Sketch $|z|\geq \text{Re}(z)+1$.

Question

I want to sketch the set $|z|\geq \text{Re}(z)+1$ and determine whether it is open, closed or neither and whether it is bounded and connected.

Let $z=x+iy,$ Then we have that the inequality is equivalent to

$$\sqrt{x^2+y^2}\geq x+1,$$

so

$$y\geq \sqrt{2x+1}.$$

So this is the closed, unbounded region above the graph of $\sqrt{2x+1}.$ I don't understand how to determine connectedness.

Is my answer so far correct?

Answer 1

Be very careful when squaring: the inequality $1\ge -2$ is true, but $1^2\ge(-2)^2$ is false.

The inequality $A\ge B$ is equivalent to $A^2\ge B^2$ provided $A\ge0$ and $B\ge0$.

If $\operatorname{Re}z<-1$, then the inequality $|z|\ge\operatorname{Re}z+1$ is surely true. Therefore all complex numbers $x+yi$ with $x<-1$ belong to the set.

If $x\ge-1$, then you can indeed square and the inequality becomes $$ x^2+y^2\ge x^2+2x+1 $$ that is, $$ y^2\ge 2x+1 $$ The locus $y^2=2x+1$ is a parabola, which divides the plane into three subsets:

1. the parabola
2. the inner points (those from which you can't draw tangents to the parabola)
3. the outer points

For all inner points it either holds $y^2>2x+1$ or $y^2<2x+1$. Consider the segment joining two inner points; if in one the former inequality holds and in the other the latter, then the segment joining them has to meet the parabola. Since $(0,0)$ is an inner point and $0<2\cdot0+1$, we conclude that the inner points don't belong to the given set.

Similarly, the outer points all satisfy $y^2>2x+1$.

On the other hand, all points with $x<-1$ are outer points, as the vertex has coordinates $(-1/2,0)$, so the set is indeed the set of outer points to the parabola $y^2=2x+1$ and the points of the parabola.

Answer 2

Well, if $x\geq -1$ then we get $y^2\geq 2x+1$. Since $y^2 = 2x+1$ is a parabola it determine two connected component and $y^2\geq 2x+1$ is one of them and it is closed. Name this component $A$

But if $x<-1$ you can't square it. But in this case any point $P(x,y)$ if $x<-1$ is a solution and they determine open halfplane $B$ (left from from the line $x=-1$). But this plane is $B$ is subset of $A$, so their union is again $A$.

So this region is closed and it has one component..

Answer 3

shaded region. before squaring we need to make comdtions for right of inequality to be positive or negative