Question Number 135298 by Ahmed1hamouda last updated on 12/Mar/21
Answered by mr W last updated on 12/Mar/21
$$\mathrm{1} \\ $$$${z}={x}+{yi} \\ $$$$\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{{x}+{yi}}=\frac{{x}−{yi}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$${Re}\left(\frac{\mathrm{1}}{{z}}\right)=\frac{{x}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \geqslant\mathrm{2}{x} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \geqslant\mathrm{1} \\ $$
Commented by mr W last updated on 12/Mar/21
Answered by mr W last updated on 12/Mar/21
$$\mathrm{2} \\ $$$${z}={x}+{yi} \\ $$$$\overset{−} {{z}}={x}−{yi} \\ $$$${z}−\overset{−} {{z}}=\mathrm{2}{yi} \\ $$$$\frac{{z}−\bar {{z}}}{\mathrm{2}{i}}={y}={Im}\left({z}\right) \\ $$