Question Number 58592 by rahul 19 last updated on 25/Apr/19
$${If}\:\mid{z}−\mathrm{1}\mid=\mathrm{1},\:{then}\:{prove}\:{that}\:{arg}\left({z}\right)\:=\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{arg}\left({z}−\mathrm{1}\right). \\ $$
Answered by tanmay last updated on 26/Apr/19
$${z}={re}^{{i}\theta} ={rcos}\theta+{irsin}\theta \\ $$$${z}−\mathrm{1}=\left({rcos}\theta−\mathrm{1}\right)+{irsin}\theta \\ $$$$\mid{z}−\mathrm{1}\mid=\sqrt{\left({rcos}\theta−\mathrm{1}\right)^{\mathrm{2}} +\left({rsin}\theta\right)^{\mathrm{2}} } \\ $$$${r}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta−\mathrm{2}{rcos}\theta+\mathrm{1}+{r}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{rcos}\theta=\mathrm{0} \\ $$$${r}\left({r}−\mathrm{2}{cos}\theta\right)=\mathrm{0} \\ $$$${r}\neq\mathrm{0}\:\:{r}=\mathrm{2}{cos}\theta \\ $$$${z}=\mathrm{2}{cos}\theta×{cos}\theta+{i}\mathrm{2}{cos}\theta{sin}\theta \\ $$$${A}+{iB}=\left(\mathrm{2}{cos}^{\mathrm{2}} \theta\right)+{i}\left(\mathrm{2}{sin}\theta{cos}\theta\right) \\ $$$${arg}\left({z}\right)=\theta \\ $$$${now} \\ $$$${z}−\mathrm{1} \\ $$$$\mathrm{2}{cos}^{\mathrm{2}} \theta+{isin}\mathrm{2}\theta−\mathrm{1} \\ $$$$={cos}\mathrm{2}\theta+{isin}\mathrm{2}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{arg}\left({z}−\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\theta=\theta \\ $$
Commented by JDamian last updated on 25/Apr/19
$${There}\:{is}\:{a}\:{mistake}\:{in}\:{the}\:{fourth}\:{line}: \\ $$$${rsin}^{\mathrm{2}} \theta\:\:\:\:{should}\:{be}\:\:\:\:\boldsymbol{{r}}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta \\ $$
Commented by tanmay last updated on 26/Apr/19
$${thank}\:{you}\:{sir} \\ $$
Commented by rahul 19 last updated on 26/Apr/19
$${thank}\:{U}\:{sir}. \\ $$