Find the argument of complex number 
What is the argument of $z = (1+\cos 2a)+i(\sin 2a)$ if $\pi/2<a<3\pi/2 $?

After using the formula of $\sin2a$ and $\cos2a$, I am getting the argument as $a$ when $\pi/2<a<\pi$ and $a-2\pi$ when $\pi<a<3\pi/2$ but both the answers are incorrect
My approach 

Answer given in book 

 A: HINT
Note that
$$\frac{\sin 2a}{1+\cos 2a}=\tan a$$
A: By your work it's $$2\cos\alpha(\cos\alpha+i\sin\alpha)=-2\cos\alpha(\cos(\alpha-\pi)+i\sin(\alpha-\pi)).$$
I think by the definition in the book $$-\pi<\arg{z}\leq\pi$$ and since
$$-\frac{\pi}{2}<\alpha-\pi<\frac{\pi}{2},$$ we obtain $\arg{z}=\alpha-\pi.$
A: I am giving a different approach here, less formal but with the intention to see what is going on and why the argument cannot be $a$, as the OP understandably suggested. Without a bit "digging" into the numbers, it is understandable that an average reader (like me!) wouldn't immediately follow the last to lines of the book's answer as to why these steps are needed.
When we let $a$ run from interval $\pi/2$ to $3\pi/2$ and we calculate the arguments (I made corresponding "lists" with the TI), the corresponding interval for arg(z) is from $-\pi/2$ to $\pi/2$. When you want to relate this with input $a$, it becomes clear that $a$ is $\pi$ "too high", from which the book's suggestion $arg(z)=a-\pi$ follows.
