- #1

joypav

- 151

- 0

$$P_n (z)=\frac{1}{2^nn!}\frac{d^n}{dz^n}(z^2-1)^n$$

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!