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.

set of point represented by the inequality

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.

  • $\begingroup$ You're missing that the range $-\pi\lt\theta\lt\pi$ actually covers two cycles of $\cos(2\theta+\frac\pi4)$, so you also need to consider the case where (to choose a specific parametrization) $\frac{3\pi}2\lt(2\theta+\frac\pi4)\lt\frac{5\pi}2$. $\endgroup$ Jul 23, 2016 at 5:23
  • $\begingroup$ (And my preferred solution would be to understand what the effect of the operations $z\mapsto z^2$ and $w\mapsto (1+i)w$ on regions of the plane, particularly sectors, are and to essentially reverse-transform the 'target' region to understand the 'source' regions that map to it.) $\endgroup$ Jul 23, 2016 at 5:25
  • $\begingroup$ @Steven Stadnicki:So according to your first comment the fault that I did in my solution is that instead of considering $-\pi\lt \theta \lt \pi$, I considered $-\dfrac{\pi}{2}\lt \theta \lt \dfrac{\pi}{2}$. And at your second comment, can you elaborate some more, and what does reverse transforming the target mean. $\endgroup$
    – user350331
    Jul 23, 2016 at 5:33
  • $\begingroup$ Effectively - although what you actually did was consider $-\pi\lt(2\theta+\frac\pi4)\lt\pi$, so the section of $\theta$'s domain that you considered was really $\frac{-5\pi}{8}\lt\theta\lt\frac{3\pi}{8}$; the key point that you only wound up looking at a region of 'width' $\pi$ instead of $2\pi$ is still the same, though. As for the rest of it, I'll see if I can flesh it out into a proper answer at some point. $\endgroup$ Jul 23, 2016 at 7:20
  • $\begingroup$ The second inequality that you provided, from that I got the interval $\dfrac{5\pi}{8}\lt \theta \lt \dfrac{9\pi}{8}$, from which I got the same interval. Can you elaborate some more on the second comment about the reverse transformation. $\endgroup$
    – user350331
    Jul 23, 2016 at 7:24

2 Answers 2


what am I missing in my solution

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


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. $$-\pi\le \theta\lt -\frac78\pi\quad\text{or}\quad -\frac 38\pi\lt \theta\lt\frac{\pi}{8}\quad\text{or}\quad \frac{5}{8}\pi\lt \theta\lt\pi$$

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,


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 $(1)$, $$\begin{align}&\Re((1+i)z^2)\gt 0\\&\iff ((\sqrt 2-1)x-y)((\sqrt 2+1)x+y)\gt 0\\&\iff -(\sqrt 2+1)x\lt y\lt (\sqrt 2-1)x\quad\text{or}\quad (\sqrt 2-1)x\lt y\lt -(\sqrt 2+1)x\end{align}$$

  • $\begingroup$ Using real and imaginary parts is definitely "more elegant" here. +1 for this part. $\endgroup$
    – Did
    Jul 23, 2016 at 10:44

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*}

  • $\begingroup$ Can you tell me what does the modulo function signify in the inequality. Though I do know that it deals with the remainder, but I still don't know much about it. $\endgroup$
    – user350331
    Jul 23, 2016 at 11:44
  • $\begingroup$ It's a somewhat unorthodox use of modulo here, for the sake of concision. I mean $\theta$ has to belong to one of the intervals $\;\Bigl]-\dfrac{3\pi}8+k\pi,\dfrac\pi8+k\pi\Bigr[$ for some $k\in\mathbf Z$. $\endgroup$
    – Bernard
    Jul 23, 2016 at 12:42

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