Define complex-valued Legendre Polynomials by $P_n(z)={1\over n! 2^n}{d^n\over dz^n}(z^2-1)^n$. show $P_n(1)=1$ and $p_n(-1)=(-1)^n$.
Although this is not the first or the second time I am exposed to these polynomials, it was defined on the spot in a complex analysis exercise, along with questions about integration and winding numbers, by which I am saying that other representation of this formula are not immediate if unexplained. I wonder how I can adjust the complex analysis approach to this, or, generally, how I can avoid using formulas I haven't arrived at n my own. What I have done so far is: ${d^n\over d z^n} (z-1)^n(z+1)^n=...=n!(z+1)^n+(z-1)q_n(z) $, which could possibly work analogously for $z=-1$, except that I can't genuinely reason that differentiation as it is complex, and, although possibly correct, it should be clarified as legitimate. Can you help me with this?
$$P_n(z)={1\over n! 2^n}{d^n\over dz^n}(z^2-1)^n$$ $$ =\frac{1}{2^{n+1}\pi i}\int_C \frac{(s^2-1)^nds}{(s-z)^{n+1}}.$$
Sub $z=1$,because $$ \frac{(s^2-1)^n}{(s-1)^{n+1}}=\frac{(s-1)^n(s+1)^n}{(s-1)^{n+1}}=\frac{(s+1)^n}{s-1}$$ it reveals that
$$P_n(1)=\frac{1}{2^{n+1}\pi i}\int_C \frac{(s+1)^nds}{s-1}=\frac{1}{2^{n+1}\pi i}\cdot2\pi i\cdot(1+1)^n=1.$$
The values of $P_n(-1)=(-1)^n$ can be found in a similar way.