Menu Close

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  + .... + z_n ∣ ≤  ∣z_1 ∣ + ∣z_2 ∣ + ∣z_3 ∣ + .... + ∣z_n ∣
Provethatz1+z2+z3+.+znz1+z2+z3+.+zn
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?
IhavethisproofinmybookbutIhaveadoubtinthatproof.Cananyoneexplainthis?Whyonlyr1ismultipliedwithallotherr2,r3,.rn?Whyallthesearenotmultipliedwitheachother?
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 ∣
z1=a1+b1iz2=a2+b2iz1+z2∣=(a1+a2)2+(b1+b2)2z1∣=a12+b12z2∣=a22+b22z1+z2∣=a12+b12+a22+b22since(a1+a2)2+(b1+b2)2a12+b12+a22+b22(seebelow)⇒∣z1+z2∣⩽∣z1+z2z1+z2+z3+.+zn⩽∣z1+z2+z3+.+zn⩽∣z1+z2+z3+.+zn⩽∣z1+z2+z3+.+zn
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 ))
AB=a12+b12BC=a22+b22AC=(a1+a2)2+(b1+b2)2sinceACAB+BC(a1+a2)2+(b1+b2)2a12+b12+a22+b22
Commented by Tinkutara last updated on 12/Aug/17
Thank you very much Sir!
ThankyouverymuchSir!

Leave a Reply

Your email address will not be published. Required fields are marked *