JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Complex Numbers

From the previous question, $S$ is the sector of the circle $|z|<4$ in the region where $x>0$ and $\sqrt{3}x+y>0$.

We need to find ${\min }_{z\in S}|z-(1-3i)|$, which is the minimum distance from the point $(1,-3)$ to the region $S$.

The boundary of $S$ consists of:

1. Arc of circle $|z|=4$
2. Positive $x$-axis ($y=0,x>0$)
3. Line $\sqrt{3}x+y=0$ (or $y=-\sqrt{3}x$)

The point $(1,-3)$ is outside $S$ (since $\sqrt{3}(1)+(-3)=\sqrt{3}-3<0$).

The minimum distance is the perpendicular distance from $(1,-3)$ to the line $\sqrt{3}x+y=0$:

$d=\frac{|\sqrt{3}(1)+(-3)|}{\sqrt{3+1}}=\frac{|\sqrt{3}-3|}{2}=\frac{3-\sqrt{3}}{2}$