Is the centralizer of a semisimple element in a connected algebraic group always connected There is an exercise on page 142 of Humphreys' Linear Algebraic Groups:

Ex.1 Let $G$ be a connected algebraic group, $x \in G$ is semisimple. Must $C_G(x)$ be connected?

When $G$ is solvable, I think of another fact whose correctedness is proved on the book:

Let $H$ be a subgroup (not necessarily closed) of a connected solvable group $G$, $H$ consisting of semisimple elements. Then $C_G(H) = N_G(H)$ is connected.

So, suppose that $G$ is solvable, set $H = \langle x \rangle$. It appears that $C_G(x)$ is connected.
I think the general case could be reduced to the solvable case if for any $y \in C_G(x)$, I can find a Borel subgroup of $G$ containing both $x$ and $y$. (Then $y$ must be in $C_B(x)$ which is connected.) 
I think the Borel subgroup could be found, may be through the method of Borel variety. But I have difficulty in this.
Another exercise on the same page is:

Ex.2 Let $G$ be a connected algebraic group. If $x \in G$ has semisimple part $x_s$, then $x$ is contained in the identity component of $C_G(x_s)$.

If the answer to Ex.1 is affirmative, then Ex.2 would be obvious, since $x \in C_G(x_s)$. But I am not sure.
So, would you please tell me the answer to Ex.1, or help me with the proof or counterexample? If the centralizer could be not connected, how can I give Ex.2 a proof?
Sincere thanks.
 A: The answer to Ex.1 is, in general, "no".
(It is "yes", in the case that $G$ is reductive and simply-connected).
For a counter-example, take $G=PGL(2,\mathbb{C})$ and
$$
x= \left[ \begin{array}{cc}
1 & 0\\
0 & -1
\end{array} \right].
$$
(Note, this is not a matrix, but the class of all its non-zero complex multiples, as $PGL(2,\mathbb{C})=GL(2,\mathbb{C})/\mathbb{C}^*$).
Then, you can check directly, by computation, that $C_G(x)$ is disconnected. There is also an interesting geometric way of identifying this centralizer. First note that $PGL(2,\mathbb{C})$ is the group of Möbius transformations acting transitively on the Riemann sphere $\mathbb{C}P^1$. Since a matrix:
$$ \left[\begin{array}{cc}
a & b\\
c & d
\end{array}\right] $$
corresponds to $z\mapsto \frac{az+b}{cz+d}$, the element $x$ above corresponds to the transformation $z\mapsto -z$, whose set of fixed points is $X=\{0,\infty\}$. The centralizer of $x$ can be viewed geometrically as the subgroup of Möbius transformations that sends $X$ to itself. These are the Möbius transformations of the form:
$$z\mapsto\alpha z, \, \mbox{ or }\, z\mapsto\frac{\alpha}{z}, $$
with $\alpha\in\mathbb{C}^{*}$, which is clearly a disconnected group.
