# Prove that $\arg \frac{(z_3-z_2)}{(z_3-z_1)} = \frac12 \arg\frac{ z_2}{z_1}$

If $$|z_1|=|z_2|=|z_3|$$

I have used: $$z_1=x_1+\mathrm iy_1,z_2=x_2+\mathrm iy_2,z_3=x_3+\mathrm iy_3$$ and obtained $$\arg\frac{z_3-z_2}{z_3-z_1} = \arctan \frac{(y_3-y_2)(x_3-x_1)-(y_3-y_1)(x_3-x_2)}{(x_3-x_2)(x_3-x_1)}$$ But, then I'm stuck.

Urgent help needed.

• Why don't you use polar coordinates instead for $z_1,z_2,z_3$? – Shahab Sep 2 '15 at 0:13
• @Shahab, how would that work? using r1,theta1...etc? – Marissa Sep 2 '15 at 0:17
• Let $z_i=re^{i\theta_i}$ for $i=1,2,3$. Then $\frac{z_3-z_2}{z_3-z_1}=\frac{e^{i\theta_3}-e^{i\theta_2}}{e^{i\theta_3}-e^{i\theta_1}}=\frac{\cos\theta_3-\cos\theta_2+i(\sin\theta_3-\sin\theta_2)}{\cos\theta_3-\cos\theta_1+i(\sin\theta_3-\sin\theta_1)}$ – Shahab Sep 2 '15 at 0:20
• I think the complex-analysis tag is misleading – RowanS Sep 2 '15 at 0:22
• @RowanS it is a problem from the book introductory complex analysis by silverman. So, I believe the tag is appropiate. – Marissa Sep 2 '15 at 0:34

$arg({a \over b})=arg(a)-arg(b)$

$arg(z_3-z_2)-arg(z_3-z_1)=\frac{1}{2}(arg(z_2)-arg(z_1))$

Now let $z_k=re^{i\theta_k}=r(\cos{\theta_k+i\sin{\theta_k}})$

$arg(r(\cos{\theta_3}-\cos{\theta_2}+i(\sin{\theta_3}-\sin{\theta_2})))-arg(r(\cos{\theta_3}-\cos{\theta_1}+i(\sin{\theta_3}-\sin{\theta_1})))=\frac{1}{2}(arg(r(\cos{\theta_2+i\sin{\theta_2}}))-arg(r(\cos{\theta_1+i\sin{\theta_1}})))$

$arctan({\sin{\theta_3}-\sin{\theta_2} \over \cos{\theta_3}-\cos{\theta_2}})-arctan({\sin{\theta_3}-\sin{\theta_1} \over \cos{\theta_3}-\cos{\theta_1}})={1 \over 2}(arctan({\sin{\theta_2} \over \cos{\theta_2}})-arctan({{\sin{\theta_1} \over \cos{\theta_1}}}))$

It can be proven with trig identities that ${{\sin{x}-\sin{y}} \over {\cos{x}-\cos{y}} }={\cot({\frac{x+y}{-2}})}$

$arctan({\cot({\frac{\theta_3+\theta_2}{-2}})})-arctan({\cot({\frac{\theta_3+\theta_1}{-2}})})={1 \over 2}(arctan(\tan{\theta_2})-arctan(\tan{\theta_1})$

It can be shown that $arctan(\cot{x})={\pi \over 2}-x$

$({\pi \over 2} - \frac{\theta_3+\theta_2}{-2})-({\pi \over 2} - \frac{\theta_3+\theta_1}{-2})={1 \over 2}({\theta_2}-{\theta_1})$

$\frac{\theta_3+\theta_2}{2}-\frac{\theta_3+\theta_1}{2}={1 \over 2}({\theta_2}-{\theta_1})$

$\frac{\theta_3+\theta_2-\theta_3-\theta_1}{2}={1 \over 2}({\theta_2}-{\theta_1})$

$\frac{\theta_2-\theta_1}{2}={1 \over 2}({\theta_2}-{\theta_1})$

The statement appears to be proven, but be aware that this makes many assumptions, largest of which is that the angles all exist in the first quadrant. If the angles were all in different quadrants, the proof would be much more involved, and to make such a proof would need a lot of meticulous calculations and studying many different cases - mostly because of the arctan function and its limited range.

Good luck.

• thank you! I was trying it with polar coordinates just now and got stuck with the cotx, I was doing it backwards. Thank you. – Marissa Sep 2 '15 at 1:36

I don't think the statement is true. The equation doesn't work for $z_1=z_2=z_3=1$

• it never said the three zs were equal to 1 just equal to each other. – Marissa Sep 2 '15 at 0:35
• I think his point is that if they do not work for z1=z2=z3=1, the statement cannot be true because it does not work for all possibilities of z1, z2, and z3. – Ud779 Sep 2 '15 at 1:27