Geometrically, the set \(\{z \in \mathbf{C}:|z-2-2 i| \leq 1\}\) represents
Geometrically, the set \(\{z \in \mathbf{C}:|z-2-2 i| \leq 1\}\) represents
a closed circular disc with center at \((-2,-2)\) and with radius 1
a closed circular disc with center at \((2,2)\) and with radius 1
a closed circular disc with center at \((1,1)\) and with radius 0.5
a closed circular disc with center at \((-1,-1)\) and with radius 0.5
Solution
Given inequality is \(|z-2-2 i| \leq 1\)
Let \(z=x+i y\), then we get
\(\begin{aligned}
& & \sqrt{(x-2)^2+(y-2)^2} & \leq 1 \\
\Rightarrow & & (x-2)^2+(y-2)^2 & \leq 1
\end{aligned}\)
The above inequality represents a closed circular disc with center at \((2,2)\) and with radius 1.
Hence, option (b) is correct.