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Question Number 16540 by gourav~ last updated on 23/Jun/17

if ∣Z∣=1 Then ((1+Z)/(1+Z^� ))  is equal to...  a) Z    b)  Z^�   c) Z+Z^�   d) N.O.T

$${if}\:\mid{Z}\mid=\mathrm{1}\:{Then}\:\frac{\mathrm{1}+{Z}}{\mathrm{1}+\bar {{Z}}}\:\:{is}\:{equal}\:{to}... \\ $$$$\left.{a}\right)\:{Z}\:\: \\ $$$$\left.{b}\right)\:\:\bar {{Z}} \\ $$$$\left.{c}\right)\:{Z}+\bar {{Z}} \\ $$$$\left.{d}\right)\:{N}.{O}.{T} \\ $$

Answered by sma3l2996 last updated on 23/Jun/17

Z=a+ib⇔∣Z∣=(√(a^2 +b^2 ))=1⇔b=(√(1−a^2 ))  ((1+a+i(√(1−a^2 )))/(1+a−i(√(1−a^2 ))))=(((1+a+i(√(1−a^2 )))^2 )/((1+a)^2 −(i(√(1−a^2 )))^2 ))=(((1+a)^2 +2(1+a)i(√(1−a^2 ))−1+a^2 )/(1+a^2 +2a+1−a^2 ))  =((2a^2 +2a+2(1+a)i(√(1−a^2 )))/(2+2a))=((2(1+a)(a+i(√(1−a^2 ))))/(2(1+a)))  =a+i(√(1−a^2 ))=Z  so the answer is (a)

$${Z}={a}+{ib}\Leftrightarrow\mid{Z}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{1}\Leftrightarrow{b}=\sqrt{\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}+{a}+{i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\mathrm{1}+{a}−{i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}=\frac{\left(\mathrm{1}+{a}+{i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\right)^{\mathrm{2}} }{\left(\mathrm{1}+{a}\right)^{\mathrm{2}} −\left({i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\right)^{\mathrm{2}} }=\frac{\left(\mathrm{1}+{a}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}+{a}\right){i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }−\mathrm{1}+{a}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{1}−{a}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{2}\left(\mathrm{1}+{a}\right){i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\mathrm{2}+\mathrm{2}{a}}=\frac{\mathrm{2}\left(\mathrm{1}+{a}\right)\left({a}+{i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\right)}{\mathrm{2}\left(\mathrm{1}+{a}\right)} \\ $$$$={a}+{i}\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }={Z} \\ $$$${so}\:{the}\:{answer}\:{is}\:\left({a}\right) \\ $$

Answered by ajfour last updated on 23/Jun/17

  let  z=e^(iθ)    then   z^�  =e^(−iθ)      ((1+z)/(1+z^� ))= ((1+e^(iθ) )/(1+e^(−iθ) )) =((e^(iθ/2) /e^(−iθ/2) ))(((e^(−iθ/2) +e^(iθ/2) )/(e^(iθ/2) +e^(−iθ/2) )))   = e^(iθ)  = z .   option (a) .

$$\:\:{let}\:\:{z}={e}^{{i}\theta} \:\:\:{then}\:\:\:\bar {{z}}\:={e}^{−{i}\theta} \\ $$$$\:\:\:\frac{\mathrm{1}+{z}}{\mathrm{1}+\bar {{z}}}=\:\frac{\mathrm{1}+{e}^{{i}\theta} }{\mathrm{1}+{e}^{−{i}\theta} }\:=\left(\frac{{e}^{{i}\theta/\mathrm{2}} }{{e}^{−{i}\theta/\mathrm{2}} }\right)\left(\frac{{e}^{−{i}\theta/\mathrm{2}} +{e}^{{i}\theta/\mathrm{2}} }{{e}^{{i}\theta/\mathrm{2}} +{e}^{−{i}\theta/\mathrm{2}} }\right) \\ $$$$\:=\:{e}^{{i}\theta} \:=\:{z}\:.\:\:\:{option}\:\left({a}\right)\:. \\ $$$$ \\ $$

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