There does not exist two continuous functions that the values constitute the spectrum of the matrix $A(z) = \pmatrix{0 & z \\ 1 & 0}$ I read an example from Rajendra Bhatia's Matrix Analysis.
Let $A(z) = \pmatrix{0 & z \\ 1 & 0}$ with $z \in \mathbb C$. The eigenvalues are clearly $\pm \sqrt{z}$. Then he says these cannot be represented by two continuous functions over any domain $G$ contains $0$. I guess this should violate some basic property of complex-valued functions, which unfortunately I am not familiar with. Could someone point this out for me? Thanks.
 A: If such continuous functions $f$ and $g$ existed, then the would satisfy
$$
f=-g \quad \& \quad f^2(z)=g^2(z)=z.
$$
Thus for every $z\ne 0$, we have that $f(z)\ne 0$, and hence
$$
\frac{f(z+h)-f(z)}{h}=\frac{1}{f(z+h)+f(z)}\cdot\frac{f^2(z+h)-f^2(z)}{h}=
\frac{1}{f(z+h)+f(z)}\cdot\frac{z+h-z}{h}\\=\frac{1}{f(z+h)+f(z)}\to \frac{1}{2f(z)}.
$$
Therefore, $f$ (and similarly $g$) is analytic in a region
$$
U=\{z\in\mathbb C: 0<|z|<R\}
$$
for some $R>0$, and as $|\,f(z)|=|z|^{1/2}\le R^{1/2}$, then $f$ is bounded in $U$ and therefore extends analytically in $0$, and as $\lim_{z\to 0}f^2(z)=0$, then 
$\lim_{z\to 0}f(z)=0$. And thus, 
$$
\hat f(z)=\left\{
\begin{array}{ccc}
f(z) & if & z\ne 0, \\
0 & if & z=0,
\end{array}
\right.
$$
is analytic in $U\cup\{0\}$, and $z=0$ is a zero of $\hat f$. Hence, there is a $k\in\mathbb N$, such that
$$
\hat f(z)=z^kh(z),
$$
where $h(z)$ is analytic and $h(0)\ne 0$. Thus
$$
z=\hat f^2(z)=z^{2k}h^2(z),
$$
and finally
$$
1=\hat f^2(z)=z^{2k-1}h^2(z),
$$
for all $z\in U$. This is impossible, since the right hand side tends to zero, as $z\to 0$.
