Question Number 16540 by gourav~ last updated on 23/Jun/17
$${if}\:\mid{Z}\mid=\mathrm{1}\:{Then}\:\frac{\mathrm{1}+{Z}}{\mathrm{1}+\bar {{Z}}}\:\:{is}\:{equal}\:{to}... \\ $$$$\left.{a}\right)\:{Z}\:\: \\ $$$$\left.{b}\right)\:\:\bar {{Z}} \\ $$$$\left.{c}\right)\:{Z}+\bar {{Z}} \\ $$$$\left.{d}\right)\:{N}.{O}.{T} \\ $$
Answered by sma3l2996 last updated on 23/Jun/17
$${Z}={a}+{ib}\Leftrightarrow\mid{Z}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{1}\Leftrightarrow{b}=\sqrt{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}+{a}+{i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\mathrm{1}+{a}−{i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}=\frac{\left(\mathrm{1}+{a}+{i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\right)^{\mathrm{2}} }{\left(\mathrm{1}+{a}\right)^{\mathrm{2}} −\left({i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\right)^{\mathrm{2}} }=\frac{\left(\mathrm{1}+{a}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}+{a}\right){i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }−\mathrm{1}+{a}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{2}\left(\mathrm{1}+{a}\right){i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\mathrm{2}+\mathrm{2}{a}}=\frac{\mathrm{2}\left(\mathrm{1}+{a}\right)\left({a}+{i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\right)}{\mathrm{2}\left(\mathrm{1}+{a}\right)} \\ $$$$={a}+{i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }={Z} \\ $$$${so}\:{the}\:{answer}\:{is}\:\left({a}\right) \\ $$
Answered by ajfour last updated on 23/Jun/17
$$\:\:{let}\:\:{z}={e}^{{i}\theta} \:\:\:{then}\:\:\:\bar {{z}}\:={e}^{−{i}\theta} \\ $$$$\:\:\:\frac{\mathrm{1}+{z}}{\mathrm{1}+\bar {{z}}}=\:\frac{\mathrm{1}+{e}^{{i}\theta} }{\mathrm{1}+{e}^{−{i}\theta} }\:=\left(\frac{{e}^{{i}\theta/\mathrm{2}} }{{e}^{−{i}\theta/\mathrm{2}} }\right)\left(\frac{{e}^{−{i}\theta/\mathrm{2}} +{e}^{{i}\theta/\mathrm{2}} }{{e}^{{i}\theta/\mathrm{2}} +{e}^{−{i}\theta/\mathrm{2}} }\right) \\ $$$$\:=\:{e}^{{i}\theta} \:=\:{z}\:.\:\:\:{option}\:\left({a}\right)\:. \\ $$$$ \\ $$