Prove that if |z_1|=|z_2|=|z_3| , then…

Prove that if $|z_1|=|z_2|=|z_3|$, then $\arg\dfrac{z_3-z_2}{z_3-z_1} = \dfrac12\arg\dfrac{z_2}{z_1}$.

Step 1 · Polar form

Let $r=|z_1|=|z_2|=|z_3|>0$ and write $z_1=re^{i\alpha}$, $z_2=re^{i\beta}$, $z_3=re^{i\gamma}$, where $\alpha=\arg z_1$, $\beta=\arg z_2$, $\gamma=\arg z_3$.

Step 2 · The right-hand side

$$ \frac{z_2}{z_1} = e^{i(\beta-\alpha)} \quad\Longrightarrow\quad \arg\frac{z_2}{z_1} = \beta-\alpha \pmod{2\pi} $$

so the claim to prove is $\ \arg\dfrac{z_3-z_2}{z_3-z_1} = \dfrac{\beta-\alpha}2$.

Step 3 · Factor each difference

Using $e^{i\gamma}-e^{i\beta} = e^{i\frac{\gamma+\beta}2}\left(e^{i\frac{\gamma-\beta}2}-e^{-i\frac{\gamma-\beta}2}\right) = 2i\,e^{i\frac{\gamma+\beta}2}\sin\frac{\gamma-\beta}2$:

$$ z_3-z_2 = 2ir\,e^{i\frac{\gamma+\beta}2}\sin\frac{\gamma-\beta}2, \qquad z_3-z_1 = 2ir\,e^{i\frac{\gamma+\alpha}2}\sin\frac{\gamma-\alpha}2 $$

Step 4 · Form the ratio

$$ \frac{z_3-z_2}{z_3-z_1} = e^{i\left(\frac{\gamma+\beta}2-\frac{\gamma+\alpha}2\right)}\cdot\frac{\sin\frac{\gamma-\beta}2}{\sin\frac{\gamma-\alpha}2} = e^{i\frac{\beta-\alpha}2}\cdot R, \qquad R=\frac{\sin\frac{\gamma-\beta}2}{\sin\frac{\gamma-\alpha}2} $$

The factor $R$ is a ratio of two real sines, hence $R\in\mathbb{R}$.

Step 5 · Read off the argument

Since $R$ is real, it contributes either $0$ or $\pi$ to the argument, according to its sign:

$$ \arg\frac{z_3-z_2}{z_3-z_1} = \begin{cases} \dfrac{\beta-\alpha}2 \pmod{2\pi}, & R>0 \\[4pt] \dfrac{\beta-\alpha}2+\pi \pmod{2\pi}, & R<0 \end{cases} $$

In both cases, the two possible values differ by exactly $\pi$, so:

$$ \boxed{\arg\frac{z_3-z_2}{z_3-z_1} \equiv \frac12\arg\frac{z_2}{z_1} \pmod{\pi}} $$

Step 6 · When does equality hold exactly (mod $2\pi$, not just mod $\pi$)?

This is precisely the classical distinction between the two arcs of the circle determined by $z_1,z_2$: if $z_3$ lies on the arc for which $\gamma-\beta$ and $\gamma-\alpha$ lie on the same side of $0$ with $\left|\sin\frac{\gamma-\beta}2\right|$, $\left|\sin\frac{\gamma-\alpha}2\right|$ giving $R>0$ (equivalently, $z_3$ is on the major/minor arc consistent with the usual "angle inscribed in the same arc" configuration), equality holds exactly mod $2\pi$: $\ \arg\dfrac{z_3-z_2}{z_3-z_1}=\dfrac{\beta-\alpha}2$. If $z_3$ is on the other arc, the two sides differ by exactly $\pi$ — the familiar fact that angles inscribed from opposite arcs are supplementary.

Step 7 · Geometric meaning

This is the complex-number form of the inscribed angle theorem: for three points on a circle centered at the origin, the angle subtended at $z_3$ by the chord $z_1z_2$ equals half the central angle $\beta-\alpha$ subtended by the same chord — modulo $\pi$, since the inscribed angle depends on which of the two arcs $z_3$ lies on, while the identity as an equality of directed lines (angles mod $\pi$) always holds without exception.