TITLE: $\int_C \frac{f(z)}{(z-1)^{2020}}dz$
QUESTION [7 upvotes]: Let $f(z)=u(x,y)+iv(x,y)$ be an entire function such that $au+bv\ge \ln(ab), a>1,b>1.$ Then evaluate
$$\int_C \frac{f(z)}{(z-1)^{2020}}dz,$$ where $C$ is an equilateral triangle of side $1$ with centroid at $z=1.$

It seems that I can use the Cauchy's Integral formula here and by doing so the integral would be
$$\frac{2\pi i}{2019!}f^{(2019)}(1)$$
I have no idea how to connect the first part of the question in solving this problem. Help please

REPLY [6 votes]: $a \operatorname{Re}(f) +b \operatorname{Im}(f)\ge c$ with real constants $a, b, c$ and $(a, b) \ne (0, 0)$, means  that the image of $f$ is contained in some half-plane. For an entire function that implies that $f$ is constant.
So it does not matter that $a>1$ and $b>1$ as long as not both are zero. It also does not matter that the lower bound is $c = \ln(ab)$, or that $C$ is an equilateral triangle with a given side length. These are all red herrings.
Since $f$ is constant, $\int_C \frac{f(z)}{(z-1)^n}dz = 0$ for any closed curve $C$ not going through $z=1$, and any integer $n \ne 1$.