condition for two complex number $a$ and $b$ to be $|a+b| = |a| + |b|$ If $a$ and $b$ are two complex numbers, and  $a \neq 0$, then how to show that the condition required for $|a+b| = |a| + |b|$ is $b/a$ is real and non-negative.
I did the following and I got stuck  
$ \hspace{12 mm}|a+b|^2 = (|a| + |b|)^2 \\
\implies (a+b)(\bar a + \bar b) = |a|^2 + 2|a||b| + |b|^2 \\
\implies \bar a b + a \bar b = 2|a||b|$
 A: You can take a geometric approach.
Let's consider $a$ and $b$ as two two-dimensional vectors over the reals. Then the complex norm is simply the two-dimensional euclidean norm. So you have two vectors whose norms sum up to the norm of their sum. Triangle inequality (for a Euclidean space) says you get equality only if the triangle is degenerate, that is, $a$ is a scalar multiple of $b$. Said scalar must be real and non-negative. Bam, you're done.
A: As you said we want
$$\bar{a}b+a\bar{b}=2\left|ab\right|\implies(\bar{a}b+a\bar{b})^2=4\left|ab\right|^2\implies(\bar{a}b)^2+2a\bar{a}b\bar{b}+(a\bar{b})^2=4a\bar{a}b\bar{b}\implies(\bar{a}b)^2-2a\bar{a}b\bar{b}+(a\bar{b})^2=0\implies(\bar{a}b-a\bar{b})^2=0\implies\bar{a}b=a\bar{b}$$
I think you can now complete the proof using the fact that 
$$z\in \mathbb{R}\iff z=\bar{z}$$
and that
$$\bar{a}b+a\bar{b}=2\left|ab\right|\ge 0$$
A: If $a = r_1e^{i \theta_1}$ and $b = r_2e^{i \theta_2}$, then $$\bar{a} b + a \bar{b} = r_1r_2 \left(e^{i (\theta_1 - \theta_2)} + e^{i (\theta_2 - \theta_1)}\right) = 2r_1r_2 \cos(\theta_1 - \theta_2)$$
$$2 \vert a \vert \vert b \vert = 2r_1 r_2$$
Hence, we get that $$\cos(\theta_1 - \theta_2) = 1 \implies \theta_1 = \theta_2$$
Hence, $b = k a$ where $k \in \mathbb{R}^+ \cup \{0\}$.
A: Your approach is fine.  You already have
$$
|a+b|^2=|a|^2+|b|^2 \implies 2\text{Re}(a^{*}b)=2|a^{*}b|,
$$
which holds if and only if $a^{*}b$ is real and non-negative.  Given the additional condition that $a\neq 0$, this is equivalent to $b/a=(a^{*}b)/|a|^2$ being real and non-negative.
