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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

Bijesh K.S
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1 Answers1

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$a \operatorname{Re}(f) +b \operatorname{Im}(f)\ge c$ 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 $n \ne 1$.

Martin R
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  • (+1) You might consider mentioning the Little Picard Theorem to which you are appealing. – Mark Viola May 26 '21 at 20:36
  • @MarkViola: You don't need the Little Picard Theorem, Liouville's theorem (which is more elementary stuff) is sufficient. – Martin R May 26 '21 at 20:38
  • Is it really sufficient? Liousville implies and entire function is unbounded or constant. How does a lower bound imply boundedness? – Mark Viola May 26 '21 at 20:40
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    @MarkViola: If $f$ omits a half-plane then it omits some disk $B_r(a)$. Then $g(z) = r/(f(z)-a)$ is an entire function and bounded. – Martin R May 26 '21 at 20:41
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    The image of $f$ is contained in some half-plane $H$. Choose a disk $B_r(a)$ which is completely contained in the complement of $H$. Then $|f(z)-a| \ge r$ for all $z$, so that $g(z) = r/(f(z)-a)$ is entire with $|g(z)| \le 1$. – Martin R May 26 '21 at 20:56
  • Thanks! Much appreciated. – Mark Viola May 26 '21 at 21:00