# the triangle cosine relation in a complex plane

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?

• my main issue is applying the cosine rule to get the third side. Whats its complex equivalent? – maik Mar 31 at 5:40
• What about $z_B-z_C$? – GReyes Mar 31 at 5:45
• not given, but two sides and the angle in between is known, so the remaining angles and sides can be uniquely determined, at least using trigonometry i know how to find this. – maik Mar 31 at 5:48
• You need $|z_B-z_C|^2=(z_B-z_C)\overline{(z_B-z_C)}$ and $z_B-z_C=(z_B-z_A)+(z_A-z_C)$. You know that $|z_B-z_A|^2=c^2$ and $|z_C-z_A|^2=b^2$... – GReyes Mar 31 at 5:52
• we also know $\angle BAC = \alpha$ – maik Mar 31 at 5:53

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 $$a^2=|z_B-z_C|^2=(z_B-z_C)\overline{(z_B-z_C)}=(z_B-z_A)\overline{(z_B-z_A)}+(z_A-z_C)\overline{(z_A-z_C)}+2Re[(z_B-z_A)\overline{(z_A-z_C)}]=$$ $$=c^2+b^2+2bcRe(e^{i(\alpha_1-\alpha_2-\pi)})=c^2+b^2-2bc\cos\alpha$$

• well i know i can write it using only the real parts. but i want to write it in a purely complex form. – maik Mar 31 at 6:54
• You can replace the term with the real part by $(z_B-z_A)\overline{(z_A-z_C)}+\overline{(z_B-z_A)}(z_A-z_C)$ – GReyes Mar 31 at 7:07
• Is this purely complex for you? – GReyes Mar 31 at 7:08
• with modulus , i have to take the square root. I want to avoid sauqre roos and convert to the form with has $e^{i\alpha}$ in it. Then taking the square root will only be dividing the angle in exponential by 2. And i guess it will also look much nicer. – maik Mar 31 at 11:51