Given: $\int_{1}^{z} \frac{a\sin\zeta+b\zeta\cos\zeta}{\zeta ^4}d\zeta $, prove that: $a+3b=0$ The integral:  $$\int_{1}^{z} \frac{a\sin\zeta+b\zeta\cos\zeta}{\zeta ^4}d\zeta $$
defines one-to-one function in a plane that is punctured in the origin. 
Prove that $a+3b=0$ 
I hope I translated this in an accurate way." punctured in the origin" means that we have the whole plane where the function is analytic except the origin ( if I got it right)  in the solution of this question they said that the integral defines one-to-one function if and only if the residue of the integrand: $\frac{a\sin\zeta+b\zeta\cos\zeta}{\zeta ^4}$ is zero at $z=0$. well Im not sure even I understood the phrasing of the question in order to make a linkage to the residue. what does it mean that the integral “defines” one to one function ? and what this has to do with residue? 
an addition to my question:
isn't right that for $log(z)$ ( which happens to appear in the expansion of the function that this integral defines ) to be bijective we need a branch cut? like for example: $ -\pi<Arg(z)<=\pi$? so whats the problem here? that the cut is only at one point and we need it to be including this point and continuing to a the whole left axis of $x$? which mean discluding the negative $x$ axis?
Note: I know how to find the final answer by finding the residue, no need to do that in the answer. 
 A: The Taylor series of $a\sin\zeta+b\zeta\cos\zeta$ is given by
$$ (a+b)\zeta -(a+3b)\frac{\zeta^3}{6}+(a+5b)\frac{\zeta^5}{120}+\ldots \tag{1} $$
hence by dividing by $\zeta^4$ and performing $\int_{1}^{z}(\ldots)\,d\zeta$ we get that the given function fulfills
$$f(z) =  - \frac{a+b}{z^2} - \frac{a+3b}{6}\,\log(z) + g(z)\tag{2} $$
with $g(z)$ being a holomorphic function. If $a+3b\neq 0$ such function has a branch point at the origin (a logarithmic singularity) preventing the function to be bijective on the punctured plane.
A: I try an interpretation: I note $f(\zeta)$ the function under the integral sign. Then the hypothesis is that $F(z)=\int_1^z f(\zeta)d\zeta$ is well defined; but here, the integral is the integral of $f(\zeta)$ over any curve $\gamma$ from $[0,1]$ to $\mathbb{C}$ such that $\gamma(0)=1$ and $\gamma(1)=z$, and $\gamma(t)\not = 0$ for all $t\in [0,1]$.  Now if $C^+$ is the part of the circle $C$ with center $0$, radius $1$, in $\Im{(z)}\geq 0$, and $C^-$ the part of the circle  in $\Im{(z)}\leq 0$ (starting from the point $1$, and going to the point $-1$), you have $\int_{C^+}f(\zeta)d\zeta=\int_{C^-}f(\zeta)d\zeta=F(-1)$. This show that $\int_{C}f(\zeta)d\zeta=0$; but this is $2i\pi {Res}(f,0)$ and we are done. 
