This doesn't answer your question completely, but at least it's a start.

If $G$ is a complex Lie group, then its quotient $G/H$ by a complex subgroup $H$ is always a complex manifold. In case $G$ is semisimple (and connected), then the *compact* quotients $G/H$ come from what are called parabolic subgroups $H$. For simplicity, let's assume that $G$ is simple. Then, up to conjugacy, the parabolic subgroups of $G$ lie in one-to-one correspondence with subsets of the nodes of the Dynkin diagram of $\mathfrak{g}$. It's easy to find a description of this bijection in the literature, so I will say no more about it. Let me just mention that a parabolic subgroup corresponding to omitting one node from the Dynkin diagram is called maximal parabolic.

Now suppose that $K$ is a compact real form of the complex semisimple group $G$. Then if the quotient space $G/H$ is compact (i.e., if $H$ is parabolic), a theorem of Montgomery asserts that $K$ acts transitively on $G/H$, and so we obtain an identification $G/H = K/(K\cap H)$. The a priori real manifold $K/(K\cap H)$ is thus endowed with a complex structure.

In summary, we have the following result:

Let $K$ be a compact semisimple Lie group with complexification $G$. Then, for a closed subgroup $S$ of $K$, the homogeneous space $K/S$ is complex if $S = K \cap P$ for some parabolic subgroup $P$ of $G$.

Here's an example. Let $G = \rm{SL}(n,\mathbb{C})$. In terms of the bijection between parabolic subgroups of $G$ and nodes, it's not hard to see that the maximal parabolic corresponding to deleting the $k$th node is the subgroup of $G$ preserving a $k$-dimensional subspace of $\mathbb{C}^n$. In particular, if we let $P$ denote the maximal parabolic corresponding to the deletion of the first node, we get $G/P = \mathbb{CP}^{n-1}$. Now the maximal compact subgroup of $G$ is $K = \rm{SU}(n)$. In an appropriate basis, $P$ is the subgroup of $G$ of the form
$$ \begin{pmatrix} \ast & \ast & \cdots & \ast \\ 0 & \ast & \cdots & \ast \\ \vdots & \ast & \cdots & \ast \\ 0 & \ast & \cdots & \ast \end{pmatrix}. $$
Thus $K \cap P = \rm{U}(n-1)$ (to see this quickly, just recall that unitary matrices leave invariant the perp of an invariant subspace), where I'm thinking of $\rm{U}(n-1)$ as sitting in $K$ as
$$ \begin{pmatrix} (\det A)^{-1} & 0 \\ 0 & A \end{pmatrix}. $$
So by the result stated above we obtain the identification $\rm{SU}(n)/\rm{U}(n-1) = \mathbb{CP}^{n-1}$ given in the OP. Using the other maximal parabolics we can get a similar description for the complex Grassmannian of $k$-planes in $\mathbb{C}^n$.

Here are a couple of useful references:

Wang, Hsien-Chung, *Closed manifolds with homogeneous complex structure.* Amer. J. Math. **76** (1954). 1–32.

Wolf, Joseph A, *The action of a real semisimple group on a complex flag manifold. I. Orbit structure and holomorphic arc components.* Bull. Amer. Math. Soc. **75** (1969), 1121–1237.

Both these papers address the issue of finding complex structures on *compact* homogeneous spaces (so their results apply in particular to homogeneous spaces coming from compact Lie groups). There's a standing assumption of simply connectedness in Wang, but if I remember correctly, there is no such restriction in Wolf's paper.

**Edit:** Here's some more information that might be helpful. Let $K$ and $G$ be as above. A parabolic subgroup $P$ of $G$ admits "Levi decompositions" $P=LU$, where $U$ is the unipotent radical of $P$ and $L$ is a reductive group (called a Levi factor). In particular, every Levi factor $L$ has a compact real form $L_{\mathbb{R}}$. The different Levi factors of $P$ are all conjugate. There is a choice that is compatible with the choice of maximal compact $K$, and in this case we have $P \cap K = L_{\mathbb{R}}$.

isoften considered as a complex group -- just the usual definition, with complex entries. If this is your context, there's no trouble with $G/G_0$ being complex.) – Dave Anderson Nov 18 '10 at 17:52