# Evaluating $\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$

Let $A(z_1),B(z_2),C(z_3)$ are complex numbers satisfying $|z-\sqrt{3}i|=1$ and $3z_1+\sqrt{3}i=2z_2+2z_3$. The question asks to find the value of $$\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$$ I tried to shift the origin to $\sqrt{3}i$ and transformed the original condition in this system but it does not help me. Any ideas how to proceed? Thanks.

• Use underscores to get indices: z_1 makes $z_1$ rather than $z1$. – Arthur Feb 15 '17 at 10:54
• I have slightly edited your question to include the $1, 2, 3$ as subscripts. Please review and let me know if I have unwittingly changed the question. – Kevin Feb 15 '17 at 10:54
• @Bacon Thanks for your edit. – Navin Feb 15 '17 at 10:56

Put $t_j = z_j - \sqrt{3}i$. Then $|t_j|=1$ and $3t_1 = 2t_2 + 2t_3$. Hence $\frac{3}{4}t_1 = \frac{1}{2}(t_2 + t_3)$. The points $t_j$ lie on the unit circle, with center $O$ at the origin and form a triangle $ABC$, with $A = t_1$ etc. The midpoint $D$ of the chord $BC$ is $\frac{3}{4}t_1$ and hence lies on $OA$ at a distance $\frac{3}{4}$. This means that in the right angled triangle $ODB$, $OB = 1, OD = \frac{3}{4}$. Thus if the angle $BOC$ is $2\theta$, we have $\cos\theta = \frac{3}{4}$. Now $\frac{t_2}{t_3} = \cos 2\theta + i \sin 2\theta$ and $\cos 2\theta = 2\cos^2 \theta -1 = \frac{1}{8}$ and $\sin 2\theta = \pm\sqrt{1-\cos^2 2\theta} = \pm\frac{\sqrt{63}}{8}$. Thus the required ratio is $$\frac{1}{8} \pm 3\frac{\sqrt{7}}{8} i$$

• A priori, $\sin(2\theta) = \pm \sqrt{1-\cos^2(2\theta)}$, so you have to add an extra argument. – C. Dubussy Feb 15 '17 at 11:30
• @C.Dubussy Thanks, Edited the answer – user348749 Feb 15 '17 at 11:34

We can rewrite the equation as $3(z_1 - \sqrt {3} i) = 2(z_2-\sqrt {3}i)+2(z_3-\sqrt {3}i)$

i.e. $3\vec{OZ_1} = 2\vec{OZ_2}+2 \vec{OZ_3}$

So if $\theta = \pm \arg \frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$ is the angle between $\vec{OZ_2}$ and $\vec{OZ_3}$, we have from the above equation that

$9 = 2^2+2^2+8 \cos \theta \Rightarrow \cos \theta = \frac{1}{8} \Rightarrow \sin \theta = \pm \frac{3\sqrt 7}{8}$

Since $\left| \frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}\right|=1$, we have that

$\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i} = \frac{1}{8} \pm \frac{3\sqrt 7}{8}i$