User:Phinda Hlawe/sandbox

Triangle Inequality for Complex numbers

Let z,w be complex numbers,then we have that ||z|-|w||≤|z-w|≤|z|+|w| also |z+w|≤|z|+|w|.

Proof:

Clearly |z|²+|w|²≤|z|²+2|w||y|+|w|² Since we have that 2|z||w|≥0 by defination of the modulo. So we have that |z|²+|w|²≤(|z+w|)² which implies that |z|+|w|≤|z+w|.

Now |z|²-|w|²≤|z|²-2|w||y|+|w|² Since 2|z||w|≥0 Again this leads use to conclude that |z|²-|w|²≤(|z-w|)²≤(|z+w|)² since |z|²-2|w||y|-|w|²≤|z|²+2|w||y|+|w|² then we have that |z-w|≤|z|+|w|.

Also |z|=|z-w+w|≤|z-w|+|w| (as shown above) ⇒|z|-|w|≤|z-w|, Similarly |w|=|w-z+z|≤|w-z|+|z| ⇒|w|-|z|≤|w-z| multiplying both sides by -1 gives -|z-w|≤|z|-|w| so -|z-w|≤|z|-|w|≤|z-w| ⇒||z|-|w||≤|z-w|.

||z|-|-w||≤|z-(-w)|⇒||z|-|w||≤|z+w| since |-w|=|w|. And therefore ||z|-|w||≤|z-w|≤|z|+|w|, |z+w|≤|z|+|w| has been shown.