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Let $a$ be given and let $\gamma$ be a simple closed curve oriented counterclockwise. Suppose that $\gamma$ encloses the numbers $z=\pm ai$. Show that

$\int \frac{e^z}{z^2+a^2}dz = \frac{2\pi i\text{sin}a}{a}$.

Usually with these problems I can find a way to rewrite the integrand and apply Cauchy's theorem or the deformation theorem. I'm not sure how to begin here however.

Also:

Let $f$ be analytic "on and inside" a simple closed curve $\gamma$. Suppose that $f=0$ on $\gamma$. Show that $f=0$ inside $\gamma$.

My proof:

Let $z_0$ be inside $\gamma$. Then by CIF we have

$f(z_0) = \int_{\gamma} \frac{f(z)}{z-z_0}dz$. But since $f=0$ on $\gamma$, we have $f(z_0) = \int_{\gamma} \frac{f(z)}{z-z_0}dz = 0$. Since $z_0$ was an arbitrary point inside $\gamma$, $f=0$ inside $\gamma$.

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  • $\begingroup$ It is a direct application of residue theorem. (Have you heard of it?) What is left is to calculate residues at $\pm ai$ and add them. $\endgroup$ Feb 26, 2019 at 17:28
  • $\begingroup$ I haven't learned about the residue theorem. Is there some other way to do it? $\endgroup$
    – mXdX
    Feb 26, 2019 at 17:29
  • $\begingroup$ Then maybe partial fraction is the way to do it. $\endgroup$ Feb 26, 2019 at 17:30
  • $\begingroup$ That's what I was thinking, but I wasn't sure how to decompose this fraction. Any advice? $\endgroup$
    – mXdX
    Feb 26, 2019 at 17:31

2 Answers 2

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\begin{align} \oint_{\gamma} \frac{e^z}{z^2+a^2}\mathrm dz&=\oint_{\gamma} \frac{e^z}{(z-ia)(z+ia)}\mathrm dz\\ &=\frac {1}{2ia}\oint_{\gamma} \left (\frac{e^z}{z-ia}-\frac{e^z}{z+ia}\right)\mathrm dz\\ &=\frac {1}{2ia}(2\pi i e^{ia}-2\pi i e^{-ia})\\ &=2\pi i\frac {e^{ia}-e^{-ia}}{2i\,a}\\ &= \frac{2\pi i\sin a}{a} \end{align}

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  • $\begingroup$ Does this use the information that the closed curve contains the roots? $\endgroup$
    – mXdX
    Feb 26, 2019 at 17:37
  • $\begingroup$ Would you mind explaining how you got from the 2nd line to the 3rd? Everything else makes sense to me. $\endgroup$
    – mXdX
    Feb 26, 2019 at 17:40
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    $\begingroup$ Cauchy's integral formula. $\endgroup$
    – cqfd
    Feb 26, 2019 at 17:52
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    $\begingroup$ I wasn't familiar with CIF before, but now I am after reading your link -- makes the problem very easy. Quick question about it: for the f(z) (numerator in the RHS of the formula) inside the integral, we are talking about the values of f(z) that are on the curve gamma, right? and the f(a) (LHS of the formula) is a point inside? Referring to your link for the notation. $\endgroup$
    – mXdX
    Feb 26, 2019 at 18:12
  • $\begingroup$ $f (a) $ need not be inside the region enclosed by the path. But note that $a $ is inside the contour. In some sense, CIF allows one to calculate the image of an interior point just by using the behaviour of the function at the boundaries. $\endgroup$
    – cqfd
    Feb 26, 2019 at 18:21
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Hint:

$$\int_\gamma \dfrac{e^z}{z^2+a^2}\mathrm dz=\int_\gamma\dfrac{e^z}{(z-ia)(z+ia)}\mathrm dz=2\pi i (\text{Res}_{z=ia}+\text{Res}_{z=-ia})$$


$$\dfrac{e^z}{(z-ia)(z+ia)}=\dfrac{e^z}{2ia}\left[\dfrac{(z+ia)-(z-ia)}{(z+ia)(z-ia)}\right]=\dfrac{1}{2ia}\left[\dfrac{e^z}{z-ia}-\dfrac{e^z}{z+ia}\right]$$

Doing the Partial Fraction Decomposition you can use the Cauchy Integral Formula for Derivatives to solve this Contour Integral which is stated as follows:

$$f^{(k)}(w)=\dfrac{k!}{2\pi i}\oint \dfrac{f(z)}{(z-w)^{k+1}}\mathrm dz$$

Further make use of De Moivre's result which immediately gives you the result as $2\pi i \sin a /a$.

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  • $\begingroup$ Shouldn't that last equality in the second line have a $2$ in the denominator? $\endgroup$
    – mXdX
    Feb 26, 2019 at 17:50
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    $\begingroup$ @mXdX Yes it should, I corrected it. Thanks for pointing out. $\endgroup$ Feb 26, 2019 at 17:58

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