In the middle of problem i've got inequality: Find all $z\in\mathbb{C}$ such as: $$ ||z|-1|<|z+1|$$
So i start to elaborate on that for $z=x+iy$:
$$ |\sqrt{x^2+y^2}-1|<\sqrt{(x+1)^2+y^2} $$
Now i see, that i have to split it for 2 parts: For $\sqrt{x^2+y^2}<1$ and $\sqrt{x^2+y^2}\geq1$. And Here I'm stock. After a few calculations I've got only y>0. Wolafram solutions have more specific result.
Thank you in advance for help and suggestions!
$||z|-1| < |z+1| \iff ||z|-1|^2 < |z+1|^2 \iff |z|^2 - 2|z| +1 < (z+1)(\overline{z}+1)$ $\iff |z|^2 - 2|z| +1 < |z|^2 +z+ \overline{z} +1 \iff 0< 2|z| + z+ \overline{z} \iff 0< |z| + \Re(z)$. If $z=x+iy$ this means $-x < \sqrt{x^2+y^2}$ which is always true if $y \not= 0$. If $y=0$ the inequality reads $-x < |x|$ which means $x > 0$. So, the inequality holds if and only if $z \in \mathbb{C}\setminus (-\infty,0]$.