Is this set region?

Question

$$ |z^2 - 1| < 1 $$ Hint: use polar coordinates.

the answer is not a region.

I don't know how to start. Whenever I am trying to do, it failed.

*. z is complex number.

Answer 1

Following the hint, we write $z=re^{i\theta}$ with $r\geq 0$ and $\theta\in[0,2\pi)$. Then $$ z^2-1=r^2e^{2i\theta}-1=r^2\cos(2\theta)-1+ir^2\sin(2\theta). $$ Then $$ |z^2-1|^2=(r^2\cos(2\theta)-1)^2+(r^2\sin(2\theta))^2 $$ $$ =r^4-2r^2\cos(2\theta)+1. $$ Since $|z^2-1|<1$ if and only if $|z^2-1|^2<1$, you are looking for all $r\geq 0$ and $\theta\in[0,2\pi)$ such that $$ r^4-2r^2\cos(2\theta)=r^2(r^2-2\cos(2\theta))<0. $$ This is equivalent to $r>0$ and $r^2-2\cos(2\theta)<0$, i.e $$ 0< r^2 <2\cos(2\theta).$$ Since $\cos(2\theta)$ must be positive, we need to restrict $\theta$ to $[0,\pi/4)\cup (3\pi/4,5\pi/4)\cup (7\pi/4,2\pi)$. Then $$ 0<r<\sqrt{2}\sqrt{\cos(2\theta)}. $$ For $\theta\in (3\pi/4,5\pi/4)$, we get an open connected bounded set. For $\theta\in [0,\pi/4)\cup(7\pi/4,2\pi)$, we get another open connected bounded set. But the intersection is empty. So your set is open, bounded, and has two connected components.

Answer 2

Hint also in rectangular coordinates (good'ol analytic geometry). Put $\,z=x+yi\,$ :

$$|z^2-1|^2<1^2=1\iff (x^2-y^2 -1)^2+(2xy)^2<1\iff $$

$$(x^2+y^2)^2-2(x^2-y^2)<0$$

Having a nice geometric vision the above is the difference between a circle-like region and the intersection of two straight lines and, thus not connected.

But, of course, Julien's answer is neater.