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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?

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    Have you tried starting with the binomial series for $(z^2-1)^n$? – 2017-01-07
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    I actually took it into consideration, reminded of that by analytic functions, but I got stuck and then thought it might not help me. I will get back to this. – 2017-01-07
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    Cauchy's integral formula will serve very well here – 2017-01-07
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    Thank you! I really can't tell how could I overlook that option. I will try to work with your suggestions. – 2017-01-07

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Use the Cauchy integral formula $$f^{(n)}(z)=\frac{n!}{2\pi i}\int_C \frac{f(s)ds}{(s-z)^{n+1}}.$$

$$P_n(z)={1\over n! 2^n}{d^n\over dz^n}(z^2-1)^n$$ $$ =\frac{1}{2^{n+1}\pi i}\int_C \frac{(s^2-1)^nds}{(s-z)^{n+1}}.$$

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.

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    Did you use the residue theorem? How can I check that ${(z+1)^n\over z-1}$ is holomorphic on $\Bbb{C}\setminus {1}$? – 2017-01-07
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    Write $f(z)=(z+1)^n$ It is obvious that $f$ is analytic within and on the simple closed contour $C$. $\int_C \frac{f(z)dz}{z-z_0}=2\pi i f(z_0)$ – 2017-01-07
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    Good aspect, thank you! – 2017-01-07