Using the Cauchy integral formula show that

${\oint}_{|z|=2}{\displaystyle \frac{{e}^{z}dz}{(z-1{)}^{2}(z-3)}}=-\frac{3}{2}je\pi .$

${\oint}_{|z|=2}{\displaystyle \frac{{e}^{z}dz}{(z-1{)}^{2}(z-3)}}=-\frac{3}{2}je\pi .$

Kallie Arroyo
2022-06-01
Answered

Using the Cauchy integral formula show that

${\oint}_{|z|=2}{\displaystyle \frac{{e}^{z}dz}{(z-1{)}^{2}(z-3)}}=-\frac{3}{2}je\pi .$

${\oint}_{|z|=2}{\displaystyle \frac{{e}^{z}dz}{(z-1{)}^{2}(z-3)}}=-\frac{3}{2}je\pi .$

You can still ask an expert for help

Rhett Pruitt

Answered 2022-06-02
Author has **5** answers

You have

$\frac{1}{(z-1{)}^{2}(z-3)}=-\frac{1}{4}\frac{1}{z-1}-\frac{1}{2}\frac{1}{(z-1{)}^{2}}+\frac{1}{4}\frac{1}{z-3}$

Now, you can use Cauchy's formula directly omitting the on $|z|\le 2$ holomorphic part belonging to $\frac{1}{z-3}$:

$\begin{array}{rcl}{\oint}_{|z|=2}\frac{{e}^{z}}{(z-1{)}^{2}(z-3)}dz& =& -\frac{1}{4}{\oint}_{|z|=2}\frac{{e}^{z}}{z-1}dz-\frac{1}{2}{\oint}_{|z|=2}\frac{{e}^{z}}{(z-1{)}^{2}}dz\\ & =& -\frac{1}{4}\cdot 2\pi i\cdot {e}^{1}-\frac{1}{2}\cdot 2\pi i\cdot {e}^{1}\\ & =& -\frac{3}{2}\pi ie\end{array}$

$\frac{1}{(z-1{)}^{2}(z-3)}=-\frac{1}{4}\frac{1}{z-1}-\frac{1}{2}\frac{1}{(z-1{)}^{2}}+\frac{1}{4}\frac{1}{z-3}$

Now, you can use Cauchy's formula directly omitting the on $|z|\le 2$ holomorphic part belonging to $\frac{1}{z-3}$:

$\begin{array}{rcl}{\oint}_{|z|=2}\frac{{e}^{z}}{(z-1{)}^{2}(z-3)}dz& =& -\frac{1}{4}{\oint}_{|z|=2}\frac{{e}^{z}}{z-1}dz-\frac{1}{2}{\oint}_{|z|=2}\frac{{e}^{z}}{(z-1{)}^{2}}dz\\ & =& -\frac{1}{4}\cdot 2\pi i\cdot {e}^{1}-\frac{1}{2}\cdot 2\pi i\cdot {e}^{1}\\ & =& -\frac{3}{2}\pi ie\end{array}$

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So i have this limit:

$\underset{x\to \mathrm{\infty}}{lim}x(\sqrt{{x}^{2}+4}-\sqrt{{x}^{2}+2})$

$\underset{x\to \mathrm{\infty}}{lim}x(\sqrt{{x}^{2}+4}-\sqrt{{x}^{2}+2})$

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For $f:\mathbb{R}\to \mathbb{R}$ and $\mathrm{\forall}n\in \mathbb{N}$ s. t. $n\ge 2$

$\underset{x\to 0}{lim}\{f(x)+f(nx)\}=0$

Prove/disprove :

$\underset{x\to 0}{lim}f(x)=0$

$\underset{x\to 0}{lim}\{f(x)+f(nx)\}=0$

Prove/disprove :

$\underset{x\to 0}{lim}f(x)=0$

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$\underset{h\to 0}{lim}\frac{{3}^{h}-1}{h}=\mathrm{ln}3$

evaluated?

evaluated?

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Function with compact support whose iterated antiderivatives also have compact support

Notation: If $f:\mathbb{R}\to \mathbb{R}$ is continuous, let us denote $If:\mathbb{R}\to \mathbb{R}$ its indefinite integral from 0, i.e., $(If)(x)={\int}_{0}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$, and iteratively ${I}^{k+1}f=I({I}^{k}f)$.

Remark: If f is a continuous function with support contained in the open interval ]0,1[ then If has support contained in ]0,1[ iff $(If)(1)=0$.

Main question: Does there exist a ${C}^{\mathrm{\infty}}$ function f with support contained in the open interval ]0,1[ such that ${I}^{k}f$ has support contained in ]0,1[ for every $k\ge 0$, or, equivalently, $({I}^{k}f)(1)=0$ for all $k\ge 0$?

Equivalent formulation: Does there exists a sequence $({f}_{k}{)}_{k\in \mathbb{Z}}$ of ${C}^{\mathrm{\infty}}$ functions each with support contained in the open interval ]0,1[, such that ${f}_{k-1}$ is the derivative of ${f}_{k}$?

Weaker question: Does there at least exist a continuous function f with the properties demanded in the main question?

Stronger question: Does there exist a ${C}^{\mathrm{\infty}}$ function f with compact support, whose Fourier transform vanishes identically on a nontrivial interval?

(A positive answer to the latter would imply a positive answer to the main question: rescale the function so its support is contained in ]0,1[, multiply it appropriately so its Fourier transform vanishes in a neighborhood of 0, and observe that the Fourier transform of ${I}^{k}f$ is, up to constants, ${\xi}^{k}$ times that of f.)

Edit: Before someone points out that the identically zero function fits the bill, I should add that I want my functions to not vanish identically.

Notation: If $f:\mathbb{R}\to \mathbb{R}$ is continuous, let us denote $If:\mathbb{R}\to \mathbb{R}$ its indefinite integral from 0, i.e., $(If)(x)={\int}_{0}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$, and iteratively ${I}^{k+1}f=I({I}^{k}f)$.

Remark: If f is a continuous function with support contained in the open interval ]0,1[ then If has support contained in ]0,1[ iff $(If)(1)=0$.

Main question: Does there exist a ${C}^{\mathrm{\infty}}$ function f with support contained in the open interval ]0,1[ such that ${I}^{k}f$ has support contained in ]0,1[ for every $k\ge 0$, or, equivalently, $({I}^{k}f)(1)=0$ for all $k\ge 0$?

Equivalent formulation: Does there exists a sequence $({f}_{k}{)}_{k\in \mathbb{Z}}$ of ${C}^{\mathrm{\infty}}$ functions each with support contained in the open interval ]0,1[, such that ${f}_{k-1}$ is the derivative of ${f}_{k}$?

Weaker question: Does there at least exist a continuous function f with the properties demanded in the main question?

Stronger question: Does there exist a ${C}^{\mathrm{\infty}}$ function f with compact support, whose Fourier transform vanishes identically on a nontrivial interval?

(A positive answer to the latter would imply a positive answer to the main question: rescale the function so its support is contained in ]0,1[, multiply it appropriately so its Fourier transform vanishes in a neighborhood of 0, and observe that the Fourier transform of ${I}^{k}f$ is, up to constants, ${\xi}^{k}$ times that of f.)

Edit: Before someone points out that the identically zero function fits the bill, I should add that I want my functions to not vanish identically.

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Any tip on how to calculate it?

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