Question

Show that the perpendicular distance of a point d(d is a complex number) on the Argand plane from the line $\overline{a}z+a\overline{z}+b=0$ (a is a complex number and b is real) is $\frac{|a\overline{d}+\overline{a}d+b|}{2|a|}$.

Answer 1

Equation of line is $a\overline{z}+\overline{a}z+b=0$
Put $z=x+iy$
then equation $a(x-iy)+\overline{a}(x+iy)+b=0$
$\therefore (a+\overline{a})x+(\overline{a}-a)iy+b=0$
Let $d=d_1+i{d}_2$
Distance of the line from $d=\frac{|(a+\overline{a})d_1+(\overline{a}-a)i{d}_2+b|}{{\sqrt{a+\overline{a}}}^2+\left[\right(\overline{a}-a)i{]}^2}$
$=\frac{|a(d_1-i{d}_2)+\overline{a}(d_1+i{d}_2)+b|}{\sqrt{(a+\overline{a}{)}^2-(\overline{a}-a{)}^2}}$
$=\frac{|a\overline{d}+\overline{a}d+b|}{\sqrt{4\overline{a}a}}$
$=\frac{|a\overline{d}+\overline{a}d+b|}{\sqrt{4|a{|}^2}}$
$=\frac{|a\overline{d}+\overline{a}d+b|}{2|a|}$
Ans: 1