If $\mathrm{z}=\mathrm{x}+$ iy represents a point in the Argand plane, then a point which is not in the…
If $\mathrm{z}=\mathrm{x}+$ iy represents a point in the Argand plane, then a point which is not in the region represented by
$|z-1+i| \leq 2$ is
$\frac{1-i}{2}$
$1$
$\frac{1-i}{4}$
i
Solution
$|z-1+i| \leq 2$
$\Rightarrow|\mathrm{z}-(1-\mathrm{i})| \leq 2$
(a) Let $z_1=\frac{1-1}{2}$ then $\left|z_1-(1-i)\right|$
$=\left|\frac{(1-i)}{2}\right|=\sqrt{\frac{1}{4}+\frac{1}{4}}$
$=\left|\frac{(1-i)}{2}\right|=\sqrt{\frac{1}{4}+\frac{1}{4}}$
$=\sqrt{\frac{1}{2}} \leq 2$
$\therefore \mathrm{z}_1$ is in the region.
(b) Let $z_2=1$, then $\left|z_2-(1-i)\right|=|i|=1 \leq 2$.
$\therefore \mathrm{z}_2$ is in the region.
(c) Let $\mathrm{z}_3=\frac{1-\mathrm{i}}{4}$, then $\left|\mathrm{z}_3-(1-\mathrm{i})\right|$
$=\left|-\frac{3}{4}(1-i)\right|=\frac{3}{4} \sqrt{2} \leq 2$
$\begin{aligned} & \text { (d) Let } z_4=i \text { then }\left|z_4-(1-i)\right|=|i-1+i|=|2 i-1| \\ & =\sqrt{4+1}\end{aligned}$
$=\sqrt{5}>2$
$\therefore \mathrm{i}$ is not in the region.