Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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) ✓

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com