Prove the triangle inequality involving complex numbers. Our eventual goal in this problem is to prove the triangle inequality involving complex numbers.
(a) Show that for every $z ∈ C$,
$|Re(z)| ≤ |z|$ and $|Im(z)| ≤ |z|$.
￼(b) Given $z$, $w ∈ C$, show that
$|z+w|^2 =|z|^2 +|w|^2 +2Re(zw')$.
(c) Using parts (a) and (b), prove the triangle inequality
$|z + w| ≤ |z| + |w|$.
This is what I got.
(a) By definition for a complex number $z = x + yi$,
$$|z|^2 = x^2 + y^2 = Re(z)^2 + Im(z)^2$$
From here,
$$|z|^2 ≥ Re(z)^2 \text{ and } |z|^2 ≥ Im(z)^2$$
And, finally,
$$|z| ≥ |Re(z)| \text{ and } |z| ≥ |Im(z)|$$
(b) $|z + w|^2 = (z + w)·(z + w)'$
$$= (z + w)·[z' + w']$$
$$= zz' + [zw' + z'w] + ww'$$
$$= |z|^2 + 2Re[zw'] + |w|^2$$
$$≤ |z|^2 + 2|zw'| + |w|^2$$
$$= |z|^2 + 2|z||w| + |w|^2$$
$$= (|z| + |w|)^2.$$
(c) Since both $|z+w|$ and $|z| + |w|$ are non-negative,
$$|z + w| ≤ |z| + |w|$$
 A: Looks solid! Sadly, I'm only seeing it well after the OP, but to cut down on unanswered questions, here we go!
A: Let $z,w$ be complex numbers,then we have that  $||z|-|w|| \leq |z+w|$  and $||z|-|w|| \leq |z-w| $ furthermore $|z-w| \leq |z|+|w| $ also $ |z+w| \leq |z|+|w|$.
Proof:
Clearly $ |z|^{2}+|w|^{2} \leq |z|^{2}+2|w||y|+|w|^{2} $ Since we have that $2|z||w| \leq 0$ by definition of the modulo. So we have that $|z|^{2}+|w|^{2} \leq (|z+w|)^{2}$ which implies that $|z|+|w| \leq |z+w|$.
Now $|z|^{2}-|w|^{2} \leq |z|^{2}-2|w||y|+|w|^{2}$ .Since  $2|z||w| \leq 0$ Again this leads use to conclude that $|z|^{2}-|w|^{2} \leq (|z-w|)^{2} \leq (|z+w|)^{2}$ since $|z|^{2}-2|w||y|-|w|^{2} \leq |z|^{2}+2|w||y|+|w|^{2}$ then we have that $|z-w| \leq |z|+|w|$.
Also $|z|=|z-w+w| \leq |z-w|+|w|$ (as shown above) $⇒|z|-|w|\leq|z-w|$, Similarly $|w|=|w-z+z|\leq|w-z|+|z| ⇒|w|-|z|\leq|w-z|$ multiplying both sides by -1 gives $-|z-w|\leq|z|-|w|$ so $-|z-w|\leq|z|-|w|\leq|z-w| ⇒||z|-|w||\leq|z-w|$.
$||z|-|-w||\leq|z-(-w)|⇒||z|-|w||\leq|z+w|$ since $|-w|=|w|$. And therefore $ ||z|-|w|| \leq |z-w|\leq|z|+|w|$ , $|z+w|\leq|z|+|w|$ has been shown.
