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Use complex numbers to prove the Law of Cosine: Let $\triangle A B C$ be a triangle with
|B C|=& a,|C A|=b,|A B|=c \text { and } \angle B C A=\theta . \text { Then } \\
& a^{2}+b^{2}-2 a b \cos \theta=c^{2}
Hint: Place $C$ at the origin, $B$ at $z_{1}$ and $A$ at $z_{2}$. Prove that
z_{1} \bar{z}_{2}+z_{2} \bar{z}_{1}=2\left|z_{1} z_{2}\right| \cos \theta
in Mathematics by Gold Status (10,261 points) | 12 views

1 Answer

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

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}
by Gold Status (10,261 points)

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