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Compute $$ \int_{|z|=1} \frac{2z}{z^2+2} dz $$ How is it holomorphic inside the circle? And how is the Cauchy theorem applied?

Any help for this question will be appreciated. I am not sure how to check if a rational function is holomorphic

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A rational function is holomorphic on open sets not containing its poles. – JSchlather Oct 21 '12 at 18:13
do you mean singularities by poles? – Khanak Oct 21 '12 at 18:18
Poles are one type of singularities, but not the only. – busman Oct 21 '12 at 18:22
up vote 3 down vote accepted

Cauchy theorem states that if $f(z)$ is holomorphic on an open set $\Omega \subset \mathbb{C}$, and if $\gamma$ is a rectifiable closed path lying inside $\Omega$, then $$\int_{\gamma} f(z) dz = 0$$

In general, if you have $f(z)$ as a rational function i.e. $f(z) = \dfrac{p(z)}{q(z)}$, then the only points where the function is not holomorphic are the roots of $q(z)$ i.e. all $z_0$'s such that $q(z_0) = 0$.

In your case, the only points at which the integrand is non-analytic is when the denominator is $0$ i.e. $q(z) = z^2 + 2 = 0$ i.e. $z^2 = -2 \implies z = \pm \sqrt{2}i$. But this lies outside the integration contour i.e. in your case you can choose $\Omega$ as $\mathbb{C} \backslash \{\pm \sqrt{2}i \}$.

Hence, Cauchy theorem is applicable to give you an answer $0$.

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Cauchy's theorem is for simply connected sets, or more generally curves which are null-homotopic. – JSchlather Oct 21 '12 at 20:29

$\dfrac{2z}{z^2+2}=\dfrac{2z}{(z-\sqrt{2}i)(z+\sqrt{2}i)}.$ Denominator has zeroes at points $z_1=\sqrt{2}i, \,\, z_2=-\sqrt{2}i,$ which are located outside the unit disc $D=\{z\colon \,\, |z|<1\},$ because $|\pm \sqrt{2}i|=\sqrt{2}>1.$
Thus $f(z)$ is analytical (holomorhic) in $D$, and $\displaystyle \int\limits_{|z|=1} \dfrac{2z}{z^2+2} dz=0.$

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It is holomorphic since the roots of the denominator are $\sqrt{2} i$ and $-\sqrt{2} i$ which lie outisde the unit circle. Then, by the Argument Principle, $$2 \pi i \int_{|z| = 1} \frac{2z}{z^2 +2}dz = 0.$$ Note that the numerator is the derivative of the denominator, so the theorem holds. In this case, the integral is $0$.

EDIT: The Argument Principle states that $$2\pi i\int_{\Gamma} \frac{f^{\prime} (z)}{f(z)}dz = Z_{f(z)} - P_{f(z)},$$ where $Z_{f(z)}$ and $P_{f(z)}$ denote the zeroes and poles of $f(z)$ in the interior of $\Gamma$, a closed path such that $\Gamma \sim 0$ (in the homotopy sense). In your case, $f(z) = z^2+2$, which has no zeroes nor poles inside the unit circle.

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