Describing the following region in the complex plane $|z-1|<\Im(z)$ I was considering the region shown in the title
$$|z-1|<\Im(z)$$
I took the approach of considering $z\in \mathbb{C}$ and taking $z=x+iy$ then breaking down the inequality as follows
$$|z-1|=\sqrt{(x-1)^2+y^2}<y$$
squaring both sides
$$(x-1)^2+y^2<y^2$$
but this gives $(x-1)^2<0$ which is impossible therefore I said that the set of points satisfying the inequality would be empty, but my professor keeps telling me that the solution is wrong am I missing something
 A: Suppose we consider $|z - 1| \le r$ in the complex plane. Then let $z = a + bi$ and so we have $\sqrt{(a - 1)^2 + b^2} \le r$ or $(a - 1)^2 + b^2 \le r^2$. This is the set of points that have a circle with center $(a + 1, b)$ and radius $r$ as the boundary. In other words, you may think of it as concentric circles with that center and radii from $0$ to $r$, inclusive. If we change the inequality to being strict, then it the set of concentric circles with radii $0$ to $r$ without including $r$.
Clearly, this is perfectly fine for $r > 0$, but if $r < 0$, then this does not make sense. So consider $\text{Im}(z) > 0$, since we may exclude the other case entirely. If $\text{Im}(z) > 0$, then we proceed with your steps and find that $(a - 1)^2 + b^2 < b^2$, which indeed also has no solutions.
There was mistake when you squared both sides of the inequality since you assumed that $y > 0$.
Since neither $\text{Im}(z) > 0$ or $< 0$ has solutions, we can conclude there are no solutions at all. (note $b = 0$ immediately gives $(a - 1)^2 < 0 $)
