Alternative triangle inequality proof for complex numbers

Question

I tried to prove the triangle inequality for $\mathbb{C}$ starting with complex numbers as $z=r_1(\cos\theta+i\sin\theta)$ and $w=r_2(\cos\phi+i\sin\phi)$ :

$$\begin{align}|z+w|&=|r_1\cos\theta+r_2\cos\phi+i(r_1\sin\theta+r_2\sin\phi)|\\&=r_1^2(\cos^2\theta+\sin^2\theta)+r^2_2(\cos^2\phi+\sin^2\phi)+2r_1r_2(\sin\theta\sin\phi+\cos\theta\cos\phi)\\&=r^2_1+r^2_2+2r_1r_2\cos(\theta-\phi)\end{align}$$

I know this works when considering $$|z+w|^2 = (z+w)(\bar z + \bar w) = |z|^2 + z\bar w + \bar z w + |w|^2$$ instead, can this trigonometric form lead to the triangle inequality too?

Answer 1

$|z+w|^\color{red}{2} = \ldots = r^2_1+r^2_2+2r_1r_2\cos(\theta-\phi)$

I know this works when considering $|z+w|^2 = \ldots = |z|^2 + z\bar w + \bar z w + |w|^2$

That's in fact the same as the expression you derived. Consider that $|z|^2=r_1^2\,$, $|w|^2=r_2^2\,$, and $z\bar w + \bar z w = z\bar w + \overline{z \bar w} = 2 \operatorname{Re}(z \bar w) = 2 \,|z|\,|w| \cos\big(\arg(z) - arg(w)\big) = 2r_1r_2\cos(\theta-\phi)\,$.