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 $|A|$. 
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 ?
 A: If the given equation defines two intersecting lines then we can translate them such that they  intersect at the origin. (Replace $z$ in the equation  by $z+p+iq$, and determine $p$ and $q$ such that there are no linear terms in $z$, $\bar z$ appearing. The constant term will then automatically be zero, or we would not have two lines.)
The new equation is
$$z\bar z+\bar A z^2+A\bar z^2=0,\qquad A=a\, e^{i\alpha}\ne0\ .\tag{1}$$
The map $$T:\quad{\mathbb C}\to{\mathbb C}, \qquad z\mapsto w=e^{-i\alpha/2}\, z$$ is a rotation. It  moves the two lines into the $w$-plane without changing the angle of the intersection. The equation there is obtained from $(1)$ by letting $z:=e^{i\alpha/2}\,w$, hence is
$$w\bar w+a(w^2+\bar w^2)=0\ .$$ With $w=u+iv$ this means $(u^2+v^2)+2a(u^2-v^2)=0$, or
$$v=\pm\sqrt{{2a+1\over2a-1}}\>u\ .$$
These are two lines symmetric to the $u$-axis. It follows that
$$\tau:=\tan{\theta\over2}=\sqrt{{2a+1\over2a-1}}\ ,$$
so that
$$\cos\theta={1-\tau^2\over1+\tau^2}=-{1\over2a}\ .$$
This implies
$$|A|=a=-{1\over2\cos\theta}={1\over2\cos\theta'}\ ,$$
where $\theta'$ is the angle $<{\pi\over2}$ between the two lines.
A: 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
$$2|\cos\theta'||A|=2|\Re(\omega_1\overline{\omega_2})||A|=1.$$
A: The equation of a line when it's slope is known can be written in the form of complex numbers as
$$z (m+i) + \bar z(m -i) + 2c = 0$$
(you can derive it from $y=mx +c$ form, and for clarity I didn't assume $m-i =A$ and didn't write $2c=B$)
So, it is given to us that
$$
|z|^2+\bar{A}z^2+A(\bar{z})^2+B\bar{z}+\bar{B}z+C=0
$$
represents two straight lines, therefore that equation can be broken down in the form
$$
\left[z(m_1 +i) + \bar z (m_1 - i) + 2c_1\right] \times \left[z(m_2 +i) + \bar z (m_2 -i) + 2c_2\right]=0~~~~~~~~~~~~~~~~~(i)$$
Where $m_1$ and $c_1$ are the slope and y-intercept of first line, $m_2$ and $c_2$ are the slope and y-intercept of the second line. Multiplying that out, we get
$$
z^2(m_1 +i) (m_2 + i)  + \bar{z}^2 (m_1 - i )(m_2 -i) + |z|^2 \left[(m_1+i)(m_2-i) + (m_1 -i)(m_2 +i)\right] \cdots = 0$$
By comparing the coefficients, we have
$$
\bar{A} = (m_1+i) (m_2+i) \\
A = (m_1 - i)(m_2 -i) \\
\\
(m_1 +i) (m_2 -i) + (m_1 -i)(m_2 +i) =1 \\
\\
\tan \theta = \frac{m_1 -m_2}{1+m_1 m_2}$$
Now, we can obtain something by doing some algebra:
$$
(m_1 +i) (m_2 -i) + (m_1 -i)(m_2 +i)=1 \\
m_1m_2 = -\frac{1}{2} ~~~~~~~~~~~~~~~~~~~~~~(1)$$
$$\tan \theta = \frac{m_1 -m_2}{1+m_1 m_2}\\
\tan \theta + \tan \theta (m_1 m_2) = m_1 - m_2 \\
\left(\frac{1}{2} \tan \theta\right)^2 = m_1^2 + m_2^2 - 2m_1 m_2 \\
\frac{\tan^2 \theta}{4} -1 = m_1^2 + m_2^2~~~~~~~~~~~~~~~~~~~(2)$$
$$
|A|^2 = A \bar{A} = m_1 ^2 + m_2^2 + (m_1m_2)^2 + 1 \\
\text{from eqaution (1) and (2)} \\
|A|^2 = \frac{tan^2 \theta }{4} + -1 +1 \\
|A|=\frac{\sec \theta}{2}
$$
Hope it helps!
To some it may seem unjust to assume equation (i) to be exactly the same as the given in question, but there is nothing that prohibits that assumption. If you have any query, they are all welcome.
A: Hint:
Compare the quadratic coefficients of the equation with those of $$(y-m_1x)(y-m_2x)$$ and use $$\tan^2\theta=\left(\frac{m_1-m_2}{1+ m_1m_2}\right)^2=\frac{(m_1+m_2)^2-4m_1m_2}{(1+m_1m_2)^2}.$$
