Complex inequality $|a+b|\le |a|+|b|$ My textbook says if $a$ and $b$ are two complex numbers, then $$|a+b|\le |a|+|b|,$$ and the equality holds if and only if $a\bar{b} \ge 0$.
How can we say the equality holds if and only if $a\bar b \ge 0$? I think $a\bar b$ is a complex number and complex numbers do not have order.
If we square both sides and cancel some terms, then we can see that the equality holds if Re$(a\bar b) = |a||b|$. 
It is on page 9 of Ahlfors' Complex Analysis.
 A: Note that 
$$
\operatorname{Re}(a\bar b)=|a||b|=|ab|=|a\bar b|
$$
if and only if $\operatorname{Im}(a\bar b)=0$ and $a\bar b\ge 0$, so $a\bar b$ has to be real.
A: The inequality holds if and only if $|a+b|^2\le(|a|+|b|)^2$ holds, which becomes
$$
|a|^2+a\bar{b}+\bar{a}b+|b|^2\le |a|^2+2|a|\,|b|+|b|^2
$$
We can cancel real terms from both sides and still get an equivalent inequality:
$$
a\bar{b}+\bar{a}b\le2|a|\,|b| \tag{*}
$$
This surely holds and is strict if $a\bar{b}+\bar{a}b<0$ (following Ahlfors's convention that $c>0$ means that $c$ is real and positive).
Let's suppose $a\bar{b}+\bar{a}b\ge0$. Then squaring is allowed and yields an equivalent inequality:
$$
a^2\bar{b}^2+2a\bar{a}b\bar{b}+\bar{a}^2b^2\le 4a\bar{a}b\bar{b}
$$
that is, moving the (real) right-hand side to the left
$$
(a\bar{b}-\bar{a}b)^2\le0
$$
which is true, because $a\bar{b}-\bar{a}b$ is purely imaginary.
A necessary condition for having an equality is that $a\bar{b}=\bar{a}b$, that is $a\bar{b}$ is real. Together with (*) we obtain $a\bar{b}=2|a|\,|b|\ge0$.
A: Think of $a$ and $b$ as being planar vectors.  Draw the parallelogram and apply the parallelogram law.  You will see the geometric meaning of this immediately.
