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 $7 \pi/12$.

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 $\arg(z) = \pi/3 + \pi/4 = 7\pi/12$.
====
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 $\tan^{-1}(-2 - \sqrt{3}) \ne 7\pi/12$?
