TITLE: If $|z|^2+\bar{A}z^2+A(\bar{z})^2+B\bar{z}+\bar{B}z+c=0$ represents a pair of intersecting lines.... find the value of $|A|$.
QUESTION [5 upvotes]: If $|z|^2+\bar{A}z^2+A(\bar{z})^2+B\bar{z}+\bar{B}z+c=0$ represents a pair of intersecting lines with angle of intersection $'\theta'$ then find the value of |A|.

I tried using general equation of straight line in complex form as $a\bar{z}+\bar{a}z+b=0$ where b is real,where slope of line is $\frac{-a}{\bar{a}}$ and multiplying two such lines and then compairing coefficient.I found that c is real and used $$\tan(\theta)=\big|\left(\frac{\text{slope}_1-\text{slope}_2}{1+\text{slope}_1*\text{slope}_2}\right)\big|$$
one thought was also that the use of $|z_1+z_2|^2=(z_1+z_2)(\bar{z_1}+\bar{z_2})=|z_1|^2+|z_2|^2+\bf{z_1\bar{z_2}+z_2\bar{z_1}}$
But this is becoming very long and I think this is also not much effective and there should be some good way, is there a better way out ?

REPLY [2 votes]: Let $L$ be a straight line on the Argand plane. Since we can rotate $L$ to obtain a vertical straight line and the equation of a vertical straight line on the closed right half plane is given by $\Re(z)=x$ for some constant $x\ge0$, the equation of a general straight line is given by $\Re(\omega z)=x$ for some complex number $\omega$ on the unit circle. It follows that if the angle between two lines $\Re(\omega_1 z)=x_1$ and $\Re(\omega_2 z)=x_2$ is $\theta'$, then $|\cos\theta'|=|\Re(\omega_1\overline{\omega_2})|$.
Now suppose $z$ lies on the union of the two lines $\Re(\omega_1 z)=x_1$ or $\Re(\omega_2 z)=x_2$. Then
$$(\omega_1 z+\overline{\omega_1 z}-2x_1)
(\omega_2 z+\overline{\omega_2 z}-2x_2)=0.$$
Expand the LHS, we obtain
$$2\Re(\omega_1\overline{\omega_2})|z|^2
+\omega_1\omega_2 z^2
+\overline{\omega_1\omega_2}\,\overline{z}^2
+pz+\overline{pz}
+r=0,$$
where $p$ and $r$ are some constants. If this equation is equivalent to
$$|z|^2
+\overline{A}z^2
+A\overline{z}^2
+\overline{B}z
+B\overline{z}
+c=0,$$
by comparing the coefficients of $|z|^2$ and $\overline{z}^2$ in both equations, we have $2\Re(\omega_1\overline{\omega_2})A=\overline{\omega_1\omega_2}$. Hence
$$2|\cos\theta'||A|=2|\Re(\omega_1\overline{\omega_2})||A|=1.$$