Question

How to demostrate this property of complex?

Answer 1

\[\left| z+w \right|\le \left| z \right| + \left| w \right|\]

Answer 2

I've demostrated lot's of properties...but I have no idea of how to demostrate this

Answer 3

the length of z +length of w >= length of z+w

Answer 4

only time they're equal is if they point in the same direction

Answer 5

make sense?

Answer 6

Sure! The shortest path is always in diagonal :P

Answer 7

thanks

Answer 8

Any way to demostrate this without graphics?

Answer 9

I think I just got it :) but thanks

Answer 10

yeah
let z = a+bi
let w = c + di
then z+w = (a+c) + (b+d)i
\[|z+w|=\sqrt{(a+c)^2+(b+d)^2}=\sqrt{(a+c)^2+(b+d)^2}\]
\[|z|=\sqrt{(a)^2+(b)^2}\text{ and }|w|=\sqrt{(c)^2+(d)^2}\]

\[|z+w|^2=(z+w)^2=z^2+2zw+w^2=|z|^2+2zw+|w|^2\le|z|^2+2|z||w|+|w|^2\]
\[=(|z|+|w|)^2\Rightarrow |z+w|^2 \le (|z|+|w|)^2 \Rightarrow |z+w| \le |z|+|w|\]