I don't understand how to sketch a set in the Complex Plane. So i would appreciate if someone could explain it to me. What do i have to know ? Which formulas do i need ? Can someone explain how a open ball looks like in $\mathbb{C}$ ?
Sketch the following set in the complex plane:
- $U_{1}=B_{\frac{1}{2}}(i)$
- $U_{2}=B_{1}(i)$
- $U_{3}=B_{1}(0)\cap \left\{ z\in \mathbb{C}: \operatorname{Im} z>0 \right\}$
- $U_{4}=\left\{ z\in \mathbb{C}:\left| \operatorname{Re}z \right| \le 1 \wedge -1 \le \operatorname{Im}z \le 0 \right\} $
Definition
Given $z\in \mathbb{K}$ and $\epsilon\in \mathbb{R}$ we call the set $B_{\epsilon}(z):=\left\{ w\in \mathbb{K}:\left| w-z \right| \le \epsilon \right\}$ the $\epsilon$-neighborhood of $z$ or the open $\epsilon$-ball with center $z$ (in fact, for $\mathbb{K}=\mathbb{C}$, $B_{\epsilon}(z)$ represents an open disk in the complex plane with center $z$ and radius $\epsilon$, whereas, for $\mathbb{K}=\mathbb{R}, B_{\epsilon}(z)=]z- \epsilon,z+ \epsilon[$ is the
open interval with center $z$ and length $2 \epsilon$) .
Best Answer
1,2. $U_1 = B_{\frac 12} (i)$, means, as it was already said, open neighborhood, or a circle around the point $z = i$ with radius of $r = \frac 12$. Since this set is open, which means boundary points are not taken into account. Algebraicly it's inequality. Same for the 2 question (circle with radius $r = 1$). $$ \|z-i\| < \frac 12 \\ \|z-i\| < 1 $$ It looks as follows. Smaller dashed circle is for 1, bigger is for 2.
3 This one is intersection of $B_1(0)$ which is circle and the region for all those points $z$ which have positive imaginary part. Obviously, it's upper half plane without border, since inequality is strict.
4 This one is an intersection of two stripes - horizontal and vertical. Verical stripe is for all those $z$ real part of which is between $-1$ and $1$. And horizontal stripe for all $z$ imaginary part of which is between $-1$ and $0$. This one include boundaries since inequalities are not strict.
PS
If you use closed neighborhood as definition of $B_\epsilon(z)$, juts replace all dashes for circles with solid lines.
Update
I've just realized that it might be unclear which rectangle in 4 represents which stripe. So I changed them to fix that. So, solid black line edge means that it's at finite location and actually there.