complex number relations I can structure a solution using geometry but I am struggling to find on using algebra.
In an Argand diagram, the complex numbers $z, w$ and $z+w$ are represented by the points $P, Q$ and $S$ respectively.
Show that:
$$| z + w | ≥ | z | − | w |$$
Thanks for any help
 A: We have that 
$$|z|=|z+w-w|\le |z+w|+|w|\implies |z|-|w|\le |z+w|.$$
A: Let $z=x+yi$ and $w=a+bi$, where $\{x,y,a,b\}\subset\mathbb R$.
Thus, we need to prove that
$$\sqrt{(x+a)^2+(y+b)^2}+\sqrt{a^2+b^2}\geq\sqrt{x^2+y^2}$$ or
$$\sqrt{(a^2+b^2)((x+a)^2+(y+b)^2)}\geq-a^2-b^2-ax-by.$$
If $-a^2-b^2-ax-by<0$ then the inequality is obviously true, but for
$-a^2-b^2-ax-by\geq0$ it's enough to prove that
$$(a^2+b^2)((x+a)^2+(y+b)^2)\geq(a^2+b^2+ax+by)^2,$$ which is C-S:
$$(a^2+b^2)((x+a)^2+(y+b)^2)\geq(a(x+a)+b(y+b))^2.$$
Done!
If you don't like the C-S then for the proof of the last inequality we need to prove that
$$(a^2+b^2)(a^2+b^2+2(ax+by)+x^2+y^2)\geq(a^2+b^2)^2+2(a^2+b^2)(ax+by)+(ax+by)^2$$ or
$$(a^2+b^2)(x^2+y^2)\geq(ax+by)^2$$ or
$$(ay-bx)^2\geq0.$$
A: Other answers are perfectly correct, here is an answer using cosine rule:
$$\begin{align}|z-w|^2 &=(z-w)\overline{(z-w)} \\ 
&=|z|^2 +|w|^2 - z\overline{w} - \overline{z}w \\
&= |z|^2 + |w|^2 - (z\overline{w} + \overline{z\overline{w}} )\\
&= |z|^2 + |w|^2 - 2\Re{(z\overline{w})} \\
&\ge |z|^2 - |w|^2 -2|z\overline{w}| \\
&\ge (|z| - |w|)^2
\end{align}$$
$|\Re(z)| \le|z|$ as $\Re^2(z) + \Im^2(z) =|z|^2$
Now since $$|z|^2 +|w|^2 + z\overline{w} + \overline{z}w \ge |z|^2 +|w|^2 - z\overline{w} - \overline{z}w$$
$$\implies |z+w|^2 \ge|z-w|^2$$
