Triangle Inequality. (1 Viewer)

[ProdigyInspired]

The question is:

Using the parallelogram method along with a diagram, why:

a. |z₁ + z₂| <= |z₁| + |z₂|

What conditions need to occur for |z₁ + z₂| = |z₁| + |z₂|?

Similarly
b. |z₁ - z₂| <= |z₁| - |z₂|


What conditions need to occur for |z₁ - z₂| = |z₁| - |z₂|?

[HR][/HR]

I've given a try for a.

Drew a parallelogram, with z1 and z2 as the sides. From the point where the vectors originate, I drew a diagonal, naming it z1+z2.

Now with triangle properties, it is known that one side cannot be larger than the sum of the other two.

Therefore, a triangle with the sides z1 and z2, and the length side (z1 + z2), would have |z1 + z2| being less than |z1| + |z2|.

For LHS = RHS, all the vectors have to be the same length, therefore z1 must be = z2.

 

[porcupinetree]

The question is:

Using the parallelogram method along with a diagram, why:

a. |z₁ + z₂| <= |z₁| + |z₂|

What conditions need to occur for |z₁ + z₂| = |z₁| + |z₂|?

Similarly
b. |z₁ - z₂| <= |z₁| - |z₂|


What conditions need to occur for |z₁ - z₂| = |z₁| - |z₂|?

[HR][/HR]

I've given a try for a.

Drew a parallelogram, with z1 and z2 as the sides. From the point where the vectors originate, I drew a diagonal, naming it z1+z2.

Now with triangle properties, it is known that one side cannot be larger than the sum of the other two.

Therefore, a triangle with the sides z1 and z2, and the length side z1 + z2, would have |z1 + z2| being less than |z1| + |z2|.

For LHS = RHS, all the vectors have to be the same length, therefore z1 must be = z2.

This is incorrect - in order for |z1 + z2| to equal |z1| + |z2|, the vectors must point in the same direction, i.e., z2 = k*z1, where k is a positive constant.

 

[InteGrand]

The question is:

Using the parallelogram method along with a diagram, why:

a. |z₁ + z₂| <= |z₁| + |z₂|

What conditions need to occur for |z₁ + z₂| = |z₁| + |z₂|?

Similarly
b. |z₁ - z₂| <= |z₁| - |z₂|


What conditions need to occur for |z₁ - z₂| = |z₁| - |z₂|?

[HR][/HR]

I've given a try for a.

Drew a parallelogram, with z1 and z2 as the sides. From the point where the vectors originate, I drew a diagonal, naming it z1+z2.

Now with triangle properties, it is known that one side cannot be larger than the sum of the other two.

Therefore, a triangle with the sides z1 and z2, and the length side z1 + z2, would have |z1 + z2| being less than |z1| + |z2|.

For LHS = RHS, all the vectors have to be the same length, therefore z1 must be = z2.

For a), the equality occurs if and only if at least one of the complex numbers is 0 or both complex numbers have the same orientation in their before representation (i.e. equal argument).

Edit: Mentioned above (though the k above doesn't need to be strictly positive, it can also be 0, as in one of the complex numbers can be 0 and the equality then holds).

 

[ProdigyInspired]

This is incorrect - in order for |z1 + z2| to equal |z1| + |z2|, the vectors must point in the same direction, i.e., z2 = k*z1, where k is a positive constant.

For a), the equality occurs if and only if at least one of the complex numbers is 0 or both complex numbers have the same orientation in their before representation (i.e. equal argument).

Edit: Mentioned above (though the k above doesn't need to be strictly positive, it can also be 0, as in one of the complex numbers can be 0 and the equality then holds).

So for in order the equality to occur where LHS = RHS:
arg (z1) = arg (z2) OR z1 or z2 = 0

Thus the equality would work in the following:
z1 = 2 + i, z2 = 1 + 2i
z1 = 7 + 2i, z2 = 8 + 6i <- so as long as they are in the first quadrant.

So, is my explanation for a) fine?

What about b)?

 

[InteGrand]

Thus the equality would work in the following:
z1 = 2 + i, z2 = 1 + 2i
z1 = 7 + 2i, z2 = 8 + 6i <- so as long as they are in the first quadrant.

Neither of those pairs have equal argument (or a 0 complex number), so there wouldn't be equality.

 

[ProdigyInspired]

Neither of those pairs have equal argument (or a 0 complex number), so there wouldn't be equality.

Ok so what porcupinetree was talking about as in the same direction, does that mean the vectors essentially has to have the same argument?

I was interpreting it as the same general direction.

So it would work for these right?

z1 = 1+i, z2 = 2+2i
z1 = 0, z2 = 0

If I understand correctly, if one of the complex numbers = 0, then the other one also has to equal 0 for the equality to work.
So with z2 = k * z1, where k is a positive integer or 0, the first example would be z1 = 2 * z2, and the second one is z1 = 0 * z2, which is 0 = 0.

What about with only one equals to 0?
Like: z1 = 0, z2 = 2 + i.

Would the equality work?

How would I exactly prove this?

 

[InteGrand]

Ok so what porcupinetree was talking about as in the same direction, does that mean the vectors essentially has to have the same argument?

So it would work for these right?

z1 = 1+i, z2 = 2+2i
z1 = 0, z2 = 0

Yes

 

[InteGrand]

So with z2 = k * z1, where k is a positive integer or 0, the first example would be z1 = 2 * z2, and the second one is z1 = 0 * z2, which is 0 = 0.

What about with only one equals to 0?
Like: z1 = 0, z2 = 2 + i.

Would the equality work?

How would I exactly prove this?

 

[InteGrand]

The question is:

Using the parallelogram method along with a diagram, why:

a. |z₁ + z₂| <= |z₁| + |z₂|

What conditions need to occur for |z₁ + z₂| = |z₁| + |z₂|?

Similarly
b. |z₁ - z₂| <= |z₁| - |z₂|


What conditions need to occur for |z₁ - z₂| = |z₁| - |z₂|?

[HR][/HR]

I've given a try for a.

Drew a parallelogram, with z1 and z2 as the sides. From the point where the vectors originate, I drew a diagonal, naming it z1+z2.

Now with triangle properties, it is known that one side cannot be larger than the sum of the other two.

Therefore, a triangle with the sides z1 and z2, and the length side (z1 + z2), would have |z1 + z2| being less than |z1| + |z2|.

For LHS = RHS, all the vectors have to be the same length, therefore z1 must be = z2.

 

[ProdigyInspired]

Thank you.

So to summarise.
Similarly to a normal triangle inequality,
the reverse triangle inequality also requires at least one of the complex no.'s being 0 OR, z1 & z2 have the same orientation.
The different condition is that arg(z1) = -arg(z2) could also work.

May I ask, when adding vectors, which are in the opposite direction, what method would you use?
Using the polygon/triangle/head-to-tail method creates an awkward triangle where the longest length isn't the resultant.

Doing basic vector addition is simple in all cases for me, however, having two vectors which are diagonally flipped i.e. 1st & 3rd quadrant, 2nd & 4th quadrant confuses me as they simply create one straight diagonal line.

The question I am figuring out is for (2+i) + (-5 - 3i).


Through normal calculations I found the answer is -3-2i, however I don't know how to achieve this graphically.

 

[InteGrand]

The different condition is that arg(z1) = -arg(z2) could also work.

This part is not the case (I think I know what you mean though). You meant that the complex numbers can be "flipped" from each other. But this doesn't mean the arg's are negatives of each other (this is the case for conjugates), but rather that the arguments differ by π.

 

[InteGrand]

Thank you.

So to summarise.
Similarly to a normal triangle inequality,
the reverse triangle inequality also requires at least one of the complex no.'s being 0 OR, z1 & z2 have the same orientation.
The different condition is that arg(z1) = -arg(z2) could also work.

May I ask, when adding vectors, which are in the opposite direction, what method would you use?
Using the polygon/triangle/head-to-tail method creates an awkward triangle where the longest length isn't the resultant.

Doing basic vector addition is simple in all cases for me, however, having two vectors which are diagonally flipped i.e. 1st & 3rd quadrant, 2nd & 4th quadrant confuses me as they simply create one straight diagonal line.

The question I am figuring out is for (2+i) + (-5 - 3i).
View attachment 32840

Through normal calculations I found the answer is -3-2i, however I don't know how to achieve this graphically.

 

[ProdigyInspired]

This part is not the case (I think I know what you mean though). You meant that the complex numbers can be "flipped" from each other. But this doesn't mean the arg's are negatives of each other (this is the case for conjugates), but rather that the arguments differ by π.

I realised after testing it. So rather than what I wrote, the correct condition is: arg (z2) = -180 + arg(z1)

Yes, seems simple enough. I did Prelim Physics and doing this seems the right way, but graphically solving it on the Argand diagram gives me a headache and gives me this monstrosity.


How is it possible to do the triangle method with this when it points backwards in a straightline?

 

[InteGrand]

I realised after testing it. So rather than what I wrote, the correct condition is: arg (z2) = -180 + arg(z1)



Yes, seems simple enough. I did Prelim Physics and doing this seems the right way, but graphically solving it on the Argand diagram gives me a headache and gives me this monstrosity.
View attachment 32841

How is it possible to do the triangle method with this when it points backwards in a straightline?

It's easy to mess up if doing it by hand like that, so I'd prefer to do it using the component-by-component method.