If $\arg \frac{z-z_1}{z-z_2}=\pi /6$ then find the value of $|z-7-9i|$ Let $z_1=6+3i$ and $z_2=2+3i$. If $z$ be a complex number such that $\arg \frac{z-z_1}{z-z_2}=\pi /6$ then find the value of $|z-7-9i|$
I was trying to do it using $z=x+iy$ but that resulted in hectic calculation. By $\arg \frac{z-z_1}{z-z_2}=\pi /6$ can we say that $z$ lies on a circular arc? And can we use this to obtain required value?
 A: HINT:


*

*When $\text{z}\in\mathbb{C}$ and $\arg(\text{z})=\frac{\pi}{6}$:
$$\text{z}=\Re[\text{z}]+\Im[\text{z}]i$$
Where $\Re[\text{z}]>0$ and $\Im[\text{z}]>0$.

*When $\text{z}\in\mathbb{C}$ and $\arg(\text{z})=\frac{\pi}{6}$, we can set:
$$\arg(\text{z})=\arg(\Re[\text{z}]+\Im[\text{z}]i)=\arctan\left(\frac{\Im[\text{z}]}{\Re[\text{z}]}\right)$$
Where $\Re[\text{z}]>0$ and $\Im[\text{z}]>0$.

*$$\arctan\left(\frac{\Im[\text{z}]}{\Re[\text{z}]}\right)=\frac{\pi}{6}\Longleftrightarrow\frac{\Im[\text{z}]}{\Re[\text{z}]}=\frac{1}{\sqrt{3}}$$



So, in your problem (when $\text{z}\in\mathbb{C}$):


*

*$$\Re\left[\frac{\text{z}-(6+3i)}{\text{z}-(2+3i)}\right]=1+\frac{8-4\Re[\text{z}]}{13+\Re[\text{z}](\Re[\text{z}]-4)+\Im[\text{z}](\Im[\text{z}]-6)}$$

*$$\Im\left[\frac{\text{z}-(6+3i)}{\text{z}-(2+3i)}\right]=\frac{4(\Im[\text{z}]-3)}{13+\Re[\text{z}](\Re[\text{z}]-4)+\Im[\text{z}](\Im[\text{z}]-6)}$$

*$$\frac{\Im\left[\frac{\text{z}-(6+3i)}{\text{z}-(2+3i)}\right]}{\Re\left[\frac{\text{z}-(6+3i)}{\text{z}-(2+3i)}\right]}=\frac{4(\Im[\text{z}]-3)}{21+\Re[\text{z}](\Re[\text{z}]-8)+\Im[\text{z}](\Im[\text{z}]-6)}$$


So, you've to solve:
$$\frac{4(\Im[\text{z}]-3)}{21+\Re[\text{z}](\Re[\text{z}]-8)+\Im[\text{z}](\Im[\text{z}]-6)}=\frac{1}{\sqrt{3}}$$
We want to find:
$$\left|\text{z}-7-9i\right|=\sqrt{\left(\Re[\text{z}]-7\right)^2+\left(\Im[\text{z}]-9\right)^2}$$
