I will follow route #3, via the generating function. You are correct, this is indeed a hypergeometric function. Maple quickly returns $$ { }_{2}F_{1}\left(\left[1-iz,1+iz\right], \left[1\right] | y\right) $$. More useful than this though is the ODE which this satisfies. gfun quickly returns $$ (y^2-y)w''(y) + (3y-1)w'(y) + (1+z^2)w(y), w(0)=1 $$ Translating this to a recurrence on the coefficients (i.e. u(p)y^p), we get $$-(p^2+2p+1)u(p+1) + (p^2+2p+z^2+1)u(p), u(0)=1$$ which has $$\frac{\Gamma(p+1-iz)\Gamma(p+1+iz)}{\Gamma(1-iz)\Gamma(1+iz)\Gamma(p+1)^2}$$ as solution. Seems you should be able to go from there to obtain your answer.

I will follow route #3, via the generating function. You are correct, this is indeed a hypergeometric function. Maple quickly returns $ { }_{2}F_{1}\left(\left[1-iz,1+iz\right], \left[1\right] | y\right) $. More useful than this though is the ODE which this satisfies. gfun quickly returns $$ (y^2-y)w''(y) + (3y-1)w'(y) + (1+z^2)w(y), w(0)=1 $$ Translating this to a recurrence on the coefficients (i.e. u(p)y^p), we get $$-(p^2+2p+1)u(p+1) + (p^2+2p+z^2+1)u(p), u(0)=1$$ which has $$\frac{\Gamma(p+1-iz)\Gamma(p+1+iz)}{\Gamma(1-iz)\Gamma(1+iz)\Gamma(p+1)^2}$$ as solution. Seems you should be able to go from there to obtain your answer.