Determine all complex numbers $z$ for an inequality

Question

Determine all complex numbers $z$ for which $|2z-1| \le 2|z-i|$.

Can someone please help me with this question? I'm not sure how to start it off. Its not a homework question, its on an exam preparation question sheet and I really need help! Thanks!

Answer 1

The inequality $|2z-1| \le 2|z-i|$ is equivalent with $|2z-1|^2 \le 4|z-i|^2$.

Now use, that for a complex number $w$ we have $|w|^2=w \overline{w}$.

Can you take it from here ?

Answer 2

let $z=a+bi$ then you will get $$|2(a+bi)-1|\le 2|a+bi-i|$$ or $$\sqrt{(2a-1)^2+4b^2}\le 2\sqrt{a^2+(b-1)^2}$$ simplifying we get $$3+4a-8b\geq 0$$