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Question Number 9836 by tawakalitu last updated on 06/Jan/17
Prove that.  ∣z_1  + z_2 ∣ ≤ ∣z_1 ∣ ∣z_2 ∣
$$\mathrm{Prove}\:\mathrm{that}. \\ $$$$\mid\mathrm{z}_{\mathrm{1}} \:+\:\mathrm{z}_{\mathrm{2}} \mid\:\leqslant\:\mid\mathrm{z}_{\mathrm{1}} \mid\:\mid\mathrm{z}_{\mathrm{2}} \mid \\ $$
Commented by FilupSmith last updated on 07/Jan/17
∣z_1 ∣∣z_2 ∣≤∣z_1 z_2 ∣      ∣z_1 +z_2 ∣≤∣z_1 z_2 ∣  z_2 =z_1 +n  ∣z_1 +z_2 ∣=∣2z_1 +n∣≤∣z_1 z_2 ∣     ∣2z_1 +n∣≤∣z_1 z_2 ∣  working  thinking
$$\mid{z}_{\mathrm{1}} \mid\mid{z}_{\mathrm{2}} \mid\leqslant\mid{z}_{\mathrm{1}} {z}_{\mathrm{2}} \mid\: \\ $$$$\: \\ $$$$\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid\leqslant\mid{z}_{\mathrm{1}} {z}_{\mathrm{2}} \mid \\ $$$${z}_{\mathrm{2}} ={z}_{\mathrm{1}} +{n} \\ $$$$\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid=\mid\mathrm{2}{z}_{\mathrm{1}} +{n}\mid\leqslant\mid{z}_{\mathrm{1}} {z}_{\mathrm{2}} \mid \\ $$$$\: \\ $$$$\mid\mathrm{2}{z}_{\mathrm{1}} +{n}\mid\leqslant\mid{z}_{\mathrm{1}} {z}_{\mathrm{2}} \mid \\ $$$${working} \\ $$$${thinking} \\ $$
Commented by prakash jain last updated on 08/Jan/17
z_1 =1  z_2 =0  ∣z_1 +z_2 ∣=∣1+0∣=1  ∣z_1 ∣∣z_2 ∣=0  ∣z_1  + z_2 ∣  ≰  ∣z_1 ∣ ∣z_2 ∣
$$\mathrm{z}_{\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{z}_{\mathrm{2}} =\mathrm{0} \\ $$$$\mid\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} \mid=\mid\mathrm{1}+\mathrm{0}\mid=\mathrm{1} \\ $$$$\mid\mathrm{z}_{\mathrm{1}} \mid\mid\mathrm{z}_{\mathrm{2}} \mid=\mathrm{0} \\ $$$$\mid\mathrm{z}_{\mathrm{1}} \:+\:\mathrm{z}_{\mathrm{2}} \mid\:\:\nleqslant\:\:\mid\mathrm{z}_{\mathrm{1}} \mid\:\mid\mathrm{z}_{\mathrm{2}} \mid \\ $$
Commented by prakash jain last updated on 08/Jan/17
The question should be  ∣z_1 +z_2 ∣≤∣z_1 ∣+∣z_2 ∣
$$\mathrm{The}\:\mathrm{question}\:\mathrm{should}\:\mathrm{be} \\ $$$$\mid\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} \mid\leqslant\mid\mathrm{z}_{\mathrm{1}} \mid+\mid\mathrm{z}_{\mathrm{2}} \mid \\ $$

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