MHB Complex Variables - Legendre Polynomial

[joypav]

We define the Legendre polynomial $P_n$ by
$$P_n (z)=\frac{1}{2^nn!}\frac{d^n}{dz^n}(z^2-1)^n$$
Let $\omega$ be a smooth simple closed curve around z. Show that
$$P_n (z)=\frac{1}{2i\pi}\frac{1}{2^n}\int_\omega\frac{(w^2-1)^n}{(w-z)^{n+1}}dw$$

What I have:

We know $(w^2-1)^n$ is analytic on and inside $\omega$ and z lies within $\omega$.
So, by Cauchy's Formula,
$$\frac{1}{2i\pi}\frac{1}{2^n}\int_\omega\frac{(w^2-1)^n}{(w-z)^{n+1}}dw$$
$$=\frac{1}{2i\pi}\frac{1}{2^n}\frac{2i\pi}{n!}\frac{d^n}{dw^n}(w^2-1)^n\rvert_{w=z}$$
$$=\frac{1}{2^n}\frac{1}{n!}\frac{d^n}{dw^n}(w^2-1)^n\rvert_{w=z}$$
$$=\frac{1}{2^nn!}\frac{d^n}{dz^n}(z^2-1)^n=P_n(z)$$

Now, take $\omega$ to be the circle of radius $\sqrt{|z^2-1|}$ centered at z. Show that
$$P_n(z)=\frac{1}{2\pi}\int_0^{2\pi}\left(z+\sqrt{z^2-1}\cos(\theta)\right)^{\!n}\,d\theta$$

What I have:

I am confused about how to apply the previous problem. If I could get a push in the right direction I would be very appreciative!

 

[Ackbach]

So, we have
\begin{align*}
\omega&= z+\sqrt{|z^2-1|} \, e^{i\theta} \\
\omega - z &=\sqrt{|z^2-1|} \, e^{i\theta} \\
d\omega &=\sqrt{|z^2-1|} \cdot i \cdot e^{i\theta} \, d\theta \\
\omega^2 &=z^2+2z\sqrt{|z^2-1|} \, e^{i\theta} + |z^2-1| \, e^{2i\theta}.
\end{align*}
Now we have the necessary ingredients to plug into what you showed before, namely, $\displaystyle P_n(z)=\frac{1}{2\pi i} \, \frac{1}{2^n} \, \oint_{\omega}\frac{(\omega^2 - 1)^n}{(\omega-z)^{n+1}} \, d\omega,$ to obtain
$$
P_n(z)=\frac{1}{2\pi i} \, \frac{1}{2^n} \, \int_{0}^{2\pi}\frac{\displaystyle\left(z^2+2z\sqrt{|z^2-1|} \, e^{i\theta}+|z^2-1| \, e^{2i\theta} - 1\right)^{n}}{\displaystyle\left(\sqrt{|z^2-1|} \, e^{i\theta}\right)^{n+1}}
\cdot\sqrt{|z^2-1|}\cdot i \cdot e^{i\theta} \, d\theta.
$$
There's definitely some simplification possible here. The hope is that, once you do simplify, you will get the desired result above. It does not seem impossible that you could get equality.

 

[joypav]

Thank you!

(for the sake of completion)
$$
P_n(z)=\frac{1}{2\pi i} \, \frac{1}{2^n} \, \int_{0}^{2\pi}\frac{\displaystyle\left(z^2+2z\sqrt{|z^2-1|} \, e^{i\theta}+|z^2-1| \, e^{2i\theta} - 1\right)^{n}}{\displaystyle\left(\sqrt{|z^2-1|} \, e^{i\theta}\right)^{n+1}}
\cdot\sqrt{|z^2-1|}\cdot i \cdot e^{i\theta} \, d\theta
$$
$$
=\frac{1}{2\pi} \, \frac{1}{2^n} \, \int_{0}^{2\pi}\frac{\displaystyle\left(z^2+2z\sqrt{|z^2-1|} \, e^{i\theta}+|z^2-1| \, e^{2i\theta} - 1\right)^{n}}{\displaystyle\left(\sqrt{|z^2-1|} \, e^{i\theta}\right)^{n}}d\theta
$$
$$
=\frac{1}{2\pi} \, \frac{1}{2^n} \, \int_{0}^{2\pi}\frac{\displaystyle\left(z^2+2z\sqrt{|z^2-1|} \, (\cos(\theta)+i\sin(\theta))+|z^2-1| \, (\cos(2\theta)+i\sin(2\theta)) - 1\right)^{n}}{\displaystyle\left(\sqrt{|z^2-1|} \, (\cos(\theta)+i\sin(\theta))\right)^{n}}d\theta
$$
$$
=\frac{1}{2\pi} \, \frac{1}{2^n} \, \int_{0}^{2\pi}\frac{\displaystyle\left(z^2+2z\sqrt{|z^2-1|} \, (\cos(\theta)+i\sin(\theta))+|z^2-1| \, (2\cos^2(\theta)-1+2i\sin(\theta)\cos(\theta)) - 1\right)^{n}}{\displaystyle\left(\sqrt{|z^2-1|} \, (\cos(\theta)+i\sin(\theta))\right)^{n}}d\theta
$$
$$
=\frac{1}{2\pi} \, \frac{1}{2^n} \, \int_{0}^{2\pi}\frac{\displaystyle\left((2)^n(\sqrt{|z^2-1|}) \, (\cos(\theta)+i\sin(\theta))(z+\sqrt{z^2-1}\cos(\theta) \, \right)^{n}}{\displaystyle\left(\sqrt{|z^2-1|} \, (\cos(\theta)+i\sin(\theta))\right)^{n}}d\theta
$$
$$
=\frac{1}{2\pi} \,\int_{0}^{2\pi}{\displaystyle\left(z+\sqrt{z^2-1}\cos(\theta) \, \right)^{n}}d\theta.
$$