Suppose that $b_1 \neq 0$ and $l_1(t) = a_1 + tb_1, l_2(t) = a_2 + tb_2, a_i, b_i \in \mathbb{C}, i = 1, 2$ are two complex lines. My CA book states that the angle between the two lines is given by $\mathrm{arg}(b_2/b_1)$, but I can't seem to derive this result rigorously from the form of the two lines $l_1, l_2$.
Attempts: (1) From calculus we know that the angle between two (differentiable) curves $\gamma_1, \gamma_2$ is $\mathrm{arccos}\left(\frac{\left<\gamma_1'(t), \gamma_2'(t)\right>}{||\gamma_1'(t)||||\gamma_2'(t)||}\right)$, but the $\mathrm{arg}$ function uses the $\mathrm{arctan}$ function.
(2) Since the angle between any two members of the two families of curves $\mathcal{C}_1 = \{a + tb_1\mid a \in \mathbb{C}\}, \mathcal{C}_2 = \{a + tb_2\mid a \in \mathbb{C}\}$ are (intuitively) equal, we can consider the case $a_1 = a_2 = 0$, from which we obtain the claimed equality quite easily.
Question: Without resorting to "conceptual" thinking by considering the families of curves, how can I show that the angle between $l_1$ and $l_2$ is given by $\mathrm{arg}(b_2/b_1)?$
You can get it directly from the calculus formula using $\cos^{-1}(a/b) = \tan^{-1}(\sqrt{b^2 - a^2}/a)$, $\langle a,b\rangle = \mathrm{Re}(\bar{a} b)$ and $$ |a|^2|b|^2 = a\bar{a}b\bar{b} = \frac{(a\bar{b}+b\bar{a})^2 -(a\bar{b}-b\bar{a})^2}{4} = \mathrm{Re}[\bar{a}b]^2 + \mathrm{Im}[\bar{a}b]^2. $$ So we have \begin{multline} \cos^{-1}\left(\frac{\left<\gamma_1'(t), \gamma_2'(t)\right>}{||\gamma_1'(t)||||\gamma_2'(t)||}\right) = \cos^{-1}\left(\frac{\mathrm{Re}[\bar{b_1}b_2]}{|b_1||b_2|}\right) = \tan^{-1}\left(\frac{\sqrt{|b_1|^2|b_2|^2 - \mathrm{Re}[\bar{b_1}b_2]^2}}{\mathrm{Re}[\bar{b_1}b_2]}\right) \\= \tan^{-1}\left(\frac{\mathrm{Im}[\bar{b_1}b_2]}{\mathrm{Re}[\bar{b_1}b_2]} \right) = \mathrm{arg}(\bar{b_1}b_2) = \mathrm{arg}\left(\frac{b_2}{b_1}\right) \end{multline}