Complex Numbers (1 Viewer)

[untouchablecuz]

Show that:

mod (|z1| - |z2|) ≤ mod (z1 + z2)

AND state the condition for equality to hold.

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Please help.

 

[gurmies]

hmm, I don't like this question, solution to it is quite difficult to follow =(

 

[AMorris]

So the triangle inequality (the simplest version) states that:

|w1| + |w2| >= |w1 + w2|

Now let w1 = z1 + z2 and let w2 = -z2. Then we have:

|z1 + z2| + |-z2| >= |z1 + z2 - z2|

|z1 + z2| + |z2| >= |z1|

|z1 + z2| >= |z1| - |z2|

Similarly (just switching z2 and z1 around in our last equation):

|z1 + z2| >= |z2| - |z1|

So combining these two, we get:

|z1 + z2| >= ||z1| - |z2||

 

[Zeber]

equality when it is a straight line. (ie. arg(z1) = arg(z2))

 

[shaon0]

equality when it is a straight line. (ie. arg(z1) = arg(z2))

That can be deduced from the triangle inequality. (which i still don't get)

 

[Zeber]

any two sides of a triangle added togethr is longer than the third side.

that's what it's basically saying.

 

[shaon0]

any two sides of a triangle added togethr is longer than the third side.

that's what it's basically saying.

Yeah i get it now after doing a few questions.

 

[youngminii]

I am confused, isnt' the mod of a mod just the mod?
..Wtfock

 

[Trebla]

The modulus of a SUM of two moduli is not equal to sum of the two moduli themselves. (e.g. | |z1| + |z2| | is not necessarily equal to |z1| + |z2|) The modulus pretty much acts as a function if you like.