Complex Inequality and convergence

Question

Can anyone explain why if a Laurent series converges for $|\frac{z-i}{-i}| < 1$ then it also converges for $|z-i| < 1$ and maybe explain how to deal with complex inequalities; how to deal with the absolute value sign and what happens when you divide by $i$.

Any help is appreciated, thanks!

Answer 1

Hint: $$|wz| = |w||z|,\qquad |w/z| = \cdots$$

Answer 2

Note that $\vert i\vert=1$. Hence $\vert\dfrac{z-i}{-i}\vert=\dfrac{\vert z-i\vert}{\vert -i\vert}=\vert z-i\vert$.