Question

$Aratati~ca~daca~R(z)\ \textgreater \ 1~si~z\in~C^*~atunci~|\frac{1}{z}-\frac{1}{2}| \ \textless \ \frac{1}{2}.\\ \\ Poate~parea~prostesc,~dar~primul~meu~gand~a~fost~sa~fac~astfel:\\ \\ |\frac{1}{z}-\frac{1}{2}|\ \textless \ \frac{1}{|z|}-\frac{1}{2} \ \textless \ \frac{1}{2} .\\ \\ Si~asa~as~fi~terminat,~insa~la~sfarsit~nu~e~tocmai~asa.\\ \\ Putin~ajutor~si~o~lamurire~la~ce~tocmai~am~Incercat~eu.~De~ce~n-ar~merge~asa?~~Multumesc!$

Answer 1

$\it z=a+bi,\ \ \ Re(z)>1\Rightarrow a>1\Rightarrow a^2>1 \ \ \ \ \ (*) \\ \\ |z| =\sqrt{a^2+b^2}\ \stackrel{(*)}{\Longrightarrow} \ |z| >1 \Rightarrow \dfrac{1}{|z|} <1 \ \ \ (**) \\ \\ \ \Big|\dfrac{1}{z}-\dfrac{1}{2}\Big| \leq\dfrac{1}{|z|} -\dfrac{1}{2}\ \stackrel{(**)}{\Longrightarrow}\ \Big|\dfrac{1}{z}-\dfrac{1}{2}\Big| < 1-\dfrac{1}{2} \Rightarrow \Big|\dfrac{1}{z}-\dfrac{1}{2}\Big|<\dfrac{1}{2}$