Pages 69 and 70 in John D'Angelo's "An Introduction To Complex Analysis and Geometry" book. Hi: I'm reading John D'Angelo's textbook "an introduction to complex analysis and geometry".
I'm currently going through pages 69 and 70 and I'm pretty lost with respect to most of the material in these pages. I will write down my questions one at a time and answers/explanations to any of them are appreciated.
A) First example that I'm confused about.
Example 1.4, he states : "The following identity holds for $|z| < 1$ and $d \in 
\boldsymbol{N}$.
$\frac{d!}{(1-z)^{d+1}} = \sum_{n-0}^\infty (n +d)(n+d-1)...(n+1)z^{n}$.
John states that the result is obtained by repeatedly differentiating the geometric series and changing the index of summation. Can someone show what he means by this ?
Aside: I'm going to state a theorem here ( and not worry about the proof of it. The proof seems very tricky so I'm willing to accept it on faith atleast for now ) because it's used later in other examples that I don't understand.
Theorem 1.1: Let $p(n)$ be a polynomial of degree $d$. Then $\sum_{n=0}z^{n}$ is a polynomial $q$ in $\frac{1}{1-z}$ of degree $d+1$ with no constant term.
End of Theorem Statement.
B) Second example that I'm confused about.
Example 1.5: Let $p(n) = n^2$. First write $n^2 = (n+2)(n+1) - 3(n+1) +1$. We then have for $|z| < 1$, 
$\sum_{n=0}^{\infty} n^{2}z^{n} = \sum_{n=0}^{\infty} (n + 2)(n+1) z^{n} -
\sum_{n=0}^{\infty} 3(n+1)z^{n} + \sum_{n=0}^{\infty}z^{n}$ 
$ = \frac{2}{(1-z)^{3}} - \frac{3}{(1-z)^2} + \frac{1}{1-z}$
I understand this example except for how the first two terms of the final expression are obtained. Is Theorem 1.1 being used because the degree of the first polynomial is 2 and the degree of the second polynomial is 1 ? If that's correct, then where does the 2 in the numerator of the first term come from ?
The last confusion is an exercise:
Excercise: 4.3: In each case, find an explicit function (as a rational function) for the series and state where the formula is valid.
A) $\sum_{n=0}^\infty (3+4i)^n z^n$.
B) $\sum_{n=0}^{\infty} (n^3-1)z^n$.
I have no idea how to even attempt to solve this exercise.
End of Confusions.
Thank you very much for any explanations. Also, if anyone knows where I can find or purchase solutions to the exercises in this book, please let me know. I'm not a student so just trying to learn for the fun of it and I find that the solutions can really help because I have them for a complex analysis book by Serge and they help a lot. I have looked all over the internet with no success. Thanks again.
 A: A Use induction by $d$. The inductive step is:
$$\frac{d}{dz}( \frac{d!}{(1-z)^{d+1}} ) = \frac{d}{dz} \sum_{n=0}^\infty (n +d)(n+d-1)...(n+1)z^{n}$$
The LHS is easy to derivate. The RHS is
$$ \sum_{n=0}^\infty (n +d)(n+d-1)...(n+1)nz^{n-1}$$
Now, observe that you have $0$ when $n=0$, thus 
$$ \sum_{n=0}^\infty (n +d)(n+d-1)...(n+1)nz^{n-1}= \sum_{n=1}^\infty (n +d)(n+d-1)...(n+1)nz^{n-1}= \sum_{m=0}^\infty (m+d+1)(m+d)...(m+1)z^{m}$$
the last step being the reindexing $m=n-1$.
B
You can prove that for any polynomial $P$ of degree $d$, there exist unique constants $C_0,..,C_d$ such that
$$P(n)=C_d(n +d)(n+d-1)...(n+1)+C_{d-1}(n+d-1)...(n+1)+...+C_0$$
The RHS is helpfull because when multiplied with $z^n$ becomes the RHS of Example 1.4. 
Now, when $P(n)$ is given, you can easily find $C_0,..,C_d$, the exercise skips the calculation.
If you want to prove the above statement, this is an easy induction on $d$. There exists only one Polynomial of degree exactly $d$ on the RHS, which allows you calculate $C_d$, and then if you subtract $C_d(n +d)(n+d-1)...(n+1)$ from both sides you end up in the inductive case.
Theorem 1.1 Is wrong as stated. I think that what you want is: Then $\sum_{n=0}^\infty P(n)z^n$ is a polynomial $q$ in $\frac{1}{1-z}$ of degree $d+1$ with no constant term. 
This follows immediately from combining the above observation about $C_0,..,C_k$ with Exercise 1.4. 
Last confusion
A) Is geometric.
B) Exactly as above, find constants $C_0,C_1,C_2,C_3$ such that
$$n^3-1 =C_3(n+3)(n+2)(n+1)+C_2(n+2)(n+1)+C_1(n+1)+C_0 \,.$$
The fastest way is just pluggin in $n=-1, -2, -3$ and $0$.
A: The geometric series : $1+z+z^2+z^3+...=\sum z^n=f(z)$ defines a function for $|z|<1$
And $f(z)=\frac1{1-z}$.  $f'(z)$ can be computed in two ways:directly and differentiating the series termwise.
So $f'(z)= \frac1{(1-z)^2}=1+2z+3z^2+...$ we have shown example with $d=1$.  If  you differentiate again  you get the equation for $d=2$.And so on..
The previous example is used--notthe theorem--
a) write $w=(3+4i) z$ and recall the sum :$\sum w^n$
b)as in the last example write similarly that $p(n)=n^3=(n+3)(n+2)(n+1)+A(n+2)(n+1)+B(n+1)+C$ and follow the same steps ofthe example..
