Complex numbers- elementary question How can I calculate the argument of the complex number $z= (\frac{1}{2}+ \frac{i\sqrt{3}}{2}) \cdot (1+i)$?
I always get $\tan^{-1}(-2-\sqrt{3})$, but the book answer is $7 \pi/12$.

You're given the two numbers as a multiplication. Recall that $r_1e^{i\theta} \cdot r_2e^{i \phi} = r_1r_2e^{i(\theta + \phi)},$ which is to say that arguments get added together when multiplying.
Now, note that $\frac12 + \frac{\sqrt3}2$ has argument $\frac\pi3$, while $1 + i$ has argument $\frac\pi4$. The argument of their product, then, is $\frac\pi3 + \frac\pi4 = \frac{7\pi}{12}$.
You're not wrong when you say that the argument is $\tan^{-1}(-2 - \sqrt{3})$, because
\begin{align*}
\tan \frac{7\pi}{12} &= \tan \left(\frac\pi3 + \frac\pi4\right) \\
&= \frac{\sqrt{3} + 1}{1 - (\sqrt{3})(1)} \\
&= \ldots \\
&= -2 - \sqrt{3}
\end{align*}
using the identity for $\tan(\alpha + \beta)$. It's just that it's not at all obvious that $\tan^{-1}(-2 - \sqrt{3}) = \frac{7\pi}{12}$, unlike for nicer angles whose tangents you've probably memorized!
So in this case, the "arguments get added" approach is the way to go.
