Question Number 164546 by mathls last updated on 18/Jan/22
Commented by mathls last updated on 18/Jan/22
$${Dom}\left({g}\right)=? \\ $$
Commented by mkam last updated on 18/Jan/22
$$\boldsymbol{{D}}_{\boldsymbol{{g}}\left(\boldsymbol{{z}}\right)} \:=\:−\mathrm{2}\:<\:\boldsymbol{{z}}\:<\:\mathrm{2} \\ $$
Commented by mathls last updated on 19/Jan/22
$${no}\:{find}\:{Dom}\left({g}\right) \\ $$
Commented by mkam last updated on 19/Jan/22
$$\boldsymbol{{D}}_{\boldsymbol{{g}}\left(\boldsymbol{{z}}\right)} \:=\:\boldsymbol{{Dom}}\left(\boldsymbol{{g}}\right)\: \\ $$
Answered by mathmax by abdo last updated on 19/Jan/22
$$\mid\mathrm{z}\mid<\mathrm{2}\:\:\mathrm{if}\:\mathrm{z}\:\mathrm{from}\:\mathrm{C}\:\:\:\mathrm{and}\:−\mathrm{2}<\mathrm{z}<\mathrm{2}\:\mathrm{if}\:\mathrm{z}\:\mathrm{is}\:\mathrm{real} \\ $$$$\mathrm{z}=\mathrm{x}+\mathrm{iy}\:\rightarrow\mid\mathrm{z}\mid<\mathrm{2}\:\Leftrightarrow\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }<\mathrm{2}\:\Rightarrow\mathrm{x}^{\mathrm{2}} \:+\mathrm{y}^{\mathrm{2}} <\mathrm{4}…. \\ $$