$z$ is a complex number, $|z^2-1|<1$. Question is, to verify whether the above set is a region?
So i tried to plot the graph of the above set, to get idea of the set.
First i tried converting to, cartesian coordinates, it became, $(x^2+y^2)^2<2(x^2-y^2)$. But for me,this seemed no good.
Using polar coordinate, the inequality transformed to, $r^2< 2\cos(2θ)$, it looks a lot simpler now.
Now , i don't know how to plot such graphs, in polar coordinates.
Well, first notice that you only need to check what happens as $\theta$ varies between $0$ and $2 \pi$. First, at $\theta = 0$, we get $r^2 < 2$. Since $r$ is a non-negative real number, it implies $r < \sqrt{2}$.
From $0$ to $\frac{\pi}{2}$, $\cos$ is a decreasing function, and it reaches $0$ at $\theta = \frac{\pi}{2}$. Further, it continues decreasing and reaches $-1$ at $\theta = \pi$. After $\pi$, $\cos$ starts increasing until it reaches $0$ again at $3 \frac{\pi}{2}$. Since $r$ is a non-negative real number, you will not get any solution in the range $\left[ \frac{\pi}{2}, 3 \frac{\pi}{2} \right]$.
From $3 \frac{\pi}{2}$ to $2\pi$, the function $\cos$ continues to increase till it reaches $1$ at $\theta = 2\pi$. Here, we will get non-negative $r$ satstying our condition.
Now, we must notice that the argument of $\cos$ is $2 \theta$. Therefore, all the values that I have mentioned here will be attained "faster". So, instead of not getting a solution in $\left[ \frac{\pi}{2}, 3 \frac{\pi}{2} \right]$, you will not have solution in $\left[ \frac{\pi}{4}, 3 \frac{\pi}{4} \right] \cup \left[ 5 \frac{\pi}{4}, 7 \frac{\pi}{4} \right]$. In the other areas, we will have a solution. First, the allowed radius will decrease from $\sqrt{2}$ to $0$ as theta increases from $0$ to $\frac{\pi}{4}$. Then, there will be no region in $\left[ \frac{\pi}{4}, 3 \frac{\pi}{4} \right]$. From $\theta = 3 \frac{\pi}{4}$ to $\theta = \pi$, the allowed radius increases from $0$ to $\sqrt{2}$ and then again decreases to $0$ in the region $\left[ \pi, 5 \frac{\pi}{4} \right]$. After a region of no solution, the allowed radius again increases from $0$ to $\sqrt{2}$ in the region $\left[ 7 \frac{\pi}{4}, 2 \pi \right]$. So, you will get a graph like this:
PS: The graph is not to scale. It is only to give an idea what the region might look like.