Give a geometric interpretation of the following relations…

Give a geometric interpretation of the following relations: $$ 1^{\circ}\,\, |z-2| + |z+2| = 5; \quad 2^{\circ}\,\, |z| = \text{Re}(z) + 1; \quad 3^{\circ}\,\, \text{Re}(z) + \text{Im}(z) < 1. $$

Part $1^{\circ}$ · $|z-2|+|z+2|=5$

Step 1 · Read off the distances

$|z-2|$ is the distance from $z$ to the point $2=(2,0)$, and $|z+2|=|z-(-2)|$ is the distance from $z$ to $-2=(-2,0)$. The relation says these two distances sum to the constant $5$.

Step 2 · Recognize the ellipse

A constant sum of distances to two fixed points is exactly the definition of an ellipse with foci $F_1=2$, $F_2=-2$ and $2a=5$, so:

$$ a = \frac{5}{2} $$

Step 3 · Find $b$

The distance between the foci is $2c=|2-(-2)|=4$, so $c=2$. Using $a^2=b^2+c^2$:

$$ b^2 = a^2-c^2 = \left(\frac{5}{2}\right)^2 - 2^2 = \frac{25}{4}-4 = \frac{9}{4} \quad\Longrightarrow\quad b=\frac{3}{2} $$

Step 4 · Cartesian equation

With $z=x+iy$, center at the origin, major axis along the real axis:

$$ \boxed{\frac{x^2}{(5/2)^2}+\frac{y^2}{(3/2)^2}=1} \qquad\text{i.e.}\qquad \boxed{\frac{4x^2}{25}+\frac{4y^2}{9}=1} $$

Answer for $1^{\circ}$

An ellipse centered at the origin, with foci at $(\pm2,0)$, semi-major axis $a=\frac{5}{2}$ along the real axis, semi-minor axis $b=\frac{3}{2}$ along the imaginary axis, and vertices at $\left(\pm\frac{5}{2},0\right)$, $\left(0,\pm\frac{3}{2}\right)$.

Part $2^{\circ}$ · $|z|=\text{Re}(z)+1$

Step 1 · Substitute $z=x+iy$

$$ |z|=\sqrt{x^2+y^2}, \qquad \text{Re}(z)=x $$

so the relation becomes:

$$ \sqrt{x^2+y^2} = x+1 $$

Step 2 · Domain restriction

Since the left side is $\geq 0$, we need $x+1\geq 0$, i.e. $x\geq -1$.

Step 3 · Square both sides

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

Step 4 · Identify the parabola

Rewrite as $y^2 = 2\left(x+\frac{1}{2}\right)$. Comparing with $(y-k)^2=4p(x-h)$ gives $4p=2$, so $p=\frac{1}{2}$, $h=-\frac{1}{2}$, $k=0$. The focus is at $(h+p,k)=(0,0)$, and the directrix is $x=h-p=-1$.

$$ \boxed{y^2 = 2x+1} $$

The domain constraint from $y^2=2x+1$ is $x\geq -\frac12$, which is stronger than (and consistent with) $x\geq -1$.

Answer for $2^{\circ}$

A parabola opening to the right, with vertex $\left(-\frac12,0\right)$, focus at the origin $(0,0)$, and directrix $x=-1$: the set of points equidistant from the origin and the line $x=-1$.

Part $3^{\circ}$ · $\text{Re}(z)+\text{Im}(z)<1$

Step 1 · Substitute $z=x+iy$

$$ \text{Re}(z)+\text{Im}(z) = x+y $$

so the relation becomes $x+y<1$.

Step 2 · The boundary line

The boundary is $x+y=1$, a line with $x$-intercept $(1,0)$, $y$-intercept $(0,1)$, slope $-1$.

Step 3 · Which side?

Testing the origin: $0+0=0<1$, so the origin satisfies the inequality — the region is the open half-plane containing the origin, i.e. below-left of the line.

$$ \boxed{x+y<1} $$

Equivalently, writing $x+y-1<0$: the distance from a point to the line $x+y=1$ is $d=\dfrac{|x+y-1|}{\sqrt2}$, and points of the region satisfy $x+y-1<0$ — the side of the line opposite to the normal vector $\vec n=(1,1)$.

Answer for $3^{\circ}$

The open half-plane below the line $x+y=1$: all $(x,y)$ with $x+y<1$, excluding the boundary line, containing the origin.