the triangle cosine relation in a complex plane

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I have a triangle $ABC$ in a complex plane. The arrangement of vertices is in a counterclockwise direction. The coordinates of $A$,$B$,$C$ are $z_A$,$z_B$,$z_C$ respectively. It is given that length of side $AB = c, AC = b$ and $\angle BAC = \alpha$. I need to find the other sides and angles of the triangle. How do i do this with complex numbers?

I know with trigonometry using the sine and cosine rules, all the length and angles can be derived. but how to get it using complex numbers?

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Here is the connection with the law of cosines. You can see which quantities correspond to which.

You have $z_B-z_A=ce^{i\alpha_1}$ and $z_C-z_A=be^{i\alpha_2}$, hence $z_A-z_C=be^{i(\alpha_2+\pi)}$. Clearly $\alpha_2-\alpha_1=\alpha$. Then $z_B-z_C=(z_B-z_A)+(z_A-z_C)$ and $$ a^2=|z_B-z_C|^2=(z_B-z_C)\overline{(z_B-z_C)}=(z_B-z_A)\overline{(z_B-z_A)}+(z_A-z_C)\overline{(z_A-z_C)}+2Re[(z_B-z_A)\overline{(z_A-z_C)}]= $$ $$ =c^2+b^2+2bcRe(e^{i(\alpha_1-\alpha_2-\pi)})=c^2+b^2-2bc\cos\alpha $$