Proof. Following the hint, we let $C=0, B=z_{1}$ and $A=z_{2} .$ Then $a=\left|z_{1}\right|, b=\left|z_{2}\right|$

$$

\text { and } \begin{aligned}

c=\left|z_{2}-z_{1}\right| . \text { So } \\

\qquad \begin{aligned}

a^{2}+b^{2}-c^{2} &=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-\left|z_{2}-z_{1}\right|^{2} \\

&=\left(z_{1} \bar{z}_{1}+z_{2} \bar{z}_{2}\right)-\left(z_{2}-z_{1}\right) \overline{\left(z_{2}-z_{1}\right)} \\

&=\left(z_{1} \bar{z}_{1}+z_{2} \bar{z}_{2}\right)-\left(z_{2}-z_{1}\right)\left(\bar{z}_{2}-\bar{z}_{1}\right) \\

&=\left(z_{1} \bar{z}_{1}+z_{2} \bar{z}_{2}\right)-\left(z_{1} \bar{z}_{1}+z_{2} \bar{z}_{2}-z_{1} \bar{z}_{2}-z_{2} \bar{z}_{1}\right) \\

&=z_{1} \bar{z}_{2}+z_{2} \bar{z}_{1}

\end{aligned}

\end{aligned}

$$

Let $z_{1}=r_{1} e^{i \theta_{1}}$ and $z_{2}=r_{2} e^{i \theta_{2}} .$ Then

$$

\begin{aligned}

z_{1} \bar{z}_{2}+z_{2} \bar{z}_{1} &=r_{1} e^{i \theta_{1}} \overline{r_{2} e^{i \theta_{2}}}+r_{2} e^{i \theta_{2}} \overline{r_{1} e^{i \theta_{1}}} \\

&=\left(r_{1} e^{i \theta_{1}}\right)\left(r_{2} e^{-i \theta_{2}}\right)+\left(r_{2} e^{i \theta_{2}}\right)\left(r_{1} e^{-i \theta_{1}}\right) \\

&=r_{1} r_{2} e^{i\left(\theta_{1}-\theta_{2}\right)}+r_{1} r_{2} e^{i\left(\theta_{2}-\theta_{1}\right)} \\

&=2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=2\left|z_{1}\right|\left|z_{2}\right| \cos \theta=2 a b \cos \theta

\end{aligned}

$$

Therefore, we have

$$

a^{2}+b^{2}-c^{2}=z_{1} \bar{z}_{2}+z_{2} \bar{z}_{1}=2 a b \cos \theta

$$

and hence

$$

a^{2}+b^{2}-2 a b \cos \theta=c^{2}

$$