By differentiation or otherwise, prove that $(c(z))^2 +(s(z))^2 = 1$ Assume that the two power series $s(z) = \sum a_nz^n$ and $c(z) =\sum b_nz^n$ are convergent for all $z\in \mathbb{C},$ and that they satisfy the relations $s'(z) = c(z), c'(z) =-s(z).$ Deduce the identities
$$ a_n = -a_{n-2}/(n(n-1)) , b_n=-a_{n-2}/(n(n-1))
$$(To me, it looks like there is a typo for $a_n$)
If further $s(0)=0, c(0) =1,$ determine $s(z)$ and $c(z)$ completely. By differentiation or otherwise, prove that $(c(z))^2 +(s(z))^2 = 1$
To me it looks like $s(z) = sin(z)$ and $c(z) = cos(z)$ but I don't know what to do with this information.
Also while taking the derivate of the power series in an attempt to deduce the identities, I have
$(n+1)a_{n+1}=b_n$ and it does not seem like I have the correct identity
 A: hint
if $f'(z) = 0$ for all $z$ then $f(z) = $ a constant .
this means that for all $z, f(z) = f(0).$
let $f(z) = [s(z)]^2 + [c(z)]^2.$

Addendum 
Per request of OP 

determine $s(z)$ and $c(z)$ completely.

Actually, the necessary analysis has already been provided by the comments of leslie townes.  All I need to do is fill in all of the ground work.
First of all, I completely agree with the response of Hagen von Eitzen, so following
through with the first part of my answer (before the Addendum) is the place to start.
$f'(z) = s(z)s'(z) + c(z)c'(z) = s(z)[c(z)] + c(z)[-s(z)] = 0.$
This establishes that for all $z, f(z) = f(0) = 1,$ which completely resolves
the question in the title of the query.
Next, since $s'(z) = c(z)$ and $c'(z) = -s(z)$, you have that 
$s''(z) = c'(z) = -s(z)$ and that $c''(z) = -s'(z) = -c(z).$ 
Therefore,
$$\sum_{n=2}^\infty (n)(n-1)[a_n]z^{(n-2)} = s''(z) = -s(z) =
\sum_{n=0}^\infty -a_nz^n = \sum_{n=2}^\infty -a_{(n-2)}z^{n-2}.$$
As a result, you have that for each $n \in \{2,3,4,\cdots\}$ 
$(n)(n-1)a_n = -a_{(n-2)}.$
By identical analysis, for each $n \in \{2,3,4,\cdots\}$ 
$(n)(n-1)b_n = -b_{(n-2)}.$
All that remains is to compute $a_0, a_1, b_0, b_1.$
Since $a_0 = s(0) = 0$ and $b_0 = c(0) = 1$, 
it is immediate that $a_0 = 0$ and $b_0 = 1.$
Also, remember that for all $z$, you have that $s'(z) = c(z)$ and $c'(z) = -s(z).$ 
Therefore, since $a_1 = s'(0) = c(0) = 1$, and $b_1 = c'(0) = -s(0) = 0$, 
it is immediate that $a_1 = 1$ and $b_1 = 0.$
A: For the desired equation, you do not need the power series, but only the given relations between derivatives, $s'=c$ and $c'=-s$ (plus that the equation holds at $z=0$)
A: If $s(0) = 0$ and $c(0)=1$ then you can prove your conjecture $s = \sin$ and $c = \cos $ by considering the function
$$
 F(z) = (s(z) - \sin(z))^2 + (c(z) - \cos(z))^2 \, .
$$
We have $F(0) = 0$ and
$$
 F'(z) = 2(s(z) - \sin(z))(c(z)-\cos(z)) + 2(c(z) - \cos(z))(-s(z) + \sin(z)) = 0
$$
so that $F$ is identially zero. This implies $s(x) = \sin(x)$ and $c(x) = \cos(x)$ for all $x \in \Bbb R$, and then  $s = \sin$ and $c = \cos $ because of the identity theorem for holomorphic functions.
A: Hint : Let
$$e : z \mapsto c(z) + i s(z)$$
You get $e'(z)=ie(z)$. Can you proceed further ?
