex numbers|09FTAET5Z4]], the point $(x, y)$ with $x + iy$.

1. For $u = u_1 + i u_2$ and $w = w_1 + i w_2$, with the [[complex conjugate|0FTH2E9TAS]] $\bar u$, we have $\bar u\, w = (u_1 w_1 + u_2 w_2) + i(u_1 w_2 - u_2 w_1)$. This is $u \cdot w + i \det(u, w)$, with the [[dot product|0S52TNNTH9]] and the [[determinant|00KDZRW4C0]].
2. Let $u, w \neq 0$, and suppose $\bar u\, w = \rho\, e^{i\theta}$ with a real $\rho > 0$. Taking the [[modulus|0KW2X0TFBX]] gives $\rho = \abs{u}\abs{w}$, since $\abs{e^{i\theta}} = 1$.
3. Dividing step 1 by $\rho$ and using [[Euler's formula|04R3CY41VJ]] gives $\cos\theta = \frac{u \cdot w}{\abs{u}\abs{w}}$ and $\sin\theta = \frac{\det(u, w)}{\abs{u}\abs{w}}$. So $\theta$ represents the [[oriented angle|03YN70NK0Z]] from $u$ to $w$.
4. Let $r > 0$ be the radius of the [[circle|060W5CBZP6]]. Put $a = A - O$, $b = B - O$ and $v = V - O$, each of modulus $r$.
5. By the [[polar form|0BRXFNQ5JV]], $a = r e^{i\alpha}$, $b = r e^{i\beta}$ and $v = r e^{i\nu}$ for some real $\alpha$, $\beta$, $\nu$.
6. By the [[product of exponentials|0WM7DJCM6G]], $\bar a\, b = r^2 e^{i(\beta - \alpha)}$. By step 3, $\measuredangle AOB = \beta - \alpha$ modulo $2\pi$.
7. For real $x$ and $y$, put $m = \frac{x + y}{2}$ and $h = \frac{x - y}{2}$. Then $e^{ix} - e^{iy} = e^{im}(e^{ih} - e^{-ih}) = 2i \sin(h)\, e^{im}$ by Euler's formula.
8. By step 7, $a - v = 2ir\, s_a\, e^{i(\alpha + \nu)/2}$ and $b - v = 2ir\, s_b\, e^{i(\beta + \nu)/2}$. Here $s_a = \sin\frac{\alpha - \nu}{2}$ and $s_b = \sin\frac{\beta - \nu}{2}$.
9. Hence $\overline{(a - v)}\,(b - v) = (-2ir)(2ir)\, s_a s_b\, e^{i(\beta - \alpha)/2} = s\, e^{i(\beta - \alpha)/2}$, with the real number $s = 4r^2 s_a s_b$.
10. The vectors $a - v = A - V$ and $b - v = B - V$ are nonzero, since $V \neq A$ and $V \neq B$. So $s \neq 0$.
11. If $s > 0$, step 3 with $\rho = s$ gives $\measuredangle AVB = \frac{\beta - \alpha}{2}$ modulo $2\pi$.
12. If $s < 0$, then $s = \abs{s} e^{i\pi}$, so step 9 reads $\abs{s} e^{i((\beta - \alpha)/2 + \pi)}$. Step 3 gives $\measuredangle AVB = \frac{\beta - \alpha}{2} + \pi$ modulo $2\pi$.
13. Doubling a class modulo $2\pi$ gives a well defined class modulo $2\pi$. In both cases, $2\,\measuredangle AVB = \beta - \alpha$ modulo $2\pi$, because $2\pi \equiv 0$.
14. By step 6, $\measuredangle AOB = 2\,\measuredangle AVB$.

## Idea

On a circle, the chord from $v$ to $a$ points in the direction halfway between the angles of $a$ and $v$, up to a sign. So the turn from one chord to the other is half the turn from $a$ to $b$, up to a possible extra $\pi$. Doubling removes that extra $\pi$.