Describe the set of points z satisfying |z|=Re(z)+1

Describe the set of points $z$ satisfying $\ |z|=\text{Re}(z)+1$.

Step 1 · Set up in Cartesian coordinates

Let $z=x+iy$. The equation $\sqrt{x^2+y^2}=x+1$ requires $x+1\ge0$, i.e. $x\ge-1$ (since the left side is nonnegative).

Step 2 · Square

$$ x^2+y^2=(x+1)^2=x^2+2x+1 \quad\Longrightarrow\quad y^2=2x+1 $$

Step 3 · Identify the curve

$$ y^2=2\left(x+\frac12\right) $$

a parabola opening rightward, in standard form $(y-k)^2=4p(x-h)$ with vertex $(h,k)=\left(-\frac12,0\right)$ and $4p=2\Rightarrow p=\frac12$.

Step 4 · Locate focus and directrix

For a rightward-opening parabola, the focus is at $(h+p,k)$ and the directrix is $x=h-p$:

$$ \text{Focus: } \left(-\frac12+\frac12,\,0\right)=(0,0), \qquad \text{Directrix: } x=-\frac12-\frac12=-1 $$

The focus is exactly the origin.

Step 5 · Confirm the domain constraint is automatic

From $y^2=2x+1$: $x=\frac{y^2-1}2\ge-\frac12>-1$ always, so the required constraint $x\ge-1$ imposes no additional restriction on the parabola.

Step 6 · Direct geometric confirmation

The original equation $|z|=\text{Re}(z)+1$ says: distance from $z$ to the origin equals $\text{Re}(z)+1=x+1$. But $x+1$ is exactly the distance from $z=(x,y)$ to the vertical line $x=-1$ (when $x\ge-1$). So the equation states:

$$ \text{distance to } (0,0) \ = \ \text{distance to the line } x=-1 $$

which is precisely the focus-directrix definition of a parabola with focus $(0,0)$ and directrix $x=-1$ — confirming Steps 3–4 without needing to solve for the equation first.

Verification