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 $7 \pi/12$.
 A: 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.
A: It is $\arctan(\sqrt{3})+\arctan(1)  = \frac{\pi}{3}+\frac{\pi}{4}$.
A: I)  Tedious:  $z= (\frac{1}{2}+ \frac{3^{1/2}i}{2}) * (1+i)= (1/2 - \sqrt{3}/2) + (1/2 + \sqrt 3/2) i = $.  Arg($z$) = $\arctan (1+ \sqrt{3})/(1-\sqrt{3}) = -5\pi/12 = 7\pi/12$.
(To be fair, I made data entry errors the first six times I tried to enter this into a calculator.)
II) Easier (albeit it notation intimidating):  $\arg (1+i) = \pi/4$ so $i+1 = r*e^{\frac{\pi}4 i}$.  ($r = \sqrt 2$ but we don't give a toss.)  $\arg (1/2 + i\sqrt{3}/2) = \pi/3$ so $1/2 + i\sqrt{3}/2 = s*e^{\frac{\pi}3 i}$.  ($s = 1$ but we don't give a toss.)
So  $z= (\frac{1}{2}+ \frac{3^{1/2}i}{2}) * (1+i) = r*e^{\frac{\pi}4 i}*s*e^{\frac{\pi}3 i} = rse^{(\pi/3 + \pi/4)i}$ so $\arg(z) = \pi/3 + \pi/4 = 7\pi/12$.
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Post Script: $\frac {1 + \sqrt 3}{1 - \sqrt 3}= \frac{(1 + \sqrt 3)^2}{1 - 3} = \frac {4 + 2\sqrt 3}{-2} = -2 - \sqrt 3$ so your first answer was absolutely correct.  Why did you doubt yourself?  Why did you assume $\tan^{-1}(-2 - \sqrt{3}) \ne 7\pi/12$?
