Question

$z_1,z_2,z_3$ are three complex numbers whose moduli are$a,b,c$ respectively and are such that $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$.
If $z_1\ne z_2$, then prove that $\arg {\left(\frac{z_3-z_1}{z_2-z_1}\right)}^2=\arg \left(\frac{z_3}{z_2}\right)$

Answer 1

The given determinant is a circulant which when expanded gives $3abc-a^3-b^3-c^3=0$
or $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$
or $\frac{1}{2}(a+b+c)[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]=0$
Since $a+b+c$ being the sum of moduli cannot be zero and hence we must have
$\sum {(a-b)}^2=0\Rightarrow a-b=0,b-c=0,c-a=0$
or $a=b=c$ or $\left|z_1\right|=\left|z_2\right|=\left|z_3\right|=r$ say
$\therefore$ Let $z_1={re}^{i{\theta }_1},z_2={re}^{i{\theta }_2},z_3={re}^{i{\theta }_3}$,
Now $\frac{z_3-z_1}{z_2-z_1}$
$=\frac{\left(\cos {\theta }_3+i\sin {\theta }_3\right)-\left(\cos {\theta }_1+i\sin {\theta }_1\right)}{\left(\cos {\theta }_2+i\sin {\theta }_2\right)-\left(\cos {\theta }_1+i\sin {\theta }_1\right)}$
$=\frac{\left(\cos {\theta }_3-\cos {\theta }_1\right)+i\left(\sin {\theta }_3-\sin {\theta }_1\right)}{\left(\cos {\theta }_2-\cos {\theta }_1\right)+i\left(\left(\sin {\theta }_2-\sin {\theta }_1\right)\right)}$
$=\frac{2\sin \frac{{\theta }_3+{\theta }_1}{2}\sin \frac{{\theta }_1-{\theta }_3}{2}+i2\sin \frac{{\theta }_3-{\theta }_1}{2}\cos \frac{{\theta }_3+{\theta }_1}{2}}{2\sin \frac{{\theta }_2+{\theta }_1}{2}\sin \frac{{\theta }_1-{\theta }_2}{2}+2i\sin \frac{{\theta }_2-{\theta }_1}{2}\cos \frac{{\theta }_2+{\theta }_1}{2}}$
Now, $\sin \frac{{\theta }_1-{\theta }_3}{2}=-\sin \frac{{\theta }_3-{\theta }_1}{2}=i^2\sin \frac{{\theta }_3-{\theta }_1}{2}$
$=\frac{2\sin \frac{{\theta }_3-{\theta }_1}{2}\left[\cos \frac{{\theta }_3+{\theta }_1}{2}+i\sin \frac{{\theta }_3+{\theta }_1}{2}\right]}{2\sin \frac{{\theta }_2-{\theta }_1}{2}\left[\cos \frac{{\theta }_3+{\theta }_1}{2}+i\sin \frac{{\theta }_3+{\theta }_1}{2}\right]}$
$=\frac{\sin \frac{{\theta }_3-{\theta }_1}{2}}{\sin \frac{{\theta }_2-{\theta }_1}{2}}{e}^{[i\frac{{\theta }_3+{\theta }_1}{2}-\frac{{\theta }_2+{\theta }_1}{2}]}$
or $=\frac{z_3-z_1}{z_2-z_1}=k{e}^{i\frac{{\theta }_3-{\theta }_2}{2}}$
$\therefore {\left(\frac{z_3-z_1}{z_2-z_1}\right)}^2{k}^2{\left[e^{i\left(\frac{{\theta }_3-{\theta }_2}{2}\right)}\right]}^2=k^2{e}^{i({\theta }_3-{\theta }_2)}$
$\because {\left(e^{ix}\right)}^2=e^{i2x}$
$\therefore \arg {\left(\frac{z_3-z_1}{z_2-z_1}\right)}^2={\theta }_3-{\theta }_2$
$=\arg z_3-\arg z_2=\arg \left(\frac{z_3}{z_2}\right)$