The most general Lorentz transformation that is connected to the identity is given by the conjugation by $\exp(-A)$ where
$$ A = \frac 12 \omega_{\mu\nu} \gamma^\mu \gamma^\nu $$
and $\omega_{\mu\nu}$ is an antisymmetric tensor containing $D(D-1)/2$ parameters. The group of all such transformations is isomorphic to $Spin(D-1,1)$. If $\omega$ only contains one component $0\mu$, then it is a boost, and the nonzero numerical value of $\omega$ is the rapidity - the "hyperbolic angle" $\eta$ such that $v/c=\tanh\eta$.
If only one doubly spatial component of $\omega$ is nonzero, then this component $\omega_{\mu\nu} = -\omega_{\nu\mu}$ is obviously the angle itself. Note that the spatial-spatial terms in $A$ are anti-Hermitean, producing unitary transformations; the mixed temporal-spatial terms in $A$ are Hermitean and they don't product unitary transformations on the 4-component space of spinors (but they become unitary if they're promoted to transformations of the full Hilbert space of quantum field theory).
In 4 dimensions, a general antisymmetric matrix $4\times 4$ contains 6 independent parameters and has eigenvalues $\pm i a, \pm i b$, so in 3+1 dimensions, one can always represent a general Lorentz transformation as a rotation around an axis in the 4-dimensional space followed by a boost in the complementary transverse 2-plane. This is the counterpart of the statement that any $SU(2)$ rotations in 3 dimensions is a rotation around a particular axis by an angle.
If you allowed $A$ to contain something else than $\gamma^{\mu\nu}$ matrices which generate the Lorentz group, you could get other groups. Only for a properly chosen subset of allowed values of $\omega$, you would get a closed group from the resulting exponentials (under multiplication). In particular, if you allowed $A$ to be an arbitrary complex combination of any products of gamma matrices, well, then you would allow $A$ to be any complex $4\times 4$ matrix, and its exponentials would produce the full group $GL(4,C)$ - surprising, Carl? ;-) It's not a terribly useful groups in physics because actions are usually not invariant under this "full group", are they?
Also, there are not too many groups in between $Spin(3,1)$ and $GL(4,C)$ - I guess that there's no proper group of $GL(4,C)$ that has a proper $Spin(3,1)$ subgroup. Obviously, there are many subgroups of $Spin(3,1)$ - such as $Spin(3)$, $Spin(1,1)\times Spin(2)$, and others.
Lorentz boost is simply a Lorentz transformation which doesn't involve rotation. For example, Lorentz boost in the x direction looks like this:
\begin{equation}
\left[
\begin{array}{cccc}
\gamma & -\beta \gamma & 0 & 0 \newline
-\beta \gamma & \gamma & 0 & 0 \newline
0 & 0 & 1 & 0 \newline
0 & 0 & 0 & 1
\end{array}
\right]
\end{equation}
where coordinates are written as (t, x, y, z) and
\begin{equation}
\beta = \frac{v}{c}
\end{equation}
\begin{equation}
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
\end{equation}
This is a linear transformation which given coordinates of an event in one reference frame allows one to determine the coordinates in a frame of reference moving with respect to the first reference frame at velocity v in the x direction.
The ones on the diagonal mean that the transformation does not change the y and z coordinates (i.e. it only affects time t and distance along the x direction). For comparison, Lorentz boost in the y direction looks like this:
\begin{equation}
\left[
\begin{array}{cccc}
\gamma & 0 & -\beta \gamma & 0 \newline
0 & 1 & 0 & 0 \newline
-\beta \gamma & 0 & \gamma & 0 \newline
0 & 0 & 0 & 1
\end{array}
\right]
\end{equation}
which means that the transformation does not affect the x and z directions (i.e. it only affects time and the y direction).
In order to calculate Lorentz boost for any direction one starts by determining the following values:
\begin{equation}
\gamma = \frac{1}{\sqrt{1 - \frac{v_x^2+v_y^2+v_z^2}{c^2}}}
\end{equation}
\begin{equation}
\beta_x = \frac{v_x}{c},
\beta_y = \frac{v_y}{c},
\beta_z = \frac{v_z}{c}
\end{equation}
Then the matrix form of the Lorentz boost for velocity v=(vx, vy, vz) is this:
\begin{equation}
\left[
\begin{array}{cccc}
L_{tt} & L_{tx} & L_{ty} & L_{tz} \newline
L_{xt} & L_{xx} & L_{xy} & L_{xz} \newline
L_{yt} & L_{yx} & L_{yy} & L_{yz} \newline
L_{zt} & L_{zx} & L_{zy} & L_{zz} \newline
\end{array}
\right]
\end{equation}
where
\begin{equation}
L_{tt} = \gamma
\end{equation}
\begin{equation}
L_{ta} = L_{at} = -\beta_a \gamma
\end{equation}
\begin{equation}
L_{ab} = L_{ba} = (\gamma - 1) \frac{\beta_a \beta_b}{\beta_x^2 + \beta_y^2 + \beta_z^2} + \delta_{ab} = (\gamma - 1) \frac{v_a v_b}{v^2} + \delta_{ab}
\end{equation}
where a and b are x, y or z and δab is the Kronecker delta.
Best Answer
Rotations only change the spatial coordinates $(x^i)$; the time coordinate $(x^0)$ stays unchanged.
Now suppose you're rotating around the $z$ axis. Then the rotation matrix $(R)$ for this is:
\begin{equation} R = \begin{bmatrix} \cos{\theta} & -\sin{\theta} & 0\\ \sin{\theta} & \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix} \end{equation}
This induces a rotation of coordinates $(x \rightarrow x')$ in component form as: \begin{equation} x'^i = R^i_{\ j}x^j \end{equation}
Note that the $3 \times 3$ matrix above is the spatial part of the $4 \times 4$ Lorentz transformation matrix $\Lambda$.
Now to show Lorentz invariance under this special case of rotation around $z$ axis, we just need to show that $(x')^2 + (y')^2 = (x)^2 + (y)^2$, which is trivial.