Let $z,w\in \mathbb{C}$ where $|w|<1$ (modulus). What is the set of all $z\in \mathbb{C}$ that satisfies $|z-w|\leq|1-\bar{w}z|$?
I've tried a few things with no luck. I wrote $z,w$ are complex numbers and used the inequality, but I don't know how to find the set that satisfies the inequality. Any solutions or help is greatly appreciated.
Your inequality is the same as $|z-w|^2\leq|1-\overline{w}z|^2$, where \begin{eqnarray*} |z-w|^2&=&(z-w)(\overline{z-w})=(z-w)(\overline{z}-\overline{w})=z\overline{z}-\overline{w}z-w\overline{z}+w\overline{w}\\ &=&|z|^2+|w|^2-\overline{w}z-w\overline{z},\\ \ \\ |1-\overline{w}z|^2&=&(1-\overline{w}z)(\overline{1-\overline{w}z})=(1-\overline{w}z)(1-w\overline{z})=1-w\overline{z}-\overline{w}z+w\overline{w}z\overline{z}\\&=&1-w\overline{z}-\overline{w}z+|w|^2|z|^2. \end{eqnarray*}
Plugging this into the original inequality yields $$|z|^2+|w|^2-\overline{w}z-w\overline{z}\leq1-w\overline{z}-\overline{w}z+|w|^2|z|^2,$$ and rearranging the terms shows that $1-|z|^2-|w|^2+|w|^2|z|^2\geq0$, or equivalently $$(1-|z|^2)(1-|w|^2)\geq0.$$ Given that $|w|<1$ we also $1-|w|^2>0$, so dividing through by $1-|w|^2$ yields the inequality $$1-|z|^2\geq0\qquad\text{ or equivalently }\qquad |z|^2\leq0.$$ So the set you are looking for is the closed unit disk.