Triangle Inequality with Complex Numbers I was wondering how to prove the triangle inequality with complex numbers:
Verify that the function $d(z_1, z_2)$ is a distance funtion on $\mathbb{C}$ and also on any subdomain on $\mathbb{C}$.
I finished all but the inequality, which is what I need assistance with. I don't know how to prove the distance function inequality:
$d(z_1, z_3) \le d(z_1, z_2) + d(z_2, z_3) $
 A: Without loss of generality, we only need to show that $|z_2-z_1|\le |z_2|+|z_1|$.  We will use the notation that $z_1=|z_1|e^{i\theta_1}$ and $z_2=|z_2|e^{i\theta_2}$ and $z^{*}$ will denote the complex conjugate of $z$. 
So, we write 
$$\begin{align}
|z_2-z_1|^2&=(z_2-z_1)(z_2-z_1)^*\\\\
&=|z_2|^2+|z_1|^2-2\text{Re}\{z_1z_2^*\}\\\\
&=|z_2|^2+|z_1|^2-2|z_2||z_1|\cos (\theta_1-\theta_2)\\\\
&=(|z_2|+|z_1|)^2-2|z_2||z_1|(1+\cos (\theta_1-\theta_2))\\\\
&\le(|z_2|+|z_1|)^2
\end{align}$$
and we are done!  Just let $z_2 \to z_3-z_2$ and $z_1 \to z_1-z_2$.

A: Note that we identify $\mathbb C$ with the plane $\mathbb R^2$.  If you realize that complex addition in $\mathbb C$ is the same thing as vector addition in $\mathbb R^2$, and the absolute value in $\mathbb C$ is the same thing as the norm $\|\vec{v}\|$ in $\mathbb R^2$, then the triangle inequality in $\mathbb C$ is just exactly the same as the triangle inequality in $\mathbb R^2$.
A: The more formal proof goes as so:
Let us consider $|z_1+z_2|^2 = (z_1 +z_2)(\overline{z_1}+\overline{z_2})$
Multiplying out,
$(z_1 +z_2)(\overline{z_1}+\overline{z_2}) = z_1\overline{z_1}+z_1\overline{z_2}+\overline{z_1\overline{z_2}}+z_2\overline{z_2}$
$=|z_1|^2 + 2Re(z_1\overline{z_2})+|z_2|^2$, and it is here we note that $2Re(z_1\overline{z_2})\leq 2|z_1||z_2|$, so substituting in,
$$|z_1|^2 + 2Re(z_1\overline{z_2})+|z_2|^2 \leq |z_1|^2 + 2|z_1||z_2|+|z_2|^2 = (|z_1|+|z_2|)^2$$
So, we have shown $|z_1+z_2|^2 \leq (|z_1|+|z_2|)^2 \implies |z_1+z_2| \leq |z_1|+|z_2|$
Also, the very last step is justified since we know that the modulus is always greater than or equal to 0.
A: Using just the algebraic form. Set $u=z_1-z_2$ and $v=z_2-z_3$, so you have to prove that
$$
|u+v|\le |u|+|v|
$$
This is equivalent to
$$
|u+v|^2\le (|u|+|v|)^2
$$
Since $|z|^2=z\bar{z}$, this becomes
$$
u\bar{u}+u\bar{v}+\bar{u}v+v\bar{v}\le u\bar{u}+2|u|\,|v|+v\bar{v}
$$
and, removing the alike terms, we get the equivalent inequality
$$
u\bar{v}+\bar{u}v\le 2|u|\,|v|
$$
If the left-hand side is negative, we are done. If it is nonnegative, the inequality is equivalent to
$$
(u\bar{v}+\bar{u}v)^2\le 4|u|^2\,|v|^2
$$
that becomes
$$
u^2\bar{v}^2+2u\bar{u}v\bar{v}+\bar{u}^2v^2\le 4u\bar{u}v\bar{v}
$$
Transporting the terms to the left-hand side, we obtain
$$
(u\bar{v}-\bar{u}v)^2\le0
$$
which is true, because $z=u\bar{v}-\bar{u}v$ is purely imaginary, since $z=-\bar{z}$.
