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Question Number 38044 by solihin last updated on 21/Jun/18

Commented by math khazana by abdo last updated on 22/Jun/18

zz^− =5 and (z/z^− )=−1 +((12)/5)i ⇒∣z^ ∣^2 =5 ⇒∣z∣=(√5) (z/z^− ) =(z^2 /(zz^− )) =(z^2 /5) =−1+((12)/5)i ⇒z^2 = −5 +12i ∣−5 +12i∣=(√(25 +144))=(√(169))=13 ≠∣z∣^2 =5 so there is a error in the exercise...

$${z}\overset{−} {{z}}=\mathrm{5}\:{and}\:\frac{{z}}{\overset{−} {{z}}}=−\mathrm{1}\:+\frac{\mathrm{12}}{\mathrm{5}}{i}\:\Rightarrow\mid{z}^{} \mid^{\mathrm{2}} =\mathrm{5}\:\Rightarrow\mid{z}\mid=\sqrt{\mathrm{5}} \\ $$$$\frac{{z}}{\overset{−} {{z}}}\:=\frac{{z}^{\mathrm{2}} }{{z}\overset{−} {{z}}}\:=\frac{{z}^{\mathrm{2}} }{\mathrm{5}}\:=−\mathrm{1}+\frac{\mathrm{12}}{\mathrm{5}}{i}\:\Rightarrow{z}^{\mathrm{2}} =\:−\mathrm{5}\:+\mathrm{12}{i} \\ $$$$\mid−\mathrm{5}\:+\mathrm{12}{i}\mid=\sqrt{\mathrm{25}\:+\mathrm{144}}=\sqrt{\mathrm{169}}=\mathrm{13}\:\neq\mid{z}\mid^{\mathrm{2}} =\mathrm{5}\:{so} \\ $$$${there}\:{is}\:{a}\:{error}\:{in}\:{the}\:{exercise}... \\ $$

Answered by ajfour last updated on 21/Jun/18

r^2 =zz^� =5 ∣(z/z^� )∣=∣−1+i((12)/5)∣ ≠ 1 ?

$${r}^{\mathrm{2}} ={z}\bar {{z}}\:=\mathrm{5} \\ $$$$\mid\frac{{z}}{\bar {{z}}}\mid=\mid−\mathrm{1}+{i}\frac{\mathrm{12}}{\mathrm{5}}\mid\:\:\neq\:\mathrm{1}\:\:? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jun/18

z=x+iy z^− =x−iy zz^− =(x+iy)(x−iy=x^2 +y^2 =5 ((x+iy)/(x−iy))=−1+((12)/5)i (((x+iy)^2 )/(x^2 +y^2 ))=−1+((12)/5)i ((x^2 −y^2 +i2xy)/5)=((−5+12i)/5) x^2 −y^2 =−5 2xy=12 x^2 +y^2 =5 x^2 −y^2 =−5 2x^2 =0 x=0 y^2 =5 y=±(√5) so x=0 y=±(√5) but 2(o)((√5) )=0 but2xy not equal to12 sl pls pls check question... pls check the

$${z}={x}+{iy} \\ $$$$\overset{−} {{z}}={x}−{iy}\: \\ $$$${z}\overset{−} {{z}}=\left({x}+{iy}\right)\left({x}−{iy}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{5}\right. \\ $$$$\frac{{x}+{iy}}{{x}−{iy}}=−\mathrm{1}+\frac{\mathrm{12}}{\mathrm{5}}{i} \\ $$$$\frac{\left({x}+{iy}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=−\mathrm{1}+\frac{\mathrm{12}}{\mathrm{5}}{i} \\ $$$$\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +{i}\mathrm{2}{xy}}{\mathrm{5}}=\frac{−\mathrm{5}+\mathrm{12}{i}}{\mathrm{5}} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =−\mathrm{5} \\ $$$$\mathrm{2}{xy}=\mathrm{12} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{5} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =−\mathrm{5} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} =\mathrm{0}\:\:{x}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} =\mathrm{5}\:\:\:{y}=\pm\sqrt{\mathrm{5}}\: \\ $$$$ \\ $$$${so}\:{x}=\mathrm{0}\:\:\:{y}=\pm\sqrt{\mathrm{5}} \\ $$$${but}\:\mathrm{2}\left({o}\right)\left(\sqrt{\mathrm{5}}\:\right)=\mathrm{0}\:\:\:{but}\mathrm{2}{xy}\:{not}\:\:{equal}\:{to}\mathrm{12} \\ $$$${sl}\:{pls}\:{pls}\:{check}\:{question}... \\ $$$${pls}\:{check}\:{the} \\ $$

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