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Question Number 19349 by Tinkutara last updated on 10/Aug/17

Prove that ∣z_1 + z_2 + z_3 + .... + z_n ∣ ≤ ∣z_1 ∣ + ∣z_2 ∣ + ∣z_3 ∣ + .... + ∣z_n ∣

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \:+\:....\:+\:{z}_{{n}} \mid\:\leqslant \\ $$$$\mid{z}_{\mathrm{1}} \mid\:+\:\mid{z}_{\mathrm{2}} \mid\:+\:\mid{z}_{\mathrm{3}} \mid\:+\:....\:+\:\mid{z}_{{n}} \mid \\ $$

Commented by Tinkutara last updated on 10/Aug/17

I have this proof in my book but I have a doubt in that proof. Can anyone explain this? Why only r_1 is multiplied with all other r_2 , r_3 , .... r_n ? Why all these are not multiplied with each other?

$$\mathrm{I}\:\mathrm{have}\:\mathrm{this}\:\mathrm{proof}\:\mathrm{in}\:\mathrm{my}\:\mathrm{book}\:\mathrm{but}\:\mathrm{I}\:\mathrm{have} \\ $$$$\mathrm{a}\:\mathrm{doubt}\:\mathrm{in}\:\mathrm{that}\:\mathrm{proof}.\:\mathrm{Can}\:\mathrm{anyone} \\ $$$$\mathrm{explain}\:\mathrm{this}?\:\mathrm{Why}\:\mathrm{only}\:{r}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{multiplied} \\ $$$$\mathrm{with}\:\mathrm{all}\:\mathrm{other}\:{r}_{\mathrm{2}} ,\:{r}_{\mathrm{3}} ,\:....\:{r}_{{n}} ?\:\mathrm{Why}\:\mathrm{all} \\ $$$$\mathrm{these}\:\mathrm{are}\:\mathrm{not}\:\mathrm{multiplied}\:\mathrm{with}\:\mathrm{each} \\ $$$$\mathrm{other}? \\ $$

Commented by Tinkutara last updated on 10/Aug/17

Answered by mrW1 last updated on 11/Aug/17

z_1 =a_1 +b_1 i z_2 =a_2 +b_2 i ∣z_1 +z_2 ∣=(√((a_1 +a_2 )^2 +(b_1 +b_2 )^2 )) ∣z_1 ∣=(√(a_1 ^2 +b_1 ^2 )) ∣z_2 ∣=(√(a_2 ^2 +b_2 ^2 )) ∣z_1 ∣+∣z_2 ∣=(√(a_1 ^2 +b_1 ^2 ))+(√(a_2 ^2 +b_2 ^2 )) since (√((a_1 +a_2 )^2 +(b_1 +b_2 )^2 ))≤(√(a_1 ^2 +b_1 ^2 ))+(√(a_2 ^2 +b_2 ^2 )) (see below) ⇒∣z_1 +z_2 ∣≤∣z_1 ∣+∣z_2 ∣ ∣z_1 + z_2 + z_3 + .... + z_n ∣ ≤∣z_1 ∣ +∣z_2 + z_3 + .... + z_n ∣ ≤∣z_1 ∣ +∣z_2 ∣ + ∣z_3 + .... + z_n ∣ ...... ≤∣z_1 ∣ +∣z_2 ∣ + ∣z_3 ∣ + .... + ∣z_n ∣

$$\mathrm{z}_{\mathrm{1}} =\mathrm{a}_{\mathrm{1}} +\mathrm{b}_{\mathrm{1}} \mathrm{i} \\ $$$$\mathrm{z}_{\mathrm{2}} =\mathrm{a}_{\mathrm{2}} +\mathrm{b}_{\mathrm{2}} \mathrm{i} \\ $$$$\mid\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} \mid=\sqrt{\left(\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{b}_{\mathrm{1}} +\mathrm{b}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\mid\mathrm{z}_{\mathrm{1}} \mid=\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$$\mid\mathrm{z}_{\mathrm{2}} \mid=\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} } \\ $$$$\mid\mathrm{z}_{\mathrm{1}} \mid+\mid\mathrm{z}_{\mathrm{2}} \mid=\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} }+\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} } \\ $$$$\mathrm{since}\:\sqrt{\left(\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{b}_{\mathrm{1}} +\mathrm{b}_{\mathrm{2}} \right)^{\mathrm{2}} }\leqslant\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} }+\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} }\:\left(\mathrm{see}\:\mathrm{below}\right) \\ $$$$\Rightarrow\mid\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} \mid\leqslant\mid\mathrm{z}_{\mathrm{1}} \mid+\mid\mathrm{z}_{\mathrm{2}} \mid \\ $$$$ \\ $$$$\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \:+\:....\:+\:{z}_{{n}} \mid\: \\ $$$$\leqslant\mid{z}_{\mathrm{1}} \mid\:+\mid{z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} \:+\:....\:+\:{z}_{{n}} \mid\: \\ $$$$\leqslant\mid{z}_{\mathrm{1}} \mid\:+\mid{z}_{\mathrm{2}} \mid\:+\:\mid{z}_{\mathrm{3}} \:+\:....\:+\:{z}_{{n}} \mid\: \\ $$$$...... \\ $$$$\leqslant\mid{z}_{\mathrm{1}} \mid\:+\mid{z}_{\mathrm{2}} \mid\:+\:\mid{z}_{\mathrm{3}} \mid\:+\:....\:+\:\mid{z}_{{n}} \mid\: \\ $$

Commented by mrW1 last updated on 11/Aug/17

Commented by mrW1 last updated on 11/Aug/17

AB=(√(a_1 ^2 +b_1 ^2 )) BC=(√(a_2 ^2 +b_2 ^2 )) AC=(√((a_1 +a_2 )^2 +(b_1 +b_2 )^2 )) since AC≤AB+BC (√((a_1 +a_2 )^2 +(b_1 +b_2 )^2 ))≤(√(a_1 ^2 +b_1 ^2 ))+(√(a_2 ^2 +b_2 ^2 ))

$$\mathrm{AB}=\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$$\mathrm{BC}=\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} } \\ $$$$\mathrm{AC}=\sqrt{\left(\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{b}_{\mathrm{1}} +\mathrm{b}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{since}\:\mathrm{AC}\leqslant\mathrm{AB}+\mathrm{BC} \\ $$$$\sqrt{\left(\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{b}_{\mathrm{1}} +\mathrm{b}_{\mathrm{2}} \right)^{\mathrm{2}} }\leqslant\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} }+\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} } \\ $$

Commented by Tinkutara last updated on 12/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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