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Question Number 127400 by mohammad17 last updated on 29/Dec/20

prove that :∣∣z_1 ∣−∣z_2 ∣∣≤∣z_1 +z_2 ∣

$${prove}\:{that}\::\mid\mid{z}_{\mathrm{1}} \mid−\mid{z}_{\mathrm{2}} \mid\mid\leqslant\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid \\ $$

Commented by mohammad17 last updated on 29/Dec/20

?

$$? \\ $$

Commented by liberty last updated on 29/Dec/20

observasi the inequality ∣ z_1 +z_2 ∣ ≤ ∣ z_1 ∣ + ∣z_2  ∣  is known as triangle inequality  .Now from  the identity z_1  = z_1 +z_2 +(−z_2 )  gives ∣z_1 ∣ = ∣z_1 +z_2 +(−z_2 )∣ ≤ ∣z_1 +z_2 ∣+∣−z_2 ∣  since ∣z_2 ∣=∣−z_2 ∣ then ∣z_1 ∣−∣z_2 ∣ ≤ ∣z_1 +z_2 ∣...(i)  but because z_1 +z_2 =z_2 +z_1  , (i) can be written   in the alternative form ∣z_1 +z_2 ∣=∣z_2 +z_1 ∣ ≥ ∣z_2 ∣−∣z_1 ∣ =−(∣z_1 ∣−∣z_2 ∣)  or ∣z_2 ∣−∣z_1 ∣ = ∣∣z_1 ∣−∣z_2 ∣∣ ...(ii)  replacing z_2  by −z_2  we get ∣z_1 −z_2 ∣ ≥ ∣ ∣z_1 ∣−∣−z_2 ∣ ∣  or ∣z_1 −z_2 ∣ ≥ ∣ ∣z_1 ∣−∣z_2 ∣ ∣

$${observasi}\:{the}\:{inequality}\:\mid\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid\:\leqslant\:\mid\:{z}_{\mathrm{1}} \mid\:+\:\mid{z}_{\mathrm{2}} \:\mid \\ $$$${is}\:{known}\:{as}\:{triangle}\:{inequality}\:\:.{Now}\:{from} \\ $$$${the}\:{identity}\:{z}_{\mathrm{1}} \:=\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} +\left(−{z}_{\mathrm{2}} \right) \\ $$$${gives}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} +\left(−{z}_{\mathrm{2}} \right)\mid\:\leqslant\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid+\mid−{z}_{\mathrm{2}} \mid \\ $$$${since}\:\mid{z}_{\mathrm{2}} \mid=\mid−{z}_{\mathrm{2}} \mid\:{then}\:\mid{z}_{\mathrm{1}} \mid−\mid{z}_{\mathrm{2}} \mid\:\leqslant\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid...\left({i}\right) \\ $$$${but}\:{because}\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} ={z}_{\mathrm{2}} +{z}_{\mathrm{1}} \:,\:\left({i}\right)\:{can}\:{be}\:{written}\: \\ $$$${in}\:{the}\:{alternative}\:{form}\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid=\mid{z}_{\mathrm{2}} +{z}_{\mathrm{1}} \mid\:\geqslant\:\mid{z}_{\mathrm{2}} \mid−\mid{z}_{\mathrm{1}} \mid\:=−\left(\mid{z}_{\mathrm{1}} \mid−\mid{z}_{\mathrm{2}} \mid\right) \\ $$$${or}\:\mid{z}_{\mathrm{2}} \mid−\mid{z}_{\mathrm{1}} \mid\:=\:\mid\mid{z}_{\mathrm{1}} \mid−\mid{z}_{\mathrm{2}} \mid\mid\:...\left({ii}\right) \\ $$$${replacing}\:{z}_{\mathrm{2}} \:{by}\:−{z}_{\mathrm{2}} \:{we}\:{get}\:\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid\:\geqslant\:\mid\:\mid{z}_{\mathrm{1}} \mid−\mid−{z}_{\mathrm{2}} \mid\:\mid \\ $$$${or}\:\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid\:\geqslant\:\mid\:\mid{z}_{\mathrm{1}} \mid−\mid{z}_{\mathrm{2}} \mid\:\mid \\ $$

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