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?

  • $\begingroup$ my main issue is applying the cosine rule to get the third side. Whats its complex equivalent? $\endgroup$ – maik Mar 31 at 5:40
  • $\begingroup$ What about $z_B-z_C$? $\endgroup$ – GReyes Mar 31 at 5:45
  • $\begingroup$ 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. $\endgroup$ – maik Mar 31 at 5:48
  • $\begingroup$ 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 $|z_C-z_A|^2=b^2$... $\endgroup$ – GReyes Mar 31 at 5:52
  • $\begingroup$ we also know $\angle BAC = \alpha$ $\endgroup$ – 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 $$ a^2=|z_B-z_C|^2=(z_B-z_C)\overline{(z_B-z_C)}=(z_B-z_A)\overline{(z_B-z_A)}+(z_A-z_C)\overline{(z_A-z_C)}+2Re[(z_B-z_A)\overline{(z_A-z_C)}]= $$ $$ =c^2+b^2+2bcRe(e^{i(\alpha_1-\alpha_2-\pi)})=c^2+b^2-2bc\cos\alpha $$

  • $\begingroup$ well i know i can write it using only the real parts. but i want to write it in a purely complex form. $\endgroup$ – maik Mar 31 at 6:54
  • $\begingroup$ You can replace the term with the real part by $(z_B-z_A)\overline{(z_A-z_C)}+\overline{(z_B-z_A)}(z_A-z_C)$ $\endgroup$ – GReyes Mar 31 at 7:07
  • $\begingroup$ Is this purely complex for you? $\endgroup$ – GReyes Mar 31 at 7:08
  • $\begingroup$ 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. $\endgroup$ – maik Mar 31 at 11:51

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