Find $\arg(z)$, where $z = -\frac{\sqrt{7}(1+i)}{\sqrt{3}+i}$

Question

This is my work so far:

I got the expression into the form $a+bi$ through some tedious algebra. I'll just list the final result here:

$-\frac{\sqrt{7}(\sqrt{3}+1)}{4} + \frac{\sqrt{7}(1-\sqrt{3})}{4}i$

When I solve for $|z|$ I get $\frac{\sqrt{14}}{2}$, which matches the answer in my textbook. However, when I solve for $arg(z)$ I get $\frac{11\pi}{12}$, while the book lists it as $\frac{13\pi}{12}$. I got my answer by solving for $cos (\theta) = \frac{a}{|z|}$, where $\theta = arg(z)$. Nothing seems wrong, so not sure if the answer in the book is a typo or I'm making a mistake.

Edit: I forgot a negative sign, I understand how the answer is derived now. Thanks!

Answer 1

$$Arg(z) = Arg(\sqrt{7}(1+i))-Arg(\sqrt{3}+i)=\frac{\pi}{4}-\frac{\pi}{6}=\pi/12.$$

Answer 2

$$\textrm{Arg} (z) = \textrm{Arg}(-1)+ \textrm{Arg}{\sqrt{7}}+\textrm{Arg}(1+i)-\textrm{Arg}(\sqrt{3}+i) = \pi+ 0+\frac{\pi}{4}-\frac{\pi}{6}=\frac{13\pi}{12}.$$