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

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4 Answers

up vote 1 down vote accepted

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 $\pi ∤ (t_3 -t_1)$

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

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Thankyou @lab bhattacharjee –  Mykolas Sep 22 '12 at 11:29
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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 $$ \frac12\arg\left(\frac{z_2}{z_1}\right)=\frac\pi4 $$

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thankyou @robjon –  Mykolas Sep 22 '12 at 18:56
    
Rob, what did you use to make the image? –  Igäria Mnagarka Sep 23 '12 at 16:38
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@GustavoBandeira: I used Intaglio, which is available only on Mac OS. –  robjohn Sep 24 '12 at 22:03
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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 $\arg z_3$.

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thankyou @Karolis Juodelė –  Mykolas Sep 22 '12 at 11:29
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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.

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