Prove-that-z-1-z-2-z-3-z-n-z-1-z-2-z-3-z-n-

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

Prove that ∣ z_{1} + z_{2} + z_{3} + \dots . + z_{n} ∣ ⩽

∣ z_{1} ∣ + ∣ z_{2} ∣ + ∣ z_{3} ∣ + \dots . + ∣ z_{n} ∣

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}, \dots . r_{n} ? Why all

these are not multiplied with each

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} ∣= \sqrt{{(a_{1} + a_{2})}^{2} + {(b_{1} + b_{2})}^{2}}

∣ z_{1} ∣= \sqrt{a_{1}^{2} + b_{1}^{2}}

∣ z_{2} ∣= \sqrt{a_{2}^{2} + b_{2}^{2}}

∣ z_{1} ∣ + ∣ z_{2} ∣= \sqrt{a_{1}^{2} + b_{1}^{2}} + \sqrt{a_{2}^{2} + b_{2}^{2}}

since \sqrt{{(a_{1} + a_{2})}^{2} + {(b_{1} + b_{2})}^{2}} ⩽ \sqrt{a_{1}^{2} + b_{1}^{2}} + \sqrt{a_{2}^{2} + b_{2}^{2}} (see below)

\Rightarrow∣ z_{1} + z_{2} ∣⩽∣ z_{1} ∣ + ∣ z_{2} ∣

∣ z_{1} + z_{2} + z_{3} + \dots . + z_{n} ∣

⩽∣ z_{1} ∣ + ∣ z_{2} + z_{3} + \dots . + z_{n} ∣

⩽∣ z_{1} ∣ + ∣ z_{2} ∣ + ∣ z_{3} + \dots . + z_{n} ∣

\dots \dots

⩽∣ z_{1} ∣ + ∣ z_{2} ∣ + ∣ z_{3} ∣ + \dots . + ∣ z_{n} ∣

Commented by mrW1 last updated on 11/Aug/17

Commented by mrW1 last updated on 11/Aug/17

AB = \sqrt{a_{1}^{2} + b_{1}^{2}}

BC = \sqrt{a_{2}^{2} + b_{2}^{2}}

AC = \sqrt{{(a_{1} + a_{2})}^{2} + {(b_{1} + b_{2})}^{2}}

since AC ⩽ AB + BC

\sqrt{{(a_{1} + a_{2})}^{2} + {(b_{1} + b_{2})}^{2}} ⩽ \sqrt{a_{1}^{2} + b_{1}^{2}} + \sqrt{a_{2}^{2} + b_{2}^{2}}

Commented by Tinkutara last updated on 12/Aug/17

Thank you very much Sir!