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.