Expected number of leaves of a planar tree with $n$ nodes I want to find out the expected number of leaves of a planar tree with $n$ nodes using the bivariate generating function for the number of leaves in a planar tree.
For the (univariate) generating function we have $$P(z)=\frac{z}{1-P(z)}$$Solving the quadratic equation gives us $$P(z)=\frac{1-\sqrt{1-4z}}{2}$$

Does this imply that the bivariate generating function is given by $$P(z,u)=\frac{z}{1-u\frac{1-\sqrt{1-4z}}{2}}$$?


And for the expected number of leave I have to evaluate $$\frac{[z^n](\frac{\partial}{\partial u}P(z,u))\big |_{u=1}}{[z^n]P(z,1)}$$?

 A: The combinatorial class for plane trees with leaves marked is
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{P} = \mathcal{Z} \times \mathcal{U} +
\mathcal{Z}\times \textsc{SEQ}_{\ge 1}(\mathcal{P})$$
This gives the functional equation
$$P(z) = uz + z \frac{P(z)}{1-P(z)}$$
or
$$z = \frac{P(z)(P(z)-1)}{u(P(z)-1)-P(z)}.$$
Now applying the residue operator for $n\ge 1$ we get
(the constant term of $P(z)$ is zero)
$$[z^n] P(z) = \frac{1}{n} [z^{n-1}] P'(z)
= \frac{1}{n}
\; \underset{z}{\mathrm{res}} \;
\frac{1}{z^n} P'(z).$$
We put $w=P(z)$ and find
$$\frac{1}{n}
\; \underset{w}{\mathrm{res}} \;
\frac{(u(w-1)-w)^{n}}{w^{n} (w-1)^{n}}.$$
Extracting the coefficient on $[u^k]$ where we must have $1\le k\le n-1$
we obtain
$$\frac{1}{n} {n\choose k}
\; \underset{w}{\mathrm{res}} \;
\frac{(w-1)^k (-1)^{n-k} w^{n-k}}
{w^{n} (w-1)^{n}}
\\ = \frac{1}{n} {n\choose k}
\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{k} (1-w)^{n-k}}
\\ = \frac{1}{n} {n\choose k} {k-1+n-1-k\choose k-1}
= \frac{1}{n} {n\choose k} {n-2\choose k-1}
= \frac{1}{k} {n-1\choose k-1} {n-2\choose k-1}.$$
As a sanity check we get for the sequence of leaf counts with $k$
increasing from $k=1$ to $n-1$ the following  values
$$1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 20, 10, 1, 
1, 15, 50, 50, 15, 1, \ldots $$
which is OEIS A001263, the Narayana  numbers.
What we have is $N(n-1,k).$  When we look this up in turn at Wikipedia
on Narayana
numbers
we find that the number of ordered rooted trees with $n+1$ vertices and
$k$ leaves is $N(n,k)$ which means we have the correct value.
We continue with the expectation where we first need the total count of
these trees which is
$$\frac{1}{n} \sum_{k=1}^{n-1} {n\choose k} {n-2\choose k-1}
= \frac{1}{n} \sum_{k=1}^{n-1} {n\choose k} {n-2\choose n-k-1}
= \frac{1}{n} {2n-2\choose n-1}$$
by Vandermonde.  These are the  familiar Catalan numbers $C_{n-1}.$ We
get for the sum total of leaves
$$\sum_{k=1}^{n-1} {n-1\choose k-1} {n-2\choose k-1}
= \sum_{k=0}^{n-2} {n-1\choose k} {n-2\choose k}
\\ = \sum_{k=0}^{n-2} {n-1\choose k} {n-2\choose n-2-k}
= {2n-3\choose n-2} = {2n-3\choose n-1}.$$
again by Vandermonde. This will produce for $n\ge 2$
$$\frac{n-1}{2n-2} {2n-2\choose n-1}
= \frac{1}{2} {2n-2\choose n-1}.$$
Looking it up we find OEIS A088218
where it says "total number of leaves in all rooted ordered trees with
$n$ edges" is $\frac{1}{2} {2n\choose n}.$ As before this is $n+1$
vertices so we definitely have the right answer.
 We are ready at last to compute the queried expectation of the 
number of leaves for $n\ge 2$
$$\frac{(2n-3)!}{(n-2)! \times (n-1)!}
\times n \times \frac{(n-1)! \times (n-1)!}{(2n-2)!}
\\ = \frac{1}{2n-2} \times n \times (n-1).$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2} n.}$$
Remark. In  the above  we have used  OP's functional  equation for
$P(z)$ to  deduce the  type of  tree being  sought (ordered  rooted or
rooted plane tree).
