JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Complex Numbers

Set A: $\mathrm{Im}z\ge 1$ represents all points on or above the line $y=1$ in the complex plane.

Set B: $|z-2-i|=3$ is a circle with center $(2,1)$ and radius 3.

Set C: For $z=x+iy$:

So $C:x+y=\sqrt{2}$, a straight line.

**Finding $A\cap B$**: Points on circle $B$ with $y\ge 1$. The circle has center $(2,1)$ and radius 3, so it extends from $y=1-3=-2$ to $y=1+3=4$. The upper semicircle (including the diameter at $y=1$) is in $A$.

**Finding $A\cap B\cap C$**: We need points on the circle $(x-2{)}^2+(y-1{)}^2=9$ with $y\ge 1$ and $x+y=\sqrt{2}$.

Substitute $x=\sqrt{2}-y$ into the circle equation:



Let $a=2-\sqrt{2}$:




This is a quadratic in $y$. The discriminant determines the number of solutions. Given the geometry (a line intersecting a circle), there can be 0, 1, or 2 intersection points. The constraint $y\ge 1$ further restricts this.

From the solution provided, there is exactly 1 point in $A\cap B\cap C$.

Answer: (B)