# Using the complex logarithm to find the sum of angles in a triangle.

Suppose you have a triangle with vertices $a$, $b$, and $c$. I asked earlier how you can define the angles in a triangle based on the $\log$ function. I received the answer that, for instance, the angle at $a$ is found as $\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|$.

Can this be used to show that the sum of angles in a triangle is $\pi$? I summed the angles as $$\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|+\left|\Im\log\left(\frac{a-b}{c-b}\right)\right|+\left|\Im\log\left(\frac{a-c}{b-c}\right)\right|.$$

I noticed that $\left|\Im\log\left(\frac{c-a}{b-a}\frac{a-b}{c-b}\frac{b-c}{a-c}\right)\right|=\left|\Im\log(-1)\right|=\pi$, when evaluating on the principal branch.

I had to cheat a bit and flip the $\frac{a-c}{b-c}$. Is there a more systematic way to prove this somehow?

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Flipping the $\frac{a-c}{b-c}$ (or any of the fractions in your angle expressions) actually isn't cheating. First, perhaps more intuitively, when you said that the measure of angle $a$ is $\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|$, swapping $b$ and $c$ should not change the measure of the angle, so you should expect $\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|=\left|\Im\log\left(\frac{b-a}{c-a}\right)\right|$. More formally, $\frac{b-a}{c-a}=(\frac{c-a}{b-a})^{-1}$, so \begin{align} \left|\Im\log\left(\frac{b-a}{c-a}\right)\right|&=\left|\Im\log\left(\left(\frac{c-a}{b-a}\right)^{-1}\right)\right| \\ &=\left|-\Im\log\left(\frac{c-a}{b-a}\right)\right| \\ &=\left|\Im\log\left(\frac{c-a}{b-a}\right)\right| \end{align} (since the factor of $-1$ only changes the sign of the imaginary part, and that sign change is wiped out by the absolute value).
I'd go about this in a slightly different way. Let's start by backing up to \begin{align} m\angle a&=\left|\Im\log\left(\frac{c-a}{b-a}\right)\right| \\ &=\left|\Im\left(\log(c-a)-\log(b-a)\right)\right| \\ &=\left|\Im\log(c-a)-\Im\log(b-a)\right|. \end{align} $\Im\log(c-a)$ and $\Im\log(b-a)$ are the directed angles from the positive real axis to the ray from $0$ to $c-a$ and $b-a$, respectively, so $\Im\log(c-a)-\Im\log(b-a)$ is the directed angle from $b-a$ to $c-a$. When I say "directed" angle, I mean that a positive angle is a counterclockwise rotation.
Now, without loss of generality, let the vertices be labeled $a$, $b$, and $c$ in a counterclockwise direction around the triangle:
Working carefully, we can ensure that we measure each angle in the positive direction, and thus avoid the absolute values: \begin{align} m\angle a&=\Im\log(c-a)-\Im\log(b-a)=\Im\log\frac{c-a}{b-a} \\ m\angle b&=\Im\log(a-b)-\Im\log(c-b)=\Im\log\frac{a-b}{c-b} \\ m\angle c&=\Im\log(b-c)-\Im\log(a-c)=\Im\log\frac{b-c}{a-c} \\ \\ m\angle a+m\angle b+m\angle c&=\Im\log\frac{c-a}{b-a}+\Im\log\frac{a-b}{c-b}+\Im\log\frac{b-c}{a-c} \\ &=\Im\log\left(\frac{c-a}{b-a}\cdot\frac{a-b}{c-b}\cdot\frac{b-c}{a-c}\right) \\ &=\Im\log\left((-1)^3\right) \\ &=\pi. \end{align}