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Question Number 192126 by universe last updated on 08/May/23

    prove that        ∣z∣ > ((∣Re(z)∣ +∣Im(z)∣)/2)  ,   ∀z∈C

$$\:\:\:\:\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$ $$\:\:\:\:\:\:\mid\boldsymbol{{z}}\mid\:>\:\frac{\mid\boldsymbol{{Re}}\left(\boldsymbol{{z}}\right)\mid\:+\mid\boldsymbol{{Im}}\left(\boldsymbol{{z}}\right)\mid}{\mathrm{2}}\:\:,\:\:\:\forall\boldsymbol{{z}}\in\mathbb{C} \\ $$

Commented byYork12 last updated on 09/May/23

sir how can I reach you out , I need to ask several questions

$${sir}\:{how}\:{can}\:{I}\:{reach}\:{you}\:{out}\:,\:{I}\:{need}\:{to}\:{ask}\:{several}\:{questions} \\ $$

Answered by AST last updated on 08/May/23

Let z=x+yi;Re(z)=((z+z^− )/2);Im(z)=((z−z^− )/(2i))=(((z−z^− )i)/(−2))  ⇒∣Re(z)∣=∣x∣;∣Im(z)∣=((∣z−z^− ∣)/2)=∣y∣;∣z∣=(√(x^2 +y^2 ))  Required to prove (√(x^2 +y^2 ))>((∣x∣+∣y∣)/2)  ≡x^2 +y^2 >((∣x∣^2 +∣y∣^2 +2∣x∣∣y∣)/4)≡3(x^2 +y^2 )>2∣x∣∣y∣  It is known that x^2 +y^2 ≥2(√(x^2 y^2 ))=2∣x∣∣y∣  Hence, 3(x^2 +y^2 )>2∣x∣∣y∣⇔∣z∣>((∣Re(z)∣+∣Im(z)∣)/2)

$${Let}\:{z}={x}+{yi};{Re}\left({z}\right)=\frac{{z}+\overset{−} {{z}}}{\mathrm{2}};{Im}\left({z}\right)=\frac{{z}−\overset{−} {{z}}}{\mathrm{2}{i}}=\frac{\left({z}−\overset{−} {{z}}\right){i}}{−\mathrm{2}} \\ $$ $$\Rightarrow\mid{Re}\left({z}\right)\mid=\mid{x}\mid;\mid{Im}\left({z}\right)\mid=\frac{\mid{z}−\overset{−} {{z}}\mid}{\mathrm{2}}=\mid{y}\mid;\mid{z}\mid=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$ $${Required}\:{to}\:{prove}\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }>\frac{\mid{x}\mid+\mid{y}\mid}{\mathrm{2}} \\ $$ $$\equiv{x}^{\mathrm{2}} +{y}^{\mathrm{2}} >\frac{\mid{x}\mid^{\mathrm{2}} +\mid{y}\mid^{\mathrm{2}} +\mathrm{2}\mid{x}\mid\mid{y}\mid}{\mathrm{4}}\equiv\mathrm{3}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)>\mathrm{2}\mid{x}\mid\mid{y}\mid \\ $$ $${It}\:{is}\:{known}\:{that}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \geqslant\mathrm{2}\sqrt{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }=\mathrm{2}\mid{x}\mid\mid{y}\mid \\ $$ $${Hence},\:\mathrm{3}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)>\mathrm{2}\mid{x}\mid\mid{y}\mid\Leftrightarrow\mid{z}\mid>\frac{\mid{Re}\left({z}\right)\mid+\mid{Im}\left({z}\right)\mid}{\mathrm{2}} \\ $$

Answered by mehdee42 last updated on 08/May/23

let : z=a+ib  ∥z∥=(√(a^2 +b^2 ))≥(√a^2 )=∣a∣   (i)  ∥z∥=(√(a^2 +b^2 ))≥(√b^2 )=∣b∣   (ii)  (i)+(ii)⇒2∥z∥>∣a∣+∣b∣ ⇒∥z∥>((∣a∣+∣b∣)/2) ✓

$${let}\::\:{z}={a}+{ib} \\ $$ $$\parallel{z}\parallel=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\geqslant\sqrt{{a}^{\mathrm{2}} }=\mid{a}\mid\:\:\:\left({i}\right) \\ $$ $$\parallel{z}\parallel=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\geqslant\sqrt{{b}^{\mathrm{2}} }=\mid{b}\mid\:\:\:\left({ii}\right) \\ $$ $$\left({i}\right)+\left({ii}\right)\Rightarrow\mathrm{2}\parallel{z}\parallel>\mid{a}\mid+\mid{b}\mid\:\Rightarrow\parallel{z}\parallel>\frac{\mid{a}\mid+\mid{b}\mid}{\mathrm{2}}\:\checkmark \\ $$ $$ \\ $$

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