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?

When dealing with complex number, polar form is almost always more useful than breaking it into real and imaginary part.
$$
Re((1+i)z^{2})<0\implies \frac{\pi}{2}<\arg{((1+i)z^{2})}<\frac{3\pi}{2}
$$
Now we use the following
$$
\begin{align}
\arg{((1+i)z^{2})}&=\arg{(1+i)}+ 2\arg{(z)}\\
&=\frac{\pi}{4}+2\arg{(z)}
\end{align}
$$
Substitute this to the first equation to obtain
$$
\frac{\pi}{8}<\arg{(z)}<\frac{5\pi}{8}
$$
