Question

The closest distance of origin from the curve given by $b\overline{z}+\overline{b}z+b\overline{b}=0$ ($b$ is also a complex number) is

• A. $1$ unit
• B. $\frac{Re\left(b\right)}{\left|b\right|}$
• C. $\frac{Im\left(b\right)}{\left|b\right|}$
• D. $\frac{1}{2}\left|b\right|$

Answer 1

The correct option is D $\frac{1}{2}\left|b\right|$

The given equation when converted to Cartesian represents the line
$(b+\overline{b})x+(\overline{b}-b)iy+b\overline{b}=0$
By closest distance of origin from the line is perpendicular distance from (0,0)
$\therefore p=\frac{b\overline{b}}{\sqrt{{(b+\overline{b})}^2+i^2{(\overline{b}-b)}^2}}=\frac{{\left|b\right|}^2}{\sqrt{4b\overline{b}}}$
$p=\frac{1}{2}\left|b\right|\because \sqrt{b\overline{b}}=\sqrt{{\left|b\right|}^2}=\left|b\right|$