In the magic world of complex numbers, the proof of the **Triangle Inequality** is somewhat more elegant, and its geometrical interpretation is more straightforward. Note that for a complex number *z*=x+iy, we denote its distance from the origin by the handy formula sqrt(x^{2}+y^{2}),

and denote this value by |*z*|.

Theorem: For any *w* and *z* in **C**, |*z*+*w*| ≤ |*z*| + |*w*|.

Helpful ASCII diagram:

`
w | w+z (in accordance with the Parallelogram Law)
\ | /\
\ | / \ <-- vector w emanating from z instead of origin
\|/ \
-----+======--------
z
`

So, the idea is just that any leg of a triangle cannot be
longer than the sum of the other sides.

**Proof**: Letting

*z* = x+iy and

*w* = s+it, we have

`
|`*z*+*w*|^{2} = (x+s)^{2}+(y+t)^{2}
= x^{2}+s^{2}+y^{2}+t^{2}+2(xs+yt)
= (x^{2}+y^{2})+(s^{2}+t^{2})+2*Re(*z**w*^{*})
= |*z*|^{2}+|*w*|^{2}+2*Re(*zw*^{*})
≤ |*z*|^{2}+|*w*|^{2}+2|*zw*^{*}|
= |*z*|^{2}+|*w*|^{2}+2|*z*||*w*|
= (|*z*| + |*w*|)^{2}.

We finish off the proof by taking the square root of both sides.