using power series expansion to find a holomorphic function which solves a differential equation Using power series expansions, 
find a function $f$ which is holomorphic on the unit disk $D:=$ {$z\in\mathbb C:|z|<1$}
and solves the differential equation 
$(1-z^2)f''(z)-4zf'(z)-2f(z)=0$
for $z\in D$ along with the initial conditions
$f(0)=0$, and $f'(0)=1$
I have been given the hint "for any $c_0,c_1 \in\mathbb C$, we have 
$$\sum_{j=0}^{\infty}  (c_0z^{2j} +c_1z^{2j+1}) =\frac{c_0+c_1z}{1-z^2} $$ as long as $|z|<1$, why?"
But i still don't understand how to answer the question or what the hint means!
Any help would be much appreciated!
Many thanks
 A: To check your hint, just develop $\frac{c_0+c_1 z}{1-z^2}$ in power series.
Now to see why it helps, you have to find power series expansion of $f(z)$.
To achieve this, write $f(z)=\sum_{n=0}^{\infty} a_n z^n$, then replace $f$ with this series in the differential equation. Develop, and manage to get a recurrence equation between, $a_{n}$, $a_{n-1}$ and $a_{n-2}$, and also conditions on $a_0$ and $a_1$, using your knowledge that $f(0)=0$ and $f'(0)=1$.
It's the usual way to solve differential equations with power series. If you have difficulties with this, I'll develop a bit more.
You will have to play with indices, like
$$f'(z)=\sum_{n=1}^\infty n a_n z^{n-1}=\sum_{n=0}^\infty (n+1) a_{n+1} z^n$$
$$f''(z)=\sum_{n=2}^\infty n (n-1) a_n z^{n-2}=\sum_{n=0}^\infty (n+1) (n+2) a_{n+2} z^{n}$$
And you will also have to use the fact that power series expansion is unique (that is, if you have $\sum_{n=0}^\infty [expression \ in \ a_n, a_{n-1}, a_{n-2}] z^n = 0$, then the expression is null for all $n$, thus you have a recurrence equation to solve.

Here is a more detailed approach.
First we just replace $f$, $f'$ and $f''$ with their expressions above, in the differential equation:
$$(1-z^2)f''(z)-4zf'(z)-2f(z)=f''(z)-z^2f''(z)-4zf'(z)-2f(z)$$
$$=\left(\sum_{n=0}^\infty (n+1) (n+2) a_{n+2} z^{n}\right) - \left(\sum_{n=2}^\infty n (n-1) a_n z^{n}\right) - 4 \left( \sum_{n=0}^\infty n a_n z^{n} \right) - 2 \left(\sum_{n=0}^\infty a_{n} z^n \right)$$
$$=\left(\sum_{n=0}^\infty (n+1) (n+2) a_{n+2} z^{n}\right) - \left(\sum_{n=0}^\infty n (n-1) a_n z^{n}\right) - 4 \left( \sum_{n=0}^\infty n a_n z^{n} \right) - 2 \left(\sum_{n=0}^\infty a_{n} z^n \right)$$
In the preceding last step, I only replace $n=2$ with $n=0$ in the bound of one of the sums, which does not change it, since coefficient $n(n-1)a_n$ guarantees we add $0$. But it's necessary, in order to write the following recurrence equation for all $n \geq 0$.
Since our differential equation is supposed to hold, this whole series is everywhere null. Hence its coefficients in $z^n$ are all zero.
Now we have to put together terms in $z^n$. This yields
$$\left( (n+1) (n+2) a_{n+2} - n (n-1) a_n -4n a_n -2 a_n \right) z^n$$
So, we have the recurrence equation
$$ (n+1) (n+2) a_{n+2} - n (n-1) a_n -4n a_n -2 a_n = 0$$
Or,
$$(n+1) (n+2) a_{n+2} - (n^3 + 3n +2) a_n = 0$$
$$(n+1) (n+2) a_{n+2} - (n+1)(n+2) a_n = 0$$
Hence $a_{n+2}=a_{n}$ for $n \geq 0$. Now we have to find $a_0$ and $a_1$ to know the whole series. But since $f(0)=0$ we have $a_0=0$, and $f'(0)=1$ gives $a_1=1$.
We have therefore $a_{2n}=0$ and $a_{2n+1}=1$, which means
$$f(z)=\sum_{n=0}^\infty z^{2n+1} = z \sum_{n=0}^\infty (z^2)^n = \frac{z}{1-z^2}$$
As a final step, you always have to check that your series really exists, e.g. that its radius of convergence is not null. But here it's immediate that this radius is $1$.
Of course, you may want to check directly that $\frac{z}{1-z^2}$ is the solution of your equation, but it's less necessary (except to make sure there is no mistake in the above proof).
