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Question Number 33095 by 7991 last updated on 10/Apr/18

z=x+yi ∈ C z^− =x−y ∈ C proof ∣z∣^2 =∣z^2 ∣=zz^− , so z≠0 →(1/z)=(z^− /(∣z∣^2 ))

$${z}={x}+{yi}\:\in\:\mathbb{C} \\ $$$$\overset{−} {{z}}={x}−{y}\:\in\:\mathbb{C} \\ $$$${proof}\:\mid{z}\mid^{\mathrm{2}} =\mid{z}^{\mathrm{2}} \mid={z}\overset{−} {{z}},\:{so}\:{z}\neq\mathrm{0}\:\rightarrow\frac{\mathrm{1}}{{z}}=\frac{\overset{−} {{z}}}{\mid{z}\mid^{\mathrm{2}} } \\ $$

Answered by Rasheed.Sindhi last updated on 10/Apr/18

z=x+yi ∈ C , z^(−) =x−yi ∣z∣=∣x+yi∣=(√(x^2 +y^2 )) ∴ ∣z∣^2 =x^2 +y^2 .............A ∣z^2 ∣=∣(x+yi)^2 ∣=∣x^2 −y^2 +2xyi∣ =(√((x^2 −y^2 )^2 +(2xy)^2 )) =(√(x^4 +y^4 −2x^2 y^2 +4x^2 y^2 )) =(√(x^4 +y^4 +2x^2 y^2 )) =(√((x^2 +y^2 )^2 )) =x^2 +y^2 ...............B z.z^(−) =(x+yi)(x−yi)=(x)^2 −(yi)^2 =x^2 +y^2 .................C From A,B & C ∣z∣^2 =∣z^2 ∣=z.z^(−) z.z^(−) =∣z^2 ∣⇒z=((∣z^2 ∣)/z^− )⇒(1/z)=(z^− /(∣z^2 ∣))

$${z}={x}+{yi}\:\in\:\mathbb{C}\:,\:\overline {{z}}={x}−{yi} \\ $$$$\mid\mathrm{z}\mid=\mid{x}+{yi}\mid=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} } \\ $$$$\therefore\:\mid\mathrm{z}\mid^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} .............\mathrm{A} \\ $$$$\mid\mathrm{z}^{\mathrm{2}} \mid=\mid\left({x}+{yi}\right)^{\mathrm{2}} \mid=\mid\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} +\mathrm{2xyi}\mid \\ $$$$\:\:\:\:\:\:=\sqrt{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{2xy}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:=\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} −\mathrm{2x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} +\mathrm{4x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:=\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{y}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:=\sqrt{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} ...............\mathrm{B} \\ $$$$\mathrm{z}.\overline {\mathrm{z}}=\left({x}+{yi}\right)\left({x}−{yi}\right)=\left(\mathrm{x}\right)^{\mathrm{2}} −\left(\mathrm{yi}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} .................\mathrm{C} \\ $$$$\mathrm{From}\:\mathrm{A},\mathrm{B}\:\&\:\mathrm{C} \\ $$$$\:\:\:\:\:\mid\mathrm{z}\mid^{\mathrm{2}} =\mid\mathrm{z}^{\mathrm{2}} \mid=\mathrm{z}.\overline {\mathrm{z}} \\ $$$$\:\:\:\mathrm{z}.\overline {\mathrm{z}}=\mid\mathrm{z}^{\mathrm{2}} \mid\Rightarrow\mathrm{z}=\frac{\mid\mathrm{z}^{\mathrm{2}} \mid}{\overset{−} {\mathrm{z}}}\Rightarrow\frac{\mathrm{1}}{\mathrm{z}}=\frac{\overset{−} {\mathrm{z}}}{\mid\mathrm{z}^{\mathrm{2}} \mid} \\ $$

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