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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 ∣
provethat:∣∣z1∣−∣z2∣∣⩽∣z1+z2∣
Commented by mohammad17 last updated on 29/Dec/20
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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 ∣ ∣
observasitheinequality∣z1+z2∣⩽∣z1∣+∣z2∣isknownastriangleinequality.Nowfromtheidentityz1=z1+z2+(−z2)gives∣z1∣=∣z1+z2+(−z2)∣⩽∣z1+z2∣+∣−z2∣since∣z2∣=∣−z2∣then∣z1∣−∣z2∣⩽∣z1+z2∣…(i)butbecausez1+z2=z2+z1,(i)canbewritteninthealternativeform∣z1+z2∣=∣z2+z1∣⩾∣z2∣−∣z1∣=−(∣z1∣−∣z2∣)or∣z2∣−∣z1∣=∣∣z1∣−∣z2∣∣…(ii)replacingz2by−z2weget∣z1−z2∣⩾∣∣z1∣−∣−z2∣∣or∣z1−z2∣⩾∣∣z1∣−∣z2∣∣

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