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Prove that

$\arg \left(\frac{z_{3}-z_{2}}{z_{3}-z_{1}}\right) = \frac{1}{2} \arg\left(\frac{z_{2}}{z_{1}}\right)$

if $|z_{1}|=|z_{2}|=|z_{3}|$.

4 Answers 4

0

Note that $\arg{w_2-w_0\over w1-w_0}=\angle(w_1,w_0,w_2)$. With the help of a figure you can easily verify that the stated formula is an immediate consequence of the theorem about central and peripheral angles.

4

This is simply an application of the Inscribed Angle Theorem. Since $ \frac{z_3-z_2}{z_3-z_1}=\frac{z_2-z_3}{z_1-z_3} $ and the Inscribed Angle Theorem says that $ 2\,\arg\left(\frac{z_2-z_3}{z_1-z_3}\right)=\arg\left(\frac{z_2}{z_1}\right)\pmod{2\pi} $

$\hspace{3.5cm}$enter image description here

we get that $ 2\,\arg\left(\frac{z_3-z_2}{z_3-z_1}\right)=\arg\left(\frac{z_2}{z_1}\right)\pmod{2\pi} $ which can be rewritten as $ \arg\left(\frac{z_3-z_2}{z_3-z_1}\right)=\frac12\arg\left(\frac{z_2}{z_1}\right)\pmod{\pi} $

This is not exactly the same and points out that the relation is not always true.

For example, suppose $z_1=\frac{1-i}{\sqrt{2}}$, $z_2=\frac{1+i}{\sqrt{2}}$, and $z_3=1$. Then $ \arg\left(\frac{z_3-z_2}{z_3-z_1}\right)=\frac54\pi $ yet $ \frac12\arg\left(\frac{z_2}{z_1}\right)=\frac\pi4 $

1

Hint. Assume $|z| = 1$.

$\frac{\cos \alpha + i \sin \alpha}{\cos \beta + i \sin \beta} = (\cos \alpha + i \sin \alpha)(\cos \beta - i \sin \beta) = \cos(\alpha - \beta) + i\sin(\alpha - \beta)$

$\cos \alpha + i\sin \alpha - \cos \beta - i\sin \beta = -2 \sin(\frac{\alpha - \beta}{2})\sin(\frac{\alpha + \beta}{2}) + 2i\sin(\frac{\alpha - \beta}{2})\cos(\frac{\alpha + \beta}{2}) =\\ 2\sin(\frac{\alpha - \beta}{2})\Big(\cos(\frac{\pi}{2}+\frac{\alpha + \beta}{2}) + i\sin(\frac{\pi}{2}+\frac{\alpha + \beta}{2}) \Big)$

Deduce how you can express your equation in terms of $\arg z_1$, $\arg z_2$ and $\arg z_3$.

1

Let $z_j=R(\cos2t_j+i\sin2t_j)$ where $j=1,2,3$ and $R \neq 0$

So, $\frac{z_3-z_2}{z_3-z_1}=\frac{R(\cos2t_3+i\sin2t_3)-R(\cos2t_2+i\sin2t_2)}{R(\cos2t_3+i\sin2t_3)-R(\cos2t_1+i\sin2t_1)}$

$=\frac{(\cos2t_3-\cos2t_2)+i(\sin2t_3-\sin2t_2)}{(\cos2t_3-\cos2t_1)+i(\sin2t_3-\sin2t_1)}$

$=\frac{-2\sin(t_3-t_2)\sin(t_3+t_2)+2i\sin(t_3-t_2)\cos(t_3+t_2)}{-2\sin(t_3-t_1)\sin(t_3+t_1)+2i\sin(t_3-t_1)\cos(t_3+t_1)}$ applying $\sin C-\sin D$ and $\cos C-\cos D$

$=\frac{2i\sin(t_3-t_2)(\cos(t_3+t_2)+i\sin(t_3+t_2))}{2i\sin(t_3-t_1)(\cos(t_3+t_1)+i\sin(t_3+t_1))}$

$=\frac{\sin(t_3-t_2)e^{i(t_3+t_2)}}{\sin(t_3-t_1)e^{i(t_3+t_1)}}$ as $e^{ix}=\cos x+i\sin x$

$=\frac{\sin(t_3-t_2)}{\sin(t_3-t_1)}e^{i(t_2-t_1)}$

$=\frac{\sin(t_3-t_2)}{\sin(t_3-t_1)}(\cos(t_2-t_1)+i\sin(t_2-t_1))=X+iY(say)$

So, $X=\frac{\sin(t_3-t_2)}{\sin(t_3-t_1)}\cos(t_2-t_1)$ and $Y=\frac{\sin(t_3-t_2)}{\sin(t_3-t_1)}\sin(t_2-t_1)$

So, $\frac Y X = \tan (t_2-t_1)$ assuming $\pi ∤ (t_3 -t_1)$

$\implies \arg \left(\frac{z_{3}-z_{2}}{z_{3}-z_{1}}\right) = \arctan \left(\frac Y X\right) $ $= t_2-t_1=\frac{1}{2}(2t_2-2t_1)=\frac{1}{2}(\arg z_2 -\arg z_1)=\frac{1}{2}\arg{\frac{z_2}{z_1}}$