Question-193414

Question Number 193414 by mokys last updated on 13/Jun/23

Answered by AST last updated on 13/Jun/23

Z and Z_2 in the closed unit disk⇒∣Z∣,∣Z_2 ∣≤1 ∣Z_1 −Z_2 ∣≥1⇒(Z_1 −Z_2 )(Z_1 ^− −Z_2 ^− )≥1 ⇒∣Z_1 ∣^2 +∣Z_2 ∣^2 ≥1+(ZZ_2 ^− +Z_2 Z_1 ^− ) (since Z_i Z_i ^− =∣Z_i ∣^2 ) ⇒∣Z_1 ∣^2 +∣Z_2 ∣^2 +ZZ_2 ^− +Z_2 Z^− ≤2(∣Z∣^2 +∣Z_2 ∣^2 )−1≤2(2)−1=3 ⇒∣Z_1 +Z_2 ∣^2 ≤3⇒∣Z_1 +Z_2 ∣≤(√3)

$${Z}\:{and}\:{Z}_{\mathrm{2}} \:{in}\:{the}\:{closed}\:{unit}\:{disk}\Rightarrow\mid{Z}\mid,\mid{Z}_{\mathrm{2}} \mid\leqslant\mathrm{1} \\ $$$$\mid{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \mid\geqslant\mathrm{1}\Rightarrow\left({Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \right)\left(\overset{−} {{Z}}_{\mathrm{1}} −\overset{−} {{Z}}_{\mathrm{2}} \right)\geqslant\mathrm{1} \\ $$$$\Rightarrow\mid{Z}_{\mathrm{1}} \mid^{\mathrm{2}} +\mid{Z}_{\mathrm{2}} \mid^{\mathrm{2}} \geqslant\mathrm{1}+\left({Z}\overset{−} {{Z}}_{\mathrm{2}} +{Z}_{\mathrm{2}} \overset{−} {{Z}}_{\mathrm{1}} \right)\:\left({since}\:{Z}_{{i}} \overset{−} {{Z}}_{{i}} =\mid{Z}_{{i}} \mid^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mid{Z}_{\mathrm{1}} \mid^{\mathrm{2}} +\mid{Z}_{\mathrm{2}} \mid^{\mathrm{2}} +{Z}\overset{−} {{Z}}_{\mathrm{2}} +{Z}_{\mathrm{2}} \overset{−} {{Z}}\leqslant\mathrm{2}\left(\mid{Z}\mid^{\mathrm{2}} +\mid{Z}_{\mathrm{2}} \mid^{\mathrm{2}} \right)−\mathrm{1}\leqslant\mathrm{2}\left(\mathrm{2}\right)−\mathrm{1}=\mathrm{3} \\ $$$$\Rightarrow\mid{Z}_{\mathrm{1}} +{Z}_{\mathrm{2}} \mid^{\mathrm{2}} \leqslant\mathrm{3}\Rightarrow\mid{Z}_{\mathrm{1}} +{Z}_{\mathrm{2}} \mid\leqslant\sqrt{\mathrm{3}} \\ $$

Answered by witcher3 last updated on 13/Jun/23

∣z_1 ∣,∣z_2 ∣≤1 z_1 =pe^(ia) z_2 =p′e^(ib) ∣z_1 −z_2 ∣=∣(pcos(a)−p′cos(b)+i(psin(a)−p′sin(b)∣≥1 ⇒p^2 +p′^2 −2pp′cos(a−b)≥1 2pp′cos(a−b)≤p^2 +p′^2 −1 ∣z_1 +z_2 ∣^2 =p^2 +p′^2 +2pp′cos(a−b) ≤2p^2 +2p′^2 −1 p=∣z_1 ∣<1,p′=∣z_2 ∣<1⇒ ∣z_1 +z_2 ∣^2 ≤2+2−1=3⇒∣z_1 +z_2 ∣≤(√3)

$$\mid\mathrm{z}_{\mathrm{1}} \mid,\mid\mathrm{z}_{\mathrm{2}} \mid\leqslant\mathrm{1} \\ $$$$\mathrm{z}_{\mathrm{1}} =\mathrm{pe}^{\mathrm{ia}} \\ $$$$\mathrm{z}_{\mathrm{2}} =\mathrm{p}'\mathrm{e}^{\mathrm{ib}} \\ $$$$\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \mid=\mid\left(\mathrm{pcos}\left(\mathrm{a}\right)−\mathrm{p}'\mathrm{cos}\left(\mathrm{b}\right)+\mathrm{i}\left(\mathrm{psin}\left(\mathrm{a}\right)−\boldsymbol{\mathrm{p}}'\mathrm{sin}\left(\mathrm{b}\right)\mid\geqslant\mathrm{1}\right.\right. \\ $$$$\Rightarrow\mathrm{p}^{\mathrm{2}} +\mathrm{p}'^{\mathrm{2}} −\mathrm{2pp}'\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)\geqslant\mathrm{1} \\ $$$$\mathrm{2pp}'\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)\leqslant\mathrm{p}^{\mathrm{2}} +\mathrm{p}'^{\mathrm{2}} −\mathrm{1} \\ $$$$\mid\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} =\mathrm{p}^{\mathrm{2}} +\mathrm{p}'^{\mathrm{2}} +\mathrm{2pp}'\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right) \\ $$$$\leqslant\mathrm{2p}^{\mathrm{2}} +\mathrm{2p}'^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{p}=\mid\mathrm{z}_{\mathrm{1}} \mid<\mathrm{1},\mathrm{p}'=\mid\mathrm{z}_{\mathrm{2}} \mid<\mathrm{1}\Rightarrow \\ $$$$\mid\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} \leqslant\mathrm{2}+\mathrm{2}−\mathrm{1}=\mathrm{3}\Rightarrow\mid\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} \mid\leqslant\sqrt{\mathrm{3}} \\ $$

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