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Determine $\displaystyle\int_{\gamma}{\frac{e^{z^{2}}}{z-1}dz}$, where $\gamma$ is the rectangule given by $x=0$, $x=3$, $y=-1$ e $y=1$.

My approach: If we consider the rectangule, then each side is given by $$\gamma_{1}:=z_{1}(t)=(3-i)t-(1-t)i$$

$$\gamma_{2}:=z_{2}(t)=(3+i)t-(1-t)(3-i)$$

$$\gamma_{3}:=z_{3}(t)=it+(3+i)(1-t)$$ $$\gamma_{4}:=z_{4}(t)=(1-t)i-it$$

My idea was use the Cauchy Integral Formula, because the function $f(z)=e^{z^2}$ is analytic over the rectangule, so

$$\displaystyle\int_{\gamma}{\frac{e^{z^{2}}}{z-1}dz}=2\pi i f(1)=2\pi ie$$

But I don't know how use the contour, thanks.

julios
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  • How are you applying the Cauchy Integral Formula, precisely? What you've written down makes no sense to me — how do you evaluate an expression in $t$ at $z=1$? – Ted Shifrin Jul 05 '18 at 23:11
  • @TedShifrin Hi, I edited my question. – julios Jul 05 '18 at 23:52
  • OK, good. You will never succeed at evaluating the line integral by explicit parametrization. As @KaviRamaMurthy said, you were just supposed to realize that the rectangle winds once around your point. In fact, you need to know whether you go counterclockwise or clockwise around the rectangle. The sign of the answer will switch if you switch orientation. – Ted Shifrin Jul 06 '18 at 00:47
  • @TedShifrin Sorry, but I don't understand your argument. what implies that "realize that the rectangle winds once around your point "? – julios Jul 06 '18 at 01:25
  • If you go once around the rectangle counterclockwise, you wind once (positively) around a point in its interior. Maybe your course hasn't talked about winding numbers. If you go once around a curve $\gamma$ counterclockwise, then the Cauchy Integral Formula says that you get $f(a) = \displaystyle{\frac1{2\pi i}\int_\gamma \frac{f(z)}{z-a},dz}$ for any point $a$ inside $\gamma$. – Ted Shifrin Jul 06 '18 at 02:10

1 Answers1

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Your answer is correct. Dependence on the contour is only through the index. It is implicitly assumed that you go round the rectangle only once which makes the index of $1$ w.r.t. the contour equal to $1$.

Kavi Rama Murthy
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  • Hi, thanks for your answer, but I don't understand what's mean that "Dependence on the contour is only through the index", and why the contour is equal to 1? Thanks! – julios Jul 06 '18 at 01:27
  • If you start from one corner of the rectangle and go round it 10 times you still have a closed contour. In that case $2\pi i f(1)$ becomes $20\pi i f(1)$ in Cauchy's Formula. Except for this it doesn't matter what contour you are considering as long as $z=1$ is inside. – Kavi Rama Murthy Jul 06 '18 at 05:22