Perpendicular distance of a point from a line in complex plane Let a line be $\bar az+a\bar z+b=0$, where $a$ is a fixed complex number and $b$ is a fixed real number.
We need to find the distance of $p+iq$ from the given line.
I am converting the given line into Cartesian form.
Let $a=a_1+ia_2$ and $z=x+iy$
Therefore, $(a_1-ia_2)(x+iy)+(a_1+ia_2)(x-iy)+b=0$
Thus, $a_1x+ia_1y-ia_2x+a_2y+a_1x-ia_1y+ia_2x+a_2y+b=0$
So, I am getting $2a_1x+2a_2y+b=0$
Thus, the perpendicular distance of $(p,q)$ is $\frac{2a_1p+2a_2q+b}{\sqrt{(2a_1)^2+(2a_2)^2}}$
Is this correct?
Edit:

Find the perpendicular distance of line $(1-i)z+(1+i)\bar z+3=0$, from $(3+2i)$

By using the above Cartesian form, I am getting $\frac{13}{2\sqrt2}$
But the answer given in the book is $\frac{13}2$
Also, on sarthaks website, they have used a formula. I wonder what the proof for that is.
Edit$2$:
Seems like sarthaks website used the formula $\frac{\bar az_1+a\bar{z_1}+b}{\sqrt{|a|+|\bar a|}}$, where $z_1$ is the point whose distance is to be found.
This formula is incorrect, right?
Maybe in the formula, in the denominator, we can remove the square root and just keep $|a|+|\bar a|?$
 A: Your formula is almost right.
As Arthur said, the perpendicular distance from $p+qi$ to the line $\bar az+a\bar z+b=0$ with complex variable $z$ where $p,q,a_1,a_2,b$ are real and $a=a_1+a_2i$ should be $$\frac{|a_1p+a_2q+\frac b2|}{\sqrt{{a_1}^2+{a_2}^{2}}}.$$
The answer given by SudhirMandal is wrong
In particular, the perpendicular distance from $3+2i$ to the line  $(1-i)z+(1+i)\bar z+3=0$ is $\frac{13}{2\sqrt2}$, as given in the question. It is the distance between $3+2i$ to $z_0=\frac{-1}4+\frac{-5}4i$, which is a point on the given line.
$$|(3+2i)-z_0|=\left|\frac{13}4+\frac{13}4i\right|=\frac{13}{2\sqrt2}.$$
Since $\frac{13}{2\sqrt2}<\frac{13}2$, the wanted perpendicular distance cannot be $\frac{13}2$ since the perpendicular distance is the smallest distance between the given point $3+2i$ and a point on the given line. Hence the answer given on sarthaks.com referred to in the question is wrong.
I do not think there is a reasonable definition of "the perpendicular distance" with which that answer, $\frac{13}2$ can be correct. Apparently, the author of that answer, mistook the distance formula of real variables for the current case of complex variables (with conjugates). 
By reverse-engineering, the author was using the following formula
$$\left|\frac{cz+a\bar z+b}{\sqrt{|c|^2 + |a|^2}}\right|$$
which is not correct even in the case when $c=\bar a$ and $b$ is real. For example, let the given line be $z+\bar z=0$, which represents all imaginary numbers and $0$. So $c=a=1, b=0$. The (perpendicular) distance from $3$ to the line is $3$, attained as the distance from $3$ to $0$. However, that formula yields $\frac{|1\cdot3+1\cdot3+0|}{\sqrt{1^2+1^2}}=\frac3{\sqrt2}$. That formula will yield wrong result still if the denominator $\sqrt{|c|^2 + |a|^2}$ is replaced with $\sqrt{|a|+|c|}$ or $|a|+|c|$.
A: Except that you need absolute value signs on that numerator (the value of your fraction can be negative, but distances can never be negative, and that's the fix), this looks good to me.
