<s>[INST] On real part of the complex number $(1+i)z^2$ 
Find the set of points belonging to the coordinate plane $xy$, for which the real part of the complex number $(1+i)z^2$ is positive.


My solution:-
Lets start with letting $z=r\cdot e^{i\theta}$. Then the expression $(1+i)z^2$ becomes $$\large\sqrt2\cdot|z|^2\cdot e^{{i}\left(2\theta+\dfrac{\pi}{4}\right)}$$
Now, as $\sqrt2\cdot|z|^2\gt0$, so $\Re{((1+i)z^2)}\gt 0 \implies\cos{\left(2\theta+\dfrac{\pi}{4}\right)}\gt 0$. So, we get 
$$-\dfrac{\pi}{2}\lt\left(2\theta+\dfrac{\pi}{4}\right)\lt\dfrac{\pi}{2}
\implies-\dfrac{3\pi}{4}\lt2\theta\lt\dfrac{\pi}{4}
\implies-\dfrac{3\pi}{8}\lt\theta\lt\dfrac{\pi}{8}$$
Now, lets find the equation of the lines which would help us show these inequalities in the coordinate plane.
The inequality can be represented by 
$$\begin{equation}
y\lt \tan{\dfrac{\pi}{8}}x\implies y\lt(\sqrt2-1)x \tag{1}
\end{equation}$$
$$\begin{equation}
y\gt \tan{(-\dfrac{3\pi}{8})}x \implies y\gt-(\sqrt2+1)x \tag{2}
\end{equation}$$
So, the inequality can be represented in the coordinate plane as in the following portion of the graph with the cross-hatched part.

My deal with the question:- 
The book I am solving gives the answer as the (cross-hatched part + un-hatched part), so what am I missing in my solution. And, as always more elegant solutions are welcome.
 [/INST] I would find the general solutions of the inequation first:
\begin{align*}
\cos\Bigl(2\theta+\frac\pi4\Bigr)>0&\iff-\frac\pi2<2\theta+\frac\pi4<\frac\pi2\iff-\frac{3\pi}4<2\theta<\frac\pi4\color{red}{\mod2\pi}\\
&\iff-\frac{3\pi}8<\theta<\frac\pi8\color{red}{\mod\pi}
\end{align*}
Now that if conventionally, we choose $\;-\pi<\theta\le\pi$,
 we obtain
\begin{alignat*}{2}&\bullet\quad&-\dfrac{3\pi}8&<\theta&&<\dfrac\pi8,\\
&\bullet\quad&-\pi&<\theta&&<-\dfrac{7\pi}8,\\
&\bullet\quad&\dfrac{5\pi}8&<\theta&&<\pi.
\end{alignat*}
</s>