Sketching a set of $Re(z^2(1+i))$ on the Complex plane. So I am trying to sketch a set of complex numbers that is characterized by inequality: $ Re((1+i)z^2) < 0 $.
I know that if I take $ z = a + bi $ , then:
$$Re(z^2) \implies (a + bi)^2 = a^2 + 2abi - b^2 \implies  Re(z^2) = a^2 - b^2$$
Similarly:
$$Re(z^2(1+i)) \implies  (a^2 + 2abi - b^2)(1+i) =  a^2 + 2abi - b^2 + a^2i - 2ab - b^2i \implies Re(z^2(1+i)) =  a^2 - 2abi + b^2 = (a-b)^2 $$
So how am I supposed to sketch a set of complex numbers $(a-b)^2 $ on a 2D plane? Or did I do something wrong?

Some facts:
We require the real part to be less than zero, which means $(1 + i)z^2$ must be located on the left half of the complex plane.  In terms of the argument, this means $(1+i) z^2$ must have an argument between $\pi/2$ and $3\pi/2$, plus some integer multiple of $2\pi$.
Squaring a complex number doubles its argument.
Multiplying a complex number by $1+i$ increases its magnitude by $|1+i| = \sqrt{2}$ and rotates it by $\pi/4$; i.e., the argument is increased by $\pi/4$.
Therefore, $$\arg((1+i)z^2) = \frac{\pi}{4} + 2\arg (z)$$ and we need this to be between $\pi/2$ and $3\pi/2$, or between $5\pi/2$ and $7\pi/2$.  This gives us $$\frac{\pi}{8} < \arg (z) < \frac{5\pi}{8} \quad \text{or} \quad \frac{9\pi}{8} < \arg (z) < \frac{13\pi}{8}.$$  Using interval notation, we can also write this as $$\arg (z) \in (\tfrac{\pi}{8}, \tfrac{5\pi}{8}) \cup (\tfrac{9\pi}{8}, \tfrac{13\pi}{8}).$$
We can visualize this region in the complex plane by coloring quadrants I and III (i.e., those regions where $\Re(z)\Im(z) = 1$, and then rotating those regions by $\pi/8$ counterclockwise.

