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Question Number 33096 by 7991 last updated on 10/Apr/18

z and w ∈ C proof ∣∣z∣−∣w∣∣ ≤ ∣z−w∣ and ∣z∣−∣w∣≤ ∣z+w∣

$${z}\:{and}\:{w}\:\in\:\mathbb{C} \\ $$$${proof}\:\mid\mid{z}\mid−\mid{w}\mid\mid\:\leqslant\:\mid{z}−{w}\mid\:{and}\:\mid{z}\mid−\mid{w}\mid\leqslant\:\mid{z}+{w}\mid \\ $$

Answered by MJS last updated on 10/Apr/18

z=x+yi w=u+vi 1. ∣(√(x^2 +y^2 ))−(√(u^2 +v^2 ))∣≤(√((x−u)^2 +(y−v)^2 )) ((√(x^2 +y^2 ))−(√(u^2 +v^2 )))^2 ≤(x−u)^2 +(y−v)^2 (x^2 +y^2 )−2(√(x^2 +y^2 ))(√(u^2 +v^2 ))+(u^2 +v^2 )≤x^2 −2xu+u^2 +y^2 −2yv+v^2 −2(√(x^2 +y^2 ))(√(u^2 +v^2 ))≤−2(xu+yv) (√(x^2 +y^2 ))(√(u^2 +v^2 ))≥xu+yv if (xu+yv)≤0 this is true if (xu+yv)>0: (x^2 +y^2 )(u^2 +v^2 )≥(xu+yv)^2 x^2 u^2 +x^2 v^2 +y^2 u^2 +y^2 v^2 ≥x^2 u^2 +2xuyv+y^2 v^2 x^2 v^2 −2xuyv+y^2 u^2 ≥0 (xv−yu)^2 ≥0 true 2. if ∣z∣−∣w∣≤0 it′s already done if ∣z∣−∣w∣>0 go on like above

$${z}={x}+{y}\mathrm{i} \\ $$$${w}={u}+{v}\mathrm{i} \\ $$$$ \\ $$$$\mathrm{1}. \\ $$$$\mid\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }−\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }\mid\leqslant\sqrt{\left({x}−{u}\right)^{\mathrm{2}} +\left({y}−{v}\right)^{\mathrm{2}} } \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }−\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }\right)^{\mathrm{2}} \leqslant\left({x}−{u}\right)^{\mathrm{2}} +\left({y}−{v}\right)^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }+\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} \right)\leqslant{x}^{\mathrm{2}} −\mathrm{2}{xu}+{u}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{yv}+{v}^{\mathrm{2}} \\ $$$$−\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }\leqslant−\mathrm{2}\left({xu}+{yv}\right) \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} }\geqslant{xu}+{yv} \\ $$$$\mathrm{if}\:\left({xu}+{yv}\right)\leqslant\mathrm{0}\:\mathrm{this}\:\mathrm{is}\:\mathrm{true} \\ $$$$\mathrm{if}\:\left({xu}+{yv}\right)>\mathrm{0}: \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} \right)\geqslant\left({xu}+{yv}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} {u}^{\mathrm{2}} +{x}^{\mathrm{2}} {v}^{\mathrm{2}} +{y}^{\mathrm{2}} {u}^{\mathrm{2}} +{y}^{\mathrm{2}} {v}^{\mathrm{2}} \geqslant{x}^{\mathrm{2}} {u}^{\mathrm{2}} +\mathrm{2}{xuyv}+{y}^{\mathrm{2}} {v}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} {v}^{\mathrm{2}} −\mathrm{2}{xuyv}+{y}^{\mathrm{2}} {u}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\left({xv}−{yu}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\mathrm{true} \\ $$$$ \\ $$$$\mathrm{2}. \\ $$$$\mathrm{if}\:\mid{z}\mid−\mid{w}\mid\leqslant\mathrm{0}\:\mathrm{it}'\mathrm{s}\:\mathrm{already}\:\mathrm{done} \\ $$$$\mathrm{if}\:\mid{z}\mid−\mid{w}\mid>\mathrm{0}\:\mathrm{go}\:\mathrm{on}\:\mathrm{like}\:\mathrm{above} \\ $$

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