Find the eigenvectors of a hermitian matrix as a function of angles I have a problem which seems flawed to me.

Consider the hermitian matrix M
\begin{pmatrix}
 a/2 & b^* \\ 
 b & -a/2
 \end{pmatrix}
  with a>0 real and b complex.
  Let $\theta$,$\phi$ be the two angles given by $\tan{\theta}$=$\frac{2|b|}{a}$ and $b=|b|e^{i\phi}$. Calculate the eigenvalues of M as a function of these angles and show that $v_1=(\cos{\frac{\theta}{2}}e^{-i\phi/2},\sin{\frac{\theta}{2}}e^{i\phi/2})^T$ and $v_2=(-\sin{\frac{\theta}{2}}e^{-i\phi/2}, \cos{\frac{\theta}{2}}e^{i\phi/2})^T$ are the corresponding eigenvectors and form an orthonormal system.

I calculated the eigenvalues to 
\begin{align}
\lambda_{1,2}=\pm|b|\sqrt{1+\cot^2{x}}
\end{align}
But no matter what I try, these eigenvalues do not correspond with the given eigenvectors. I also tried calculating the eigenvalues starting with the eigenvectors but that doesnt work, too. 
Thanks in advance
 A: You have a mistake in your calculation of the eigenvalue, but I still think the problem is not right. 
The eigenvalues of $M$ are $\frac a2\pm |b|$. One then finds easily that a corresponding pair of eigenvectors is given by 
$$
(e^{-i\phi/2},-e^{i\phi/2})^T,\ \ (e^{-i\phi/2},e^{i\phi/2})^T. 
$$
The eigenspaces are spanned by each of these two vectors, so the supposed eigenvectors in the statement of the problem aren't eigenvectors (they are not scalar multiples of the above). 

Edit: after the question has been changed, here is the new answer:
When $$M=\begin{bmatrix}a/2&\bar b\\ b&-a/2\end{bmatrix},$$
the eigenvalues are $$\pm|b|\,\left(1+\frac{a^2}{4|b|^2}\right)^{1/2}
=\pm|b|\,\left(1+\frac1{\tan^2\theta}\right)^{1/2}=\pm|b|\,\frac{(1+\tan^2\theta)^{1/2}}{\tan \theta}=\pm\,\frac{|b|}{\sin \theta}.
$$
If $(x,y)^T$ is an eigenvector for $|b|/\sin\theta$, then 
$$
\left(\frac a2-\frac{|b|}{\sin\theta}\right)x+\bar b\,y=0.
$$
That is, 
$$
\left(\frac{|b|}{\tan\theta}-\frac{|b|}{\sin\theta}\right)\,x+|b|\,e^{-i\phi}\,y=0.
$$
After multiplying by $\sin\theta\,e^{i\phi/2}/|b|$,
$$
(\cos\theta -1)\,e^{i\phi/2}\,x+\sin\theta\,e^{-i\phi/2}y=0.
$$
Using the half-angle formulas,
$$
-2\sin^2\frac\theta2\,e^{i\phi/2}\,x+2\sin\frac\theta2\,\cos\frac\theta2\,e^{-i\phi/2}\,y=0.
$$
And now, multiplying by $1/(2\sin\frac\theta2)$,
$$
-\sin\frac\theta2\,e^{i\phi/2}\,x+\cos\frac\theta2\,e^{-i\phi/2}\,y=0.
$$
Then 
$$(x,y)^T=(\cos\frac\theta2\,e^{-i\phi/2},\sin\frac\theta2e^{i\phi/2})^T$$
is an eigenvector for $|b|/\sin\theta$. 
The computation for the other eigenvalue is very similar. 
