MedVision ad

triangle inequality (complex number) (1 Viewer)

lugana

New Member
Joined
Nov 22, 2011
Messages
4
Gender
Female
HSC
2012
Hi guys,

tell me how to solve these two questions please? :)

1.
Show that ||z1|+|z2||<or=|z1+z2|. State the condition for equality to hold.


2.
Show that |z1+z2+...+zn|<or=|z1|+|z2|+...+|zn|



Seems I'm in stuck with those types of questions...
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Hi guys,

tell me how to solve these two questions please? :)

1.
Show that ||z1|+|z2||<or=|z1+z2|. State the condition for equality to hold.


2.
Show that |z1+z2+...+zn|<or=|z1|+|z2|+...+|zn|



Seems I'm in stuck with those types of questions...
There is nothing to solve there. You just gave us a whole bunch of letters and numbers.
 

Nooblet94

Premium Member
Joined
Feb 5, 2011
Messages
1,044
Gender
Male
HSC
2012
Q1.


Q2.

Judging from the thread title I think that's what you mean?


Q1 - Consider the triangle formed by the origin, Z1 and Z1+Z2. We know that in any triangle the sum of any two side lengths is greater than the length of the third side. Hence, OB is less than (or equal to) OA+AB, from which you obtain the required inequality. The inequality holds when Z1=Z2
Untitled.png

Q2 - By definition, a straight line is the path with the shortest distance between two points (in this case, the modulus of the sum of numbers), and therefore any other path (the sum of the moduli) will be longer. Hence, the inequality.
 
Last edited:

lolcakes52

Member
Joined
Oct 31, 2011
Messages
286
Gender
Undisclosed
HSC
2012
Q1.


Q2.

Judging from the thread title I think that's what you mean?


Q1 - Consider the triangle formed by the origin, Z1 and Z1+Z2. We know that in any triangle the sum of any two side lengths is greater than the length of the third side. Hence, OB is less than (or equal to) OA+AB, from which you obtain the required inequality. The inequality holds when Z1=Z2
View attachment 24125

Q2 - By definition, a straight line is the path with the shortest distance between two points (in this case, the modulus of the sum of numbers), and therefore any other path (the sum of the moduli) will be longer. Hence, the inequality.
Didn't you just prove the triangle inequality using the triangle inequality?
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Didn't you just prove the triangle inequality using the triangle inequality?
Not quite sure why, but I found this comment to be very funny.

But in all seriousness, that does seem to be the case. He utilised the fact that "We know that in any triangle the sum of any two side lengths is greater than the length of the third side", which is said triangle inequality.
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
And the equality conditions stated are incorrect.
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Equality occurs when the angle between Z_1 and Z_2 is equal to pi.

Geometrically, this means equality occurs in the degenerate case where the 'triangle' is a line.
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Equality occurs when the angle between Z_1 and Z_2 is equal to pi.

Geometrically, this means equality occurs in the degenerate case where the 'triangle' is a line.
You mean an angle of 0 right?

 

largarithmic

Member
Joined
Aug 9, 2011
Messages
202
Gender
Male
HSC
2011
I was pertaining to the geometric definition of the three vectors as the sides of a triangle Z1, Z2, and Z3.

No, that isnt a situation that corresponds with the actual question. The reason is in giving the sides vectors you havent given them directions. For the angles between Z1 and Z2 to actually be alpha, you need Z3 = Z1-Z2 or Z3 = Z2-Z1. If you want Z3=Z1+Z2 (so that it actually corresponds to the question), you need Z1 and Z2 sorta going 'in the same direction' if you think about it, in which case Z1 and Z2 are gonna have the same argument and thus angle between them 0.
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
No, that isnt a situation that corresponds with the actual question. The reason is in giving the sides vectors you havent given them directions. For the angles between Z1 and Z2 to actually be alpha, you need Z3 = Z1-Z2 or Z3 = Z2-Z1. If you want Z3=Z1+Z2 (so that it actually corresponds to the question), you need Z1 and Z2 sorta going 'in the same direction' if you think about it, in which case Z1 and Z2 are gonna have the same argument and thus angle between them 0.
Yes, if we consider Z_1 + Z_2 = Z3 (thus giving them directions), then equality occurs when Arg (Z_1) - Arg (Z_2) = 0 ---> Arg (Z_1) = Arg (Z_2) , as you said.

You are correct in saying that my diagram did not correspond to the original question directly.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top