Complex conjugate of $z$ as a different variable

Question

Can a complex conjugate be represented by a different letter than $z$?

As in: Let $y$ be a complex number satisfying $|y|<1$. Find the set of all complex numbers $z$ satisfying $|z-y|\le|1-\bar{y}z|$.

Answer 1

Given $$|z-y|\leq |1-\bar{y}z|\Rightarrow |z-y|^2\leq |1-\bar{y}z|^2$$

Now Using the formula $\bullet \; |z|^2 = z\bar{z}$ and $\displaystyle \bar{\bar{z}} = z$

So we get $$(z-y)\cdot (\bar{z}-\bar{y})\leq (1-\bar{y}z)(1-y\bar{z})$$

So we get $$z\bar{z}-y\bar{z}-\bar{y}z+y\bar{y}\leq 1-\bar{y}z-y\bar{z}+y\bar{y}z\bar{z}$$

So we get $$|z|^2+|y|^2\leq 1+|z^2||y|^2\Rightarrow |z|^2-|z|^2|y|^2+|y|^2-1\leq 0$$

So we get $$|z|^2(1-|y|^2)-1(1-|y|^2)\leq 0\Rightarrow (1-|y|^2)\cdot (|z|^2-1)\leq 0$$

Now Given $$|y|<1\Rightarrow |y|^2-1<0\Rightarrow 1-|y|^2>0$$

So $$(1-|y|^2)\cdot (|z|^2-1)\leq 0$$ means $$|z|^2-1\leq 0\Rightarrow |z|\leq 1$$

So it Represent a Circle with interior part and boundry part

whose center is at $(0,0)$ are radius $1$