Determine the set of points z if |Re(az)| + |Im(az)| = |a|^2…

Determine the set of points $z$ if $|\text{Re}(az)| + |\text{Im}(az)| = |a|^2$, where $a$ is a real number.

Explanation of Method

Since $a$ is real, multiplying $z=x+iy$ by $a$ just scales both parts: $\text{Re}(az)=ax$, $\text{Im}(az)=ay$. Substituting turns the condition into $|a|(|x|+|y|)=|a|^2$. Splitting into the case $a=0$ (trivial) and $a\neq0$ (divide through by $|a|$) reduces everything to the classical equation $|x|+|y|=\text{const}$, whose graph is a square rotated $45°$.

Step 1 · Set up

Let $z=x+iy$. Since $a\in\mathbb R$: $$ az = ax+iay \quad\Longrightarrow\quad \text{Re}(az)=ax,\quad \text{Im}(az)=ay $$

Step 2 · Rewrite the condition

$$ |ax|+|ay| = |a|^2 \quad\Longrightarrow\quad |a|\,|x|+|a|\,|y| = |a|^2 $$

Step 3 · Split into cases

Case $a=0$: the equation becomes $0+0=0$, true for every $z\in\mathbb C$ — the whole plane is the solution set.

Case $a\neq0$: divide both sides by $|a|>0$:

$$ |x|+|y| = |a| $$

Step 4 · Identify the curve $|x|+|y|=|a|$

Write $r=|a|>0$ and examine each quadrant:

Quadrant I ($x\ge0,y\ge0$): $x+y=r$, a segment from $(r,0)$ to $(0,r)$.
Quadrant II ($x\le0,y\ge0$): $-x+y=r$, a segment from $(0,r)$ to $(-r,0)$.
Quadrant III ($x\le0,y\le0$): $-x-y=r$, a segment from $(-r,0)$ to $(0,-r)$.
Quadrant IV ($x\ge0,y\le0$): $x-y=r$, a segment from $(0,-r)$ to $(r,0)$.

These four segments join into a single closed curve: a square, rotated $45°$ relative to the axes, centered at the origin.

Step 5 · Properties of the square

Vertices: $z=|a|,\ i|a|,\ -|a|,\ -i|a|$ (i.e. $(\pm|a|,0)$ and $(0,\pm|a|)$).

Side length: the distance between adjacent vertices, e.g. $(|a|,0)$ to $(0,|a|)$, is $|a|\sqrt2$. Perimeter: $4|a|\sqrt2$. Area (diagonals $2|a|$ each): $\dfrac{(2|a|)(2|a|)}{2}=2|a|^2$.

In norm notation, $|x|+|y|$ is the $L^1$ (taxicab) norm of $z$, so the equation reads $\|z\|_1=|a|$ — this is the $L^1$-norm analogue of the circle $|z|=|a|$ (which uses the Euclidean, $L^2$, norm).

Step 6 · Comparing to the circle

The square $|x|+|y|=|a|$ sits between two circles centered at the origin: it touches the circle $x^2+y^2=|a|^2$ exactly at its four vertices, and touches the smaller circle $x^2+y^2=\dfrac{|a|^2}{2}$ at the midpoint of each side (e.g. midpoint $\left(\dfrac{|a|}2,\dfrac{|a|}2\right)$ has distance $\dfrac{|a|}{\sqrt2}$ from the origin).

Step 7 · Check with examples

$a=1$: $|x|+|y|=1$. Points $(1,0)$, $(0.5,0.5)$, $(0,1)$, $(-0.3,0.7)$ all give sum $1$. ✓

$a=-2$: $|a|=2$, giving the square $|x|+|y|=2$ with vertices $(\pm2,0)$, $(0,\pm2)$.

$a=0$: the equation collapses to $0=0$, satisfied everywhere.

The square $|x|+|y|=|a|$ (blue), touching the circumscribed circle $|z|=|a|$ at its vertices and the inscribed circle $|z|=|a|/\sqrt2$ at its edge midpoints (both dashed).

Result

$$ \{z : |\text{Re}(az)|+|\text{Im}(az)| = |a|^2\} = \begin{cases} \mathbb{C} & \text{if } a=0 \\[4pt] \{z : |\text{Re}(z)|+|\text{Im}(z)| = |a|\} & \text{if } a\neq0 \end{cases} $$

For $a\neq0$ this is the square with vertices $\pm|a|,\ \pm i|a|$, side length $|a|\sqrt2$, perimeter $4|a|\sqrt2$, and area $2|a|^2$ — the $L^1$-norm analogue of the circle $|z|=|a|$.