Solving an Ordinary Differential Equation for a complex function

Question

What is the solution to the ODE $$zf'(z)+\lambda f(z)-c = z(z+i)^{-(\lambda +2)}$$ where $f$ is a complex function, $\lambda$ is a complex number and c is an arbitrary constant. i is the usual complex number. Also $z\in\mathbb{U}$, the upper half of the complex plane.

Answer 1

Hint

Such function is analytic, i.e. of the form $$f(z)=\sum_{k=0}^\infty a_kz^k.$$

Answer 2

Assuming the use of the main branch of all powers $w^a=\exp(a\,{\rm Ln}(w))$ on $\Bbb U$ for uniqueness, one finds $$ (z^λf(z))'=\frac{z^λ}{(z+i)^{λ+2}}=\frac1{(z+i)^2}\left(1-\frac{i}{z+i}\right)^λ =-\frac{i}{λ+1}\frac{d}{dz}\left(1-\frac{i}{z+i}\right)^{λ+1} $$ which can now be integrated to $$ z^λf(z)=-\frac{i}{λ+1}\left(\frac{z}{z+i}\right)^{λ+1}+C. $$

The modified equation can be solved similarly, it just contains additionally the integral of $cz^{λ-1}$ on the right side.