On real part of the complex number $(1+i)z^2$

Question

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.

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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.

Answer 1

what am I missing in my solution

After having $\cos\left(2\theta+\frac{\pi}{4}\right)\gt 0$, you have

$$-\dfrac{\pi}{2}\lt\left(2\theta+\dfrac{\pi}{4}\right)\lt\dfrac{\pi}{2}$$

which is incorrect.

To make it easy to understand why this is incorrect, let $\alpha:=2\theta+\frac{\pi}{4}$.

Then, we want to solve $$\cos\alpha\gt 0\quad\text{and}\quad -\pi\le\theta\lt \pi,$$ i.e. $$\cos\alpha\gt 0\quad\text{and}\quad -\frac{7}{4}\pi\le\alpha\lt \frac{9}{4}\pi$$ which sould be easier to solve, to have $$-\frac{7}{4}\pi\le \alpha\lt -\frac{3}{2}\pi\quad\text{or}\quad -\frac{\pi}{2}\lt\alpha\lt \frac{\pi}{2}\quad\text{or}\quad \frac 32\pi\lt\alpha\lt\frac{9}{4}\pi,$$ i.e. $$-\pi\le \theta\lt -\frac78\pi\quad\text{or}\quad -\frac 38\pi\lt \theta\lt\frac{\pi}{8}\quad\text{or}\quad \frac{5}{8}\pi\lt \theta\lt\pi$$

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as always more elegant solutions are welcome

(not sure if this is more elegant, but) another solution :

Let $z=x+iy$ where $x,y\in\mathbb R$. Then,

$$\Re((1+i)z^2)=\Re((1+i)(x+iy)^2)=x^2-y^2-2xy\tag1$$

When we solve $y^2+2xy-x^2=0$ for $y$, we get $$y=-x\pm\sqrt{x^2+x^2}=-x\pm\sqrt 2\ x=(\pm\sqrt 2-1)x$$ so from $(1)$, $$\begin{align}&\Re((1+i)z^2)\gt 0\\&\iff ((\sqrt 2-1)x-y)((\sqrt 2+1)x+y)\gt 0\\&\iff -(\sqrt 2+1)x\lt y\lt (\sqrt 2-1)x\quad\text{or}\quad (\sqrt 2-1)x\lt y\lt -(\sqrt 2+1)x\end{align}$$

Answer 2

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*}