Complex Analysis: Does $\int_{C(0,10)} f(z)$ Equal 0?

[mathmari]

Hey! :giggle:

Question 1:
If $f\in O(\Delta (0,1,15))$ then does it hold that $$\int_{C(0,10)}\frac{f(z)}{(z-6+4i)^5}\, dz=2\pi i\text{Res}\left (\frac{f(z)}{(z-6+4i)^5}, 6-4i\right )+\int_{C(0,6)}\frac{f(z)}{(z-6+4i)^5}\, dz$$ Do we maybe use here Cauchy theorem and then we get $$\int_{C(0,10)}\frac{f(z)}{(z-6+4i)^5}\, dz=0$$Question 2:
Is there a sequence of holomorphic polynomials $P_n(z), n=1,2,\ldots$ such that $$P_n(z)\rightarrow \frac{z(e^{6z}-1)(e^{4z}-1)}{\sin^2\left (\frac{z}{4}\right )\sin^2\left (\frac{2z}{5}\right )}$$ as $n\rightarrow \infty$, uniformly for $z\in \Delta (0,1,3)$ ?

since the convergence is uniform we get that $$\int \frac{z(e^{6z}-1)(e^{4z}-1)}{\sin^2\left (\frac{z}{4}\right )\sin^2\left (\frac{2z}{5}\right )}\, dz=\lim_{n\rightarrow \infty}\int P_n(z)$$ Do we have tocheck if this integral is equal to $0$ ?:unsure:

 

[I like Serena]

Question 1:
If $f\in O(\Delta (0,1,15))$ then does it hold that $$\int_{C(0,10)}\frac{f(z)}{(z-6+4i)^5}\, dz=2\pi i\text{Res}\left (\frac{f(z)}{(z-6+4i)^5}, 6-4i\right )+\int_{C(0,6)}\frac{f(z)}{(z-6+4i)^5}\, dz$$ Do we maybe use here Cauchy theorem and then we get $$\int_{C(0,10)}\frac{f(z)}{(z-6+4i)^5}\, dz=0$$

Hey mathmari!

What is $O(\Delta (0,1,15))$? (Wondering)

 

[soloenergy]

Hey! For question 1, yes, you can use Cauchy's theorem to show that the integral is equal to 0. For question 2, it is not necessary to check if the integral is equal to 0. The fact that the convergence is uniform means that the limit of the integral is equal to the integral of the limit. Therefore, you can just focus on showing that the polynomials converge to the given function. Hope this helps!