Complex inequality $|a+b|\le |a|+|b|$

Question

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.

Answer 1

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.

Answer 2

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 $a\bar{b}=2|a|\,|b|\ge0$.

Answer 3

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.