# Sketch the Set in Complex Plane

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:

1. $U_{1}=B_{\frac{1}{2}}(i)$
2. $U_{2}=B_{1}(i)$
3. $U_{3}=B_{1}(0)\cap \left\{ z\in \mathbb{C}: \operatorname{Im} z>0 \right\}$
4. $U_{4}=\left\{ z\in \mathbb{C}:\left| \operatorname{Re}z \right| \le 1 \wedge -1 \le \operatorname{Im}z \le 0 \right\}$

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 $2 \epsilon$) .

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I'm pretty sure $B_r(z)$ denotes an open ball (sometimes called a neighborhood) centered at $z$. Does that sound right and familiar? What does an open ball look like in $\mathbb{C}$? Now, what does Im$z>0$ look like? If those are both complete mysteries then comment or edit to say you're not sure what open balls look like or exactly what Im$z$ means and I or someone else will give you more. –  Todd Wilcox Mar 22 '13 at 21:28
Sketching a set means to draw a diagram of the complex plane that shows which parts of it are in the set relative to the usual coordinate system with a real and an imaginary axis. Usually one fills the areas entirely occupied by the set with crosshatching and marks the boundary with a line if it is part of the set, whereas for an open set the crosshatching just stops without an explicit boundary line. –  Henning Makholm Mar 22 '13 at 21:29
Often I see the boundary of an open set drawn with a dashed or dotted line. Scroll down for a diagram of an open ball: jeremykun.com/2012/11/04/topological-spaces-a-primer –  Todd Wilcox Mar 22 '13 at 21:33

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.

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"+" for intersections!? –  Henning Makholm Mar 22 '13 at 22:14
@HenningMakholm it's just a notation that one must put both graphs into one, and then take intersection, since I mentioned that before the picture. –  Kaster Mar 22 '13 at 22:15
Thank you @Kaster for your help. You explained it very good. –  Devid Mar 22 '13 at 22:41

I made all in one graph, the $y$-axis is the imaginar part nad the $x$ axis is for the real part. Now think which one of them is which set.

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Beautiful - what tool are you using? –  Amzoti Mar 22 '13 at 22:11
1 is the red area because it has radius $\frac{1}{2}$ and $i$ is positive 2 is the green area because it has radius $1$ and $i$ is positive 3 it is the blue area because the center is $0$ with radius 1 and it is on the positive y axis ($\operatorname{Im} z>0$) 4 Can you explain why it is from $-1$ to $0$ in the $x$ axis ? –  Devid Mar 22 '13 at 22:26
Also how do i know if the boundary belongs to the set ? –  Devid Mar 22 '13 at 22:33
@Devid it depends on inequalities that define that set. If any of then strict – that boundary doesn't belong to the set. –  Kaster Mar 22 '13 at 22:34
Thank you know i get it :) –  Devid Mar 22 '13 at 22:42